## Girls, girls, girls

Posted by Chris on January 7, 2011 – 9:03 pm

From the old ToM site, but severely edited. It got 63 replies.

I have three children. If at least one of them is a girl, what is the probability that all three of them are girls?

—–

See post 130 for a link to a Monte Carlo simulator.

January 7th, 2011 at 11:58 pm

Possibilities are: BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG

If at least one of them is a girl you can eliminate the first possibility.

Therefore the probability that all 3 are girls is 1 in 7.

January 8th, 2011 at 1:05 am

I agree with SP.

And all the 8 possibilities that SP mentioned are equally likely.

January 8th, 2011 at 1:06 am

AS ONE OF THE THREE IS A GIRL POSSIBILITIES ARE

G B B

G G G

G G B

SO PROBABILITY OF THREE OF BTHEM ARE BEING GIRLS IS 1/3

January 8th, 2011 at 1:55 am

I’m probably the sucker that Chris is looking for, but here goes . . .

You have two children that you don’t know the sex of after one girl is found. So the possibilities among those two are BB, BG, GB and GG. So the chances of all three being girls is 1 in 4.

Incidentally, I believe that 48% of live human births are girls and 52% boys. This is apparently needed to compensate for the higher mortality rate of the male sex, mainly due to fighting one another. This suggests that a supreme being up there somewhere is directing the male/female birth ratios. This, in turn, suggetst another ToM question: Is there a God?

But I digress. 1 in 4 is my answer.

January 8th, 2011 at 2:28 am

I’m going with Wiz on this.

But for Wiz – follow this link for birthrates/war – it’s the Fisher Principle…

http://www.straightdope.com/columns/read/2973/are-more-male-babies-born-after-wars

Is there a God? I dunno…. but I can prove the existence of Santa Claus.

January 8th, 2011 at 3:08 am

1:7 for the reason given very clearly by SP.

For any set of 3 randomly selected children there are 7 possible and equally likely combinations where at least 1 of those children is a girl.

Only 1 of those combinations comprises of a total of three girls and no boys.

For this particular question, where the three children are siblings, this assumes that there is a 50/50 chance of either sex being born to the parents and that there is no genetic bias towards one sex or the other from those parents.

January 8th, 2011 at 4:01 am

I may be a sucker too, but I believe its 1 in 3.

Like Wiz previously said, since the first is a girl, the other two can be either BB, GB, BG, or GG, but thats not true!

Wiz, I think you shouldnt have included either GB or BG, since they both have 1 of B and 1 of G.

1 in 3

January 8th, 2011 at 7:12 am

Yeehah . We’ve now got 1/3, 1/4 and 1/7; that’s the same as found the last time.

SP got it right first. Rohan made an important point that SP should have made. Dual Aspect correctly confirmed the result.

It’s obvious that Wiz got it wrong, because Karl agreed with his answer

Wiz’s error was in making a particular child be a girl.

Random Guy made the “first” (whatever that is – the eldest perhaps) be a girl, but then went on to make a further error.

Let the two remaining kids be Sam and Chris. Sam could be a boy and Chris could be a girl, or vice versa – they are definitely not the same thing – Sam could be Samantha or Samuel, Chris could be Christopher or Christine.

If I had simply said that I’ve got three children, then the probability that all 3 were girls would have beenn 1/8. i.e. the same probability as flipping 3 heads in a row. Similarly, the probability that all 3 were boys is 1/8. So the probability that at least one is a girl is 7/8. So the probability that all 3 are girls, given that all three are not boys (i.e. at least one is a girl) is (1/8)/(7/8) = 1/7.

January 8th, 2011 at 7:18 am

Hi Nathan. You had correctly answered the pan weight problem and had correctly noted that the thickness of the pans had to be very small for it to work. It was because of the latter point that I removed the puzzle.

January 8th, 2011 at 8:42 am

Just to dig a deeper hole, now that I have reached rock bottom… you say that Wiz’s error (which means it wasn’t mine…) was that we, I mean he, made a particular child be a girl. That’s what the problem states – 1 child was a girl. there are only four possibilities after you know that 1 is a girl… so where did I, I mean Wiz, go wrong?

January 8th, 2011 at 11:00 am

Hi Karl,

Imagine that you are standing at a bus stop in a town with an equal number of randomly distributed red and green buses.

If I said to you, “Wait for three buses to come along, then tell me the probability of the second and third being red if the first one is red.” then the fact that I have stated that the first is red renders it irrelevant and the question is solely relating to the second and third buses.

In this instance the possible outcomes are RRR RRG RGR RGG (answer 1:4).

If, instead, my statement had been “”Wait for three buses to come along, then tell me the probability of all of them being red if at least one of them is.” then the possible outcomes are extended to include the alternative colour possibilities of the first bus so the combinations can then be RRR RRG RGR RGG GRR GRG GGR (answer 1:7).

This is what Chris means about making a particular child a girl, once you do that the problem becomes about 2 children rather than three. There is nothing in the question to state which child is a girl so we have to include all three in our possible outcomes.

January 8th, 2011 at 11:10 am

I might be getting the hang of probabilities…. methinks a Mediaeval History degree was not suitable training for this sort of thing!

Thanks Dual Aspect and Chris. I look forward, as I am sure you do, to me getting the next one wrong….

January 8th, 2011 at 1:01 pm

its 27,,

January 8th, 2011 at 4:40 pm

I think that DA’s post # 10 actually confirms what Karl and I have been getting at.

DA says that by considering a particular child to be a girl we were restricting the possibilities to 4 rather than 7. That’s exactly right. If one child is a girl then she has to be the first, second or third child. In each case there are still four possibilities. Say the first one is a girl, then we have GBB, GBG, GGB and GGG. Likewise if the second child is a girl: BGB, BGG, GGB, GGG, and similarly if the third child is a girl. So, the answer is always 1 in 4.

If we did a survey of all three-child families with at least one girl then a quarter of them would have all girls.

Move over, Karl, I think we both share bottom place with this one.

January 8th, 2011 at 5:25 pm

Let’s use larger figures

If 8 million families have 3 children, assuming equal likelyhood of boys and girls,

1 million families would be all boys

3 million families would have 1 girl

3 million families would have 2 girls

1 million families would have 3 girls

In choosing a family at random from those with a guaranteed girl the probailty must be 1 in 7

But there’s still 49 posts to go!

January 8th, 2011 at 5:56 pm

The Wiz is obviously correct. the probability that one of the children is a girl is 100%. the birth of each child is an independant event. There is a 50% chance of a girl being born for each birth, the probability of two girls is .5*.5=.25 or 25% It’s exactly the same as tossing a coin.

January 8th, 2011 at 7:50 pm

Hi Wiz and Al. If I had said e.g. the first was a girl, then you’re right, the probability would be 1/4. I deduce this as the 4 equally likely possibilities are that I have GBB,GBG,GGB or GGG; so that’s 1 case in 4 => a probability of 1/4.

If I had said that I had three children, then the probability that all three are girls is 1/8. That’s because out of the 8 equally likely possibilities, BBB,BBG,BGB.BGG,GBB,GBG,GGB or GGG there’s only 1 way that all three could be girls; so 1/8.

The question directly implies that I have one of the following 7 equally likely possibilities: BBG, BGB.BGG,GBB,GBG,GGB or GGG. Only 1 corresponds to 3 girls, so that’s a probability of 1/7.

A more subtle way of seeing that 1/4 isn’t right is: the probability that it was the first (or second or third) child being the girl that I was alluding to is 1/3. We now need to combine those three possibilities together. i.e if the first was the girl, or if the second was the girl or if the third was the girl. But you can’t simply say the combined probability is 3((1/3)(1/4)) = 1/4, because you’d be including each girl more than once as the three cases aren’t mutually exclusive.

Addendum (roughly 24 hours after posting the reply): Even admitting that 1/3 is wrong. Such is the power of Wiz’s argument, that I fell into a trap. But Wiz’s argument does make for a wonderful apparent paradox. It’s far better than many of the so-called paradoxes out there.

Addendum (days later): Unfortunately, I have said “the girl that was being alluded to”, several times in this blog. I should have said “a girl that was being alluded to”. Fortunately, I don’t think it is of material significance. Later – I’m wrong. If I’d fully thought that through I’d have dealt with it. See post 79.

Also. I failed to answer Wiz’s assertion. If you did a survey of all families with three children that weren’t all boys, you’d find 1/7 were 3 girls. If you simply didn’t exclude the ones that were 3 boys, then the one’s with three girls would be 1/8.

January 8th, 2011 at 8:00 pm

GBB, GGB ,GGG

There can’t be BBB because he said at least one of them is a girl. They can’t all be boys.

Also, you can’t change the order because you end up with the same results as another one. eg. GGB = GBG.

I think it’s 1:3.

January 8th, 2011 at 8:14 pm

i asked my friends and my dad, who was a math teacher, and they all gave me the same answer.

I am absolutly positive that it’s 1:3. It’s that simple.

– A =D

January 8th, 2011 at 8:27 pm

Hi Chris,

I think you misunderstand my response. I’m saying that the combined probability of the guaranteed girl being first, second or third is exactly 1, i.e. a certainty, because that’s the premise of the question.

There are then 4 cases with equal probability after you’ve got your girl, one being GG, so the chances are 1 in 4.

Putting it mathematically:

(pG1 + pG2 + pG3) * (pBB + pBG + pGB + pGG) = 1

where pG1, pG2, pG3 are the probabilities of the girl coming first, second or third, adding up to 1, and pBB is the probability of two boys out of the other two children, pBG and pGB are the probabilities of a boy and a girl, and pGG is the probability of two girls. Each of these is equal to 1/4, so pBB + pBG + pGB + pGG =1.

More pertinent to our situation is that (pG1 + pG2 + pG3) * pGG = 1/4.

I can keep going like this for another 40-odd posts (especially if Karl and Al still agree with me), but I must admit that I find Curtis’s argument in post # 14 hard to refute.

Perhaps there are parallel universes in which each of us is right.

— See post 78, Chris.

January 8th, 2011 at 10:20 pm

Hi Wiz. LOL, thanks for helping to stretch this one out.

I thought that I’d covered your points in the last paragraph of my last post. OK I didn’t explicitly say that 1/3 + 1/3 + 1/3 = 1 or that 1/4 + 1/4 + 1/4 + 1/4 = 1 (or that 1*1 = 1 )

But taking your last equation and replacing the symbols with numbers that you want to use, you get:

(1/3 + 1/3 + 1/3)(1/4) = 3(1/3)(1/4) = 1/4, and that is exactly what I wrote.

You are assigning two (or three) probabilities to chi1d 1. In one case it’s the probability that the child is a definitely a girl and that she is the girl that was being alluded to, and in the other two cases it was the probability that child 1 might be a girl and that if child 1 is a girl, she was not the girl that was being alluded to (and if she wasn’t a girl, that HE couldn’t have been the girl being alluded to). Those statements are so complex that you should be worried about using them correctly. You can only add probabilities if the associated states are mutually exclusive; in my view, that alone rules out your solution. (I’ll admit that I’ve laid it on a bit thick).

Curtis has simply tried to get rid of probability (as such) in order to make the situation more concrete. What he’s said is that if an infinite number of samples of 8 million families were taken, then the average of those samples would contain the distributions of children that he mentioned. All he’s really done is to introduce a scaling factor of 8 million, otherwise it is identical to the argument that SP, DA, Rohan (implicitly) and myself are using – I’m sure that you already knew that. I’m only spelling it out because I’m convinced that most people have not really got the hang of what probability is really about (I don’t include you in that generalisation, BTW), and I hope that that explanation helps the penny to drop (a bit) for them.

But it is easy to see why this problem is paradoxical. Your and Al’s argument look good superficially.

Notice that in the 1/7 arguments there is none of the complexity that the 1/4 arguments have introduced. The latter approach is so complex that I’ve been unable to write down the equations for it. But I will try, even if it’s only for my own satisfaction.

Addendum. Several days after posting. In every family of three children that aren’t all boys, the probability that at least one is a girl is 1. If you randomly select from such a family, you might select a boy, you can’t assume that you selected a girl. The probability that you select a boy is 3/7 and the probability that you select a girl is 4/7.

January 8th, 2011 at 11:21 pm

I don’t like everything that I wrote in my last post, but I can’t be bothered to change any of it, and I haven’t the heart to delete it, sooo…..

Although the next initially looks lke my previous argument, it is different. We must have BBG,BGB,BGG,GBB,GBG,GGB or GGG.

It is only possible in 4 of the 7 (equally likely) cases to even contemplate assigning a probability of 1/3 to the first child of being the one (girl) alluded to. So the probability of referring to the first child (as being the girl alluded to) is 4/7 * 1/3. The same for the other two children.

So we must multiply the 1/4 by the factor of 4/7 to get 1/7.

Does that do it for you guys? (I don’t seriously expect that to convince you as I’m having a hard time with that argument myself at the moment )

It’s nearly 6:30 am here, so I’m orft to bed.

January 9th, 2011 at 2:04 am

hi Wiz

i see that you like sticking to your guns but i have to go with Chris on this

if you have 3 kids then the possibilities are: BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG (1 in

if you know that (any) one of them is a girl then you can only rule out one possibility-BBB (1 in 7)

but,if you cheat and say that it’s the first child that is the known girl then you can rule out BBB,BBG,BGB,BGG (your 1 in 4)

stop cheating!

January 9th, 2011 at 2:07 am

where did that smile face come from??

i typed in (1 in “eight”)

January 9th, 2011 at 4:24 am

Hey Wiz (post #13), yes you’re right that my post #10 confirms what you and Karl have been getting at (the first scenario discussed in the post), but it also explains that that is not what the question is asking by giving the second scenario as the alternative (and correct) version.

One more approach, then I’m going to leave it to Chris as he started all this ……

If we break down the question into its component parts we can take each of the three elements individually to deduce the answer.

Part 1: “I have 3 children”

We all agree (I hope) that this means we have one of the following combinations of sexes – BBB BBG BGG BGB GBB GBG GGB GGG.

Part 2 (The ToM part): “If at least one of them is a girl”

This doesn’t change the alternative options in the way that your brain logically expects. Of the above scenarios in part 1 the only option we can now exclude is BBB.

Part 3: “What is the probability that all three of them are girls.”

Of the remaining 7 options only 1 satisfies the requirement.

Answer 1/7

January 9th, 2011 at 6:04 am

To those who say that the BBG is the same as GBB and BGB, you need to rethink. If the youngest is a girl (BBG), that is not the same as the eldest is a girl (GBB) or the middle one is a girl (BGB) – or do you believe that eldest and youngest are the same thing?

Great news! We are past the 1/3 of 63 mark and still haven’t resorted to DUHH!, you’re an idiot, etc. (that was “8″ then “)” )

I’m about to have a cognitive dissonance.

January 9th, 2011 at 7:04 am

Just to keep things stirred up.

If I have have three children, then the probability that the eldest is a girl is 1/2.

If I have three children and exactly one is a girl, then the probability that the eldest is a girl is 1/3.

If I have three children and exactly two are girls, then the probability that the eldest is a girl is 2/3.

If I have three children and all three are girls, then the probability that the eldest is a girl is 1.

If I have three children and at most one is a girl, then the probability that the eldest is a girl is 1/4.

If I have three children and at most two are girls, then the probability that the eldest is a girl is 3/7.

If I have three children and at least one is a girl, then the probability that the eldest is a girl is 4/7.

If I have three children and at least two are girls, then the probability that the eldest is a girl is 3/4.

January 9th, 2011 at 7:27 am

,,. so using the probability that e.g. the eldest is a girl given that at least one is a girl is 4/7, I can only have been alluding to the eldest on 1/3 of 4/7 of all occasions. The same for the middle being the girl and the same again for the youngest being the girl. Hence the probability of them all being girls is (given that at least one is a girl is: 3 * (1/3)(4/7) * (1/4) = 1/7.

January 9th, 2011 at 9:34 am

Hi Angel. Ask your friends and dad to explain their calculations to you. I’m almost certain that they (like Random Guy and MOHan) are going to say that there are three cases: one two or three girls, and only one is all girls, so the probability is 1/3. The fallacy in that argument is that the three cases are not equally likely. In fact if there are three children, then the probability of no girls is 1/8, of one girl is 3/8, of two girls is 3/8 and of three girls is 1/8. Note that that totals to 1 as required.

Consider the simpler and more familiar situation of flipping two coins. In a trial of two flips, the probability of flipping two heads is 1/4 and the probability of flipping two tails is 1/4 (also). In any trial, the probability of flipping some combination of heads and tails is 1 (neglecting coins landing on their edge etc.) The probability of NOT flipping HH or TT is 1 – 1/2 = 1/2. But not flipping HH or TT is the same as flipping HT or TH. So the probability of throwing an H and a T in any order is 1/2. So mixed states don’t have the same probability as unmixed states. It is also pretty obvious that P(HH) = P(HT) = P(TH) = P(TT) when the order of flipping is specified.

Back to the girls. taking the strings ABC to be the eldest, middle and youngest child, P(BBB) = P(BBG) = P(BGB) = P(BGG) = P(GBB) = P(GBG) = P(GGB) = P(GGG) = 1/8. So,

0 girls. We must have BBB, so 1/8

1 girl. We must have BBG + BGB + GBB, so 3/8

2 girls. We must have BGG + GBG + GGB, so 3/8

3 girls. We must have GGG, so 1/8

Of course I contend that that means if there is at least one girl, that leaves 7 equally likely cases of which only one is for three girls, so 1/7. Or P(3G)/(P(1G) + P(2G) + P(3G)) = (1/8)/( 3/8 +3/8 + 1/8 ) = 1/7.

January 9th, 2011 at 2:11 pm

Do all the 1 in 7 advocates out there maybe protest just a little too much?

Deep down in their souls (if they have one, otherwise soles) is there perhaps a small, still voice trying to tell them that maybe, just maybe, they could be wrong and the Wiz right, and the true answer is 1 in 4?

Is this why the 1-in-7’s come on so strongly, to cover up their inner uncertainty?

30 posts down, 33 to go . . .

January 9th, 2011 at 2:39 pm

Hi Wiz. No doubts here. Another way to see that 1/7 is reasonable, is to start with the 1/8 and realise that we’re only tweaking the balance by about 12.5%, not 50%.

I’ve easily shown that the 1/3 argument is hopelessly in error. I’ve given more than enough to show that the reasoning for the 1/4 argument is very far from safe. I believe that I have satisfactorially shown the flaw in your argument (i.e. i can only choose e.g. the eldest girl to be the one I was alluding to, 1/3*4/7 of the time, not 1/3 of the time). Your assesment of 1/3 presupposes the result of there being 3 girls at the outset, not merely there only being at least one girl.

You haven’t published a single direct reason to show where you believe the flaw is in the very simple and clear-cut logic used to determine 1/7. If you had a bullet-proof alternative argument then I’d accept that (I’m confident that no such argument is possible). The argument that you’ve offered is so full of holes that I consider it to be irreparable.

The number of arguments made is not an indicator of my doubt, it is a measure of my certitude.

Addendum (days after posting): That was bad – the eldest can only be the one I was alluding to 4/7 of the time.

January 9th, 2011 at 4:54 pm

Hi Wiz. Try this. There are 100 children in a hall. Assuming that they have been uniformly randomly selected by gender, then the probability that at least one of them is a girl is 1 – (1/2)^100 = 0.99999999999999999999999999999921 i.e. it is almost certain that at least one is a girl. i.e. I wouldn’t even have to inspect the children to be able to say that at least one is a girl. I’m incomprehensibly more likely to win the national lottery than to be wrong about that speculation.

So the difference between the probability of them all being girls given that at least one is a girl, and simply the probability of them all being girls can only differ by about 1 part in 10^32 – i.e by a totally insignificant amount.

The probability that they are all girls is (1/2)^100. I say multiply that by 1/(1 – (1/2)^100), i.e multiply the unconditional probability by 1.0000000000000000000000000000008 (that’s 1 in English) to get the conditional figure – that seems reasonable in view of the above.

Your logic, my dear upside down friend, says that the probability of them all being girls would be (1/2)^99 = 2 * (1/2)^100, i.e. a doubling of the probability has occurred.

Your call

Wooohoooh, that’s over half of 63 posts LOL.

January 9th, 2011 at 5:50 pm

Just to get another perspective on that (and to increment the post counter). If there was 1 hall for each of the 100 possible distributions of the children, then I’d need 2^100 = 1267650600228229401496703205376 halls. In exactly one of those halls, there would be 100 boys. And in exactly 1 there’d be 100 girls. I’m saying that I’m practically certain that if I randomly chose 1 of those halls, It’d contain at least one girl. I’m also saying that the probability that it contained 100 girls is 1/1267650600228229401496703205376 and the probability that it contained 100 girls, given that it contained at least 1 girl is 1/1267650600228229401496703205375. I do like silly big numbers.

The people who say 1/3 for the original problem, won’t accept any of that. They’ll say that the probability of them all being girls, gven that at least 1 is a girl is 1/100. To them I say, if they looked in to e.g. a typical mixed gender school classroom (with 30 children in it, say) they’d most likely find that it’d have around 15 boys and 15 girls. But their logic would suggest that they believe that they would be just as likely to find 0 girls and 30 boys, 1 girl and 29 boys, … or 30 girls and 0 boys. I reckon you’d need to look into around a billion such classrooms before finding one with 30 girls and 0 boys, excluding classes which intentionally are single gender.

For all those maths teachers and the like who go for 1/3, I suggest that they acquaint themselves wth the binomial distribution and then hang their heads in shame.

January 9th, 2011 at 10:15 pm

Guys. It’s not that hard. If one is definently a girl, she gets ruled out. This leaves two children. The options are 1 boy and 1 girl, 2 boys, or 2 girls. Since we want one of the three options, the answer is 1/3.

January 10th, 2011 at 12:45 am

this is post 35

just doing my part.

January 10th, 2011 at 2:04 am

I really don’t understand why everyone is making this complicated. Perhaps it’s been posed about children rather than some inanimate object like a coin. Suppose you are told that someone has flipped 3 coins and one of them came down heads and you wanted to know: what is the probability that the other two are also heads? The order in which the coins were flipped does not matter, because the outcome of each flip is completely independent. The coins don’t know the order in which they were flipped, nor do they know how many were flipped, or the results of the flip. The probability of tossing a coin and getting a head is ½, whether you have tossed one coin or a million coins. But if I told you that I tossed a coin and it landed heads. And I asked you: what is the probability that the coin landed heads? The answer is obviously 1. If I told you that a man had 3 daughters and asked: what is the probability

that he had 3 girl children? The answer whould be 1. If we are told that that a man had 3 children, and one of them was a girl, then the probability that he has a daughter is 100%. Now we are asked what is the probability, that the other two are also girls. The probability of child being born a girl is .5. regardless of how many brothers or sisters it might already have, or will have in the future. So therefore the probability that two babies are born girls. Is .5*.5 = .25. Just as the probability that two coins were tossed and landed heads is .5*.5= .25 .

January 10th, 2011 at 5:59 am

Just when I thought it was safe to dip into the waters of Lake Probability…

Meantime a real world probability question… David and Victoria Beckham have a fourth child on the way. The first three are boys. What is the probability of the fourth child being a girl?

Please factor into your answer the possibility of Victoria having gone for gender selection – she really wants a girl, and that the chances of a 4th boy are considerably higher after the first three are boys.

I will provide all answers in due course…

January 10th, 2011 at 6:23 am

Thanks Knightmare, keep up the good work

I have edited this post severely since originally publishing it.

Hi Al. It’s only getting complicated because directly dealing with fallacious arguments is difficult to do. Wiz’s argument was especially difficult to deal with. I prefer (but don’t always remember to practice it) to show directly where the fault is in the solution (method) that I claim is faulty. Simply producing another way of solving the problem can be counter-productive – it could be annoying, to the person whose solution is being challenged, if I don’t actually point out the actual fault that I claim that they’ve made. Of course, sometimes, I might think that I’ve got a solution that’s so manifestly obviously right that it’s not unreasonable to offer it, even though it hasn’t established the precise nature of the original fault. In this case, the manifestly obvious solution was given in post 1. I can’t simply repeat it.

Despite that good intention, I also keep coming up with what I think are good arguments to prove that 1/7 is the correct answer. No-one is explaining what is wrong with the reasoning, they simply ignore it and provide a demonstrably fallacious argument that concludes with the wrong answer. I on the other hand, point out the fallacy to them; no direct defense or proof that they haven’t made a fallacious argument is offered, just another fallacious argument.

I assume that you haven’t read my posts #32 and #33 (and more since then). It conclusively demonstrates that your answer is wrong. In your case, it’s because you believe that the order of the kids doesn’t matter – well I’ve shown (in several posts now) that that is a serious error of logic.

Here’s one demonstration that the order does matter, it appeals to a well known result. BTW I’m assuming that don’t object to the listing tehnique as such. If you are, please tell me what problems you have with it.

e.g. if I flip a coin 3 times, what is the probability of getting at least one heads. The easy way to calculate that is to note that it’s 1 – the probabilty of getting 0 heads. i.e. 1 – probability of getting 3 tails, i.e. 1 – (1/2)^3 = 7/8 (a well known result).

Using the same method as for the posted problem, I’d say, I could have flipped one of the following sequences (in order first flip, second flip, third flip) TTT,TTH,THT,THH,HTT,HTH or HHH. That’s 8 equally likely possible outcomes. 7 of them have at least one head, so the probabiliy of getting at least one head is 7/8.

That shows that the list way of working is reasonable (it didn’t contradict a well known result). Moreover, it inherently used the logic that the order DOES matter, otherwise the list would have been e.g. TTT,TTH,THH,HHH and then I (and you) would have concluded that the probability of getting at least one head is 3/4 in direct contradiction to the well known result.

I’m just as unemotional about analysing the three coins problem as I am about my ficticious three kids problem. If I throw three coins and at least one of them comes comes up heads, then my analysis is: there are eight equally likely possible sequences that I could flip the coins in, only seven of them meet the criterion of a having at least one head. Only one of them corresponds to being three heads, so the probability of flipping three heads, given that I definitely flip at least one head (whether it be on the first second or third flip) is 1/7.

To make this artificial situation slightly clearer, you might find it easier to imagine that someone (out of your sight) does the three flips. After he has done all three flips then if, and only if, he gets at least one head, he’ll announce that he has flipped at least one head. If after the three flips, he has three tails, he’ll say nothing and simply make a further three flips. Both of you are fully aware of what is being done. The question is, on average (over an infinite of trials), what is the ratio of the number of times that he saw three heads to the number of times he saw at least one head? That way it should be clear that you can’t assume that it was the first or the second or the third coin, and so you can’t simply say 1* 1/2 * 1/2 – because that expression implicitly means H??.

I’ve already shown, but I’ll do it again, that if all you know is that at least one coin is a head, then the probability of it being the first coin is 4/7. That’s because the only possibilities (and they are equally likely) are HTT, HTH, HHT, HHH but not THH, THT, TTH (TTT isn’t relevant as the probability of TTT is 0). Then you can follow through with (4/7)(1/2)(1/2) = 1/7.

See posts 69 and 71 for a better explanation.

January 10th, 2011 at 6:45 am

Hi Karl. If your question had been a normal probability type question, then the answer would be 1/2.

You are right, if you have children close together, then they are more likely to have the same gender.

It just so happens that I’m the eldest of four brohers (no sisters). The biggest gap between us is two years. I have two children as do two of my brothers. We each have a boy and a girl. We all left a 3+ year gap between them.

January 10th, 2011 at 10:58 am

OK you dissenters. Try this. It should establish that the order matters and that the result is 1/7 – beyond all reasonable doubt.

I have an electronic lock. It has 10 code buttons: A,B,C,D,E,F,G,H,I,J

To operate it, you must punch in a 3-letter code in the right order. e.g. if the code was ABC, then CAB won’t work. If the code was BBG, then BGB won’t work, etc. (I hope you can see where I’m going with this )

The number of possible combinations is 10^3 = 1000. That’s because there are 10 choices for the first letter, 10 for the second and 10 for the third letter.

I have decided to set the code to be based on my children being boys or girls. I do this in the sequence eldest, middle and youngest, and I use B or G depending on whether the child is a boy or a girl. So if the eldest was a boy, the middle was a girl and the youngest was a girl, the code would be BGG.

If all you knew was that the lock used a 3 letter combination, the probability of you punching in the correct code would be 1 in 1000. If you knew that I had only used the letters B and G, then you’d know the code was one of BBB,BBG,BGB,BGG,GBB,GBG,GGB or GGG. That’s eight possible (and equally likely combinations). If you accept that, you’ve already used the logic that has been used to solve the the original problem. See ***, later for a further comment on that.

But I’ll go the whole hog. If you somehow knew that my code was based on the gender and age of my children (B or G), and you also happened to know that one of my children was a girl, but didn’t know what the others were or what their relative ages were, you’d know that the combination for the lock had to include at least one G. So all you’d know is that BBB isn’t the combination. That means that you’d know that the combination was one of BBG,BGB,BGG,GBB,GBG,GGB or GGG – that’s 7 equally likely and unique possible combinations. Therefore the probability of you guessing the right combination to my lock (in the first go) is 1/7 (given that you know that it only used 3 letters, the only letters that are used are B and G, and that there was at least one G in it). NB Let’s say you had now tried BBG and it didn’t work. That’d leave you with 6 possibilities. Then the probability of you getting the right code on your next try would be 1/6. All you’ve done is cross another combination of the list, leaving you with 6 possible and equally likely combinations.

The order of the gender of my children is important: there is a one-to-one correspondence with the lock combination. The probability of opening the lock on the first go is 1/7, so the probability of me having a particular arrangemt of children must be 1/7 also. If the code for the lock was GGG, the probability of you getting it (on the first try) is 1/7, so the probability of me having three girls (given no boys) is 1/7 also.

If the order of the ages was unimportant, then the order of the code lock would be unimportant as well. But you’d find the lock won’t agree.

*** Only knowing that B and G were used, mean that there are only eight possible combinations. I’m sure you’ll all accept that (if not, you’re beyond my ability to reason with). What you’ve effectively done is to answer the problem, what are the number of possible codes given that the code only uses B and G. You could have done this by listing all 1000 possible codes: AAA,AAB,AAC … KKK, and then crossing out all the codes that contained letters other than B and G, leaving you with the eight codes listed above. That is exactly the same logic as eliminating the BBB from the list of eight combinations of B and G when you know there must be at least one G. In fact if you had known that that was the case, you’d have eliminated BBB from the list of 1000 possibilities up front.

January 10th, 2011 at 12:16 pm

On classroom gender ratios:

If I have a group of 30 children. It is equally likely that any given one is a boy or a girl, and I assign to each the position 1-30 (using seat numbers say), then the possible collections of the kids are BBBB…BBB (0G, 30B, GBBBB…BBB (1G,29B), BGBB….BBB (1G,29B), etc. Altogether, there are 2^30 different possible collections of these 30 kids.

The number of collections which contain 0 girls (and 30 boys) is 1. The number of collections with 5 girls (and 25 boys) is 30!/(25! * 5!) = 142506. The peak is when we have 15 girls (and 15 boys), and for that we have 30!/(15! * 15!) = 155117520 collections. If we had 14 girls and 16 boys (or vice-versa) the number of collections would be 145422675, etc. If we plotted the number of collections of girls versus the number of girls, it just so happens that we’d get a graph that approximately resembles the normal-distribution i.e a bell-shaped curve. [In fact the curve is the binomial ditribution, with n = 30, p = q = 1/2]. The the curve would peak at 15G-15B and drop away to 1 with 0G-30B or 30G-0B. i.e. you would find that the most likely class distribution would be near to 15 boys and 15 girls.

There are over 1000 times as many collections with 15 girls and 15 boys as collections with 5 girls and 25 boys. So the curve is fairly peaked, and typical classes would have no more than a 2 to 1 gender ratio (I pulled that number out of the air).

If GBBBBB was “the same as” BGBBBB etc, then you’d find that the graph was a flat line of height 1. You’d be just as likely to find a class with 0 girls and 30 boys as any other combination. I’m pretty sure that doesn’t happen.

January 10th, 2011 at 8:12 pm

Chris,

I believe that we disagree mainly because we are not trying to solve the same problem. Here is the problem you posed, which I believe I have solved.

“I have three children. If at least one of them is a girl, what is the probability that all three of them are girls?”

There is no mention of any order of birth. You say that there are 7 possibilities:

BBB,BBG,BGB,BGG,GBB,GBG,or GGG .But since there is no mention of order of birth in the statement of the problem., BGG , GBG and GGB are the same also BBG,GBB and BGB are the same. Also BBB is not an option, because your statement also said that one of them is a girl. That leaves you with three possibilities; GGG, GGB or GBB. Which I suppose is the basis for some of the posts whose answer was 1/3. But that too is not correct. Each birth is an INDEPENDENT EVENT. The probability of each event is .5. But we are told that one of the births is a girl. So the probability of that event is 100%. That means that that child is out of the equation. The only uncertainly is the sex of other two. And the probability, that two children are born girls (or boys for that matter) is .5*.5 = ¼.

So A clearer way of stating your problem is: “One of my children is a girl, what is the probability that the other two are girls?” The probability of a child being born a girl (from a strictly mathematical point of view) is .5. The probability of both children being born female is .5*.5= .25. Note that the problem could have been stated as: “One of my children is a Boy, what is the probability that the other two are girls?” The answer is exactly the same. Each birth is an independent event. And the order in which they occurred is totally irrelevant. The birth of baby 2, whether a boy or a girl, does not influence the birth of baby 1 or baby 3.

Also, you did NOT ask “what is the probability of three children being born as girls? (The answer to that question is .5*.5*.5 = 1/8) Your question was “ There are three children one of which is a girl. What is the probability that the other two are also girls.” That is the question that I answered, and the answer is ¼.

January 10th, 2011 at 9:32 pm

Hi Al. Re your post 42. The question the first para is the one I asked and that you haven’t answered. The question in the last para is not the one I asked, but i guess it’s the one you answered. (Also see post 86).

What do you mean by e.g. BBG, BGB and GBB are the same? Do you mean that there are (before introducing the “that at least one is a girl” condition) four equally likely arrangements of the children and that they can be described as BBB, BBG, BGG and GGG and therefore the probability of each is 1/4? If you do, then why do you also claim that the probability of having three girls (i.e. GGG) is 1/8? Is it 1/4 or is it 1/8 (it can’t be both)? If you say that you didn’t mean that, then would you explain what you do mean when you say that BBG, BGB and GBB are the same? Better still, tell me the probabilities that you associate with each of these Vulcan melded states.

I didn’t mention in the problem that the kids were born separately because twins and triplets are sufficiently rare that I didn’t want to complicate the problem with such considerations – I wasn’t seeking to get this blog’s post count into the Guinness Book of Records. However, it happens that (as long as you assume that each child is equally likely to be a boy or a girl) that has no affect on the result – the next paragraph should explain why that is so.

The ages were merely a convenient way of labelling the chiildren to enable analysing the underlying mechanics. I could just as easily have said that I had painted a red dot on one, a blue dot on another and a green dot on the last one – that takes care of the case of triplets, etc. As before, I’ve only have done this to facilitate understanding the underlying mechanics. The probability that the one with a red dot being a girl is 1/2, etc.

Do you also believe that if you flip three coins (one from year 2001, one from 2002 and one from 2003) that you’ll get a different distribution of heads/tails than if they were all from 2001. I say that you’ll get 1/8 with 3T, 3/8 with 1H2T, 3/8 with 2H1T and 1/8 with 3H regardless of which set of coins you used. I’m not really sure what probabilities you’d assign to the 2001/2002/2003 coins case. But, you have made it clear what you would expect for the 2001 coins only case. i.e. you’d say the only cases are HHH, HHT,HTT and TTT (because e.g. HHT is the same as HTH and THH) and therefore each has a probability of 1/4. How do you acount for the difference in behaviour? In particular, you say that the probability of throwing HHH with coins from different years is 1/8, but if they’re from the same year that it would rise to 1/4. Has it something to do with the aerodynamics of the date stamping ?

What would happen if we used two coins from 2001 and one from 2002?

You did not answer the question that I posted. In view of the logic you’ve used, I don’t know for sure what question you have answered. You really should have read my posts 32 and 33, then I wouldn’t have had to post this one.

Your call.

January 10th, 2011 at 11:00 pm

Hi Al.i like you coin flipping idea so let me ask you this:

i flip 3 coins and told you that 2 of the coins were heads and asked you to guess the probability of the other flip,what would you say?

i flip 3 coins and told you that the first 2 flips were heads and asked you the probability of the last flip,what would you say?

you wouldn’t repeat youself-would you?

January 10th, 2011 at 11:16 pm

Hi Knightmare. I’ve changed one word in your post (“other” was “last”) because “last” implied an order. I hope you don’t mind.

This post gets this blog to just over 71% of the 63 posts target. Just a few more and it’ll be in the “popular posts” section. I’ll soon be able to shake hands with myself.

January 10th, 2011 at 11:25 pm

Let me try again…

We are agreed, are we not, that the possible combinations of boys and girls among two children are BB, BG, GB and GG. Each has a probability of 1 in 4. Right?

We are now told that these two siblings have a long lost sister. The odds of the original two also being girls have not changed – they are still 1 in 4. The discovery of another sister changes nothing. In particular they have not suddenly changed the odds of both the original children being girls to 1 in 7. Whether the long lost sister is older, younger or in between the first two in age is irrelevant.

Chris, your resistance to this irrefutable logic is admirable, but ultimately futile. You may choose to surrender with honour or else submit to being transported in chains to the colonies.

Your call . . .

January 10th, 2011 at 11:33 pm

Hi Chris,

I’m not sure I understand what you mean about coins from 2001 etc. but I’ll try to explain my point of view.

What I meant when I wrote that BBG, BGB, and GBB are the same is that all three represent the same status with regard to the problem we are trying to solve. That is, we are looking to find the probability that all three are Girls. The three cases given above represent exactly the same situation, which is that there are two boys and one girl. As I’ve said before, I don’t see anything in the statement of the problem that would make the birth order of the children relevant, whether the boys are older than the girl or vice versa is not relevant to the problem. We want to know what is the probability that the three children are girls. If that was the way the problem was posed, than the answer would indeed be 1/8 , i.e., .5 *.5 * .5 .

But that is not the way the problem was given. We are told that at least one of them is a Girl. So the probabilities that all three are girls, with the stipulation that one of them is a girl. Is .5 (which is the probability that one of the “unknown” is a girl) * .5 (the probability that the other “unknown” is a girl) * 1 (the probability that the girl is a girl).

We therefore have .5 * .5 * 1 = ¼ . It’s really as simple as that.

January 11th, 2011 at 12:00 am

Hi Knightmare,

I’ll try to answer your questions, but I suspect that Chris won’t like my answers. But that’s ok since the Wiz has him making little ones out of big ones out in the Colonies.

Each flip of the coin is independent of every other flip. The coin has no memory. The probability of each and every flip is .5 , whether you flip one coin once or a million times, the probability that flip number 1000001 is heads or tails is still .5 (even if the first million times you got heads, the probability of which is .5^1000000 a fairly small number). And if you flipped a million coins and got heads on the first 999999 coins, the probability of coin number 1000000 being heads is still .5 .

January 11th, 2011 at 12:08 am

Maybe this will nail it . . .

I vaguely remembered the concept of conditional probability from my student days many decades ago. So I looked it up in Wikipedia and quickly got lost. However, one thing there seems relevant: if events A and B are independent, then the probability of B given A is the same as the probability of B by itself.

Applying this to the three girls problem, we can say that the probability of one child being a girl is independent of the probability of the other two both being girls. Therefore the probability of the two other children both being girls given the presence of another sister is the same as the probability of them being girls without that sister, i.e. 1 in 4.

However, they do say that a little knowledge is a dangerous thing, so it may be that someone who really understands conditional probabilities might see things differently.

January 11th, 2011 at 12:41 am

Hi Wiz. I can only assume that you are pulling my leg, especially after posts 32 and 33 and several others. But I will respond to it as if you are being serious.

You have failed to point out a specific error in my reasoning, you are simply gainsaying. I have pointed out several very worrying points in your reasoning. your logic is far from irrefutable – I have refuted it (several times in fact).

However, I agree that P(BB) = P(BG) = P(GB) = P(GG) = 1/4. The introduction of the long lost sister, as you say, changes nothing about the other two kids. That’s because you have specified a particular sister.

If you had tried to construct the question to go with that scenario in a similar vein to the one I posted then, I guess it’d have to be something like: I have three children, one left home a long time ago and is a girl. What is the probability that the other two are girls?

Well, I’d immediately answer that with 1/4. In detail I’d reason it as (using the positional notation, where the 3rd letter denotes the long lost sister) we now have the possible equally likely combinations BBG, BGG,GBG and GGG. Only one of those corresponds to all girls so 1/4.

I see no parallel between your proposed problem and the one I posted. I only see very superficial similarities. You have identified a specific girl, I haven’t.

If you were to suggest that for the problem that I posted, that I should denote the alluded to girl by the first letter say (in the positional notation) then I’d say that I cannot do that because that would lead to a polymorphic definition of the string. The positions would no longer represent a particular child and I could no longer use the logic that I’ve been using (in a reasonably useful way). That’s because the one with a red dot has only a 4/7 probability of being a girl (whether or not I was alluding to her). So the first position in the string cannot always refer to a particular girl.

{Aside: The 4/7 is obtained by examing the possible distributions of the kids (in order red dot, green dot, blue dot) from BBG,BGB,BGG,GBB,GBG,GGB,GGG that 4 cases in 7 where the child with a red dot could be a girl, 4 in 7 where the child with a green dot could be a girl and 4 cases in 7 where the child with a blue dot could be a girl.

Depending on your philosophy, you could either say the required probability is 3(1/3 * 4/7) (1/2 * 1/2) or simply (4/7)(1/2 * 1/2). Either way you get 1/7.

end aside}

What I could do though, is to use a non-positional notation. I’ll use lower case letters for that. Then (before saying at least one is a girl) the possible children combinations would be bbb,bbg,bgg,ggg and P(bbb) = P(ggg) = 1/8, P(bbg) = P(bgg) = 3/8. Then, I’d rather boringly have to use the formal statement of conditional probability:

P(A|B) = P(A & B)/P(B) which reads, prob of A given B = prob of (A logical AND B) divided by prob of B. That translates to A = ggg, B = gbb+ggb+ggg (i.e. at least one is a girl) =>

P(ggg|gbb+ggb+ggg) = P(ggg & (gbb+ggb+ggg))/(P(gbb+ggb+ggg)

= P(ggg)/(P(gbb+ggb+ggg)

= (1/8)/(1/8 + 3/8 + 3/8)

= 1/7

{Added after you last post: A and B are not independent. Independence applies to say flipping a coin twice (and usually is called the “no memory” effect; but if it were so, then P( H? & ?T) = P(H?)*P(?T). But for the girls problem, A = ggg and B = gbb+ggb+ggg. Also note that only one of gbb, ggb or ggg can be true and so A & B = ggg. (“+” and “&” here are logical OR and AND), so P(A & B) = P(A) in this case.

Just for amusement, what is the probabity of throwing a heads given that I threw a tails? P(H|T) = P(H&T)/P(T). If I was meaning throwing a heads and a tails at the same time, then I say H & T = {} (the empty set) and so P(H & T) = 0 so P(H|T) = 0, no surprise there. If I meant flip a coin twice, what is the probability that I throw a head given that I threw a tails last flip, then as those are independent P(H | T last time) = P(H & T last time)/P(T last time) = P(H)*P(T last time)/ P(T last time) = P(H), so no surprise there either.}

Now show me specifically and directly where there is even the smallest speck of a hint of a fault in the logic that I’ve used (anywhere on this page).

I guarantee that any scenario you introduce that isn’t so fiendishly complex and convoluted that it’s too difficult to discuss, but doesn’t produce the result of 1/7, that I’ll be able to at least hint at a flaw in it, if not completely demolish it.

January 11th, 2011 at 1:44 am

Oh my gosh. Al. What probabilities do you assign to bbb,bbg,bgg and ggg? Until you answer that, I can’t proceed with the discussion. You will experience a cognitive dissonace when you try to answer that question. In particular you will have to choose to go with the flow and accept that P(ggg) = 1/8 and that you order doesn’t matter is gibberish, or that you order doesn’t matter is true and P(ggg) = 1/4 and the rest of the world and yourself (as you also say 1/8) wrong.

I gave the coins different dates so there was an eldest, a middle and a youngest coin (in one case) and the dates were irrelevant in the other case. No that’s not true; I did it to illustrate that your logic is manifestly incorrect.

Your post 48 hasn’t answered either of Knightmare’s questions. You’ve simply stated well known facts.

For his first question the answer is P(H) = 1/4, P(T) = 3/4. For his second question the answer is P(H) = P(T) = 1/2.

I’m assuming that Knightmare had meant “at least two” were heads for the first question (I’ve only just noticed that omission.)

Hi Wiz. I do understand conditional probability, I’ve been demonstrating that for several weeks at least. By pure luck, my last post used the formal stuff [rats, I hadn't seen you post until after I published my last one]. I had wished to avoid it because those statements are fairly hard to translate, and they detach the reader (and the writer come to that) from the underlying mechanics. I’m not in the slightest bit interested in the answer; I’m only interested in how you derive it.

The underlying concept behind conditional probability is very simple. We use it all the time. For instance, we glibly assume that the probability of flipping a heads is 1/2 and flipping a tails is 1/2 and that flipping one of them is 1. But we should be saying e.g. the probability of flipping a heads given that the coin doesn’t land on an edge is 1/2 etc. The probability of a coin landing on it’s edge is about 1/10,000. So the probability of a coin flipping to heads is really (1/2)*(1 – 1/10,000).

But P(H GIVEN not landed on edge) = P(H AND not landed on edge)/P(not landing on edge)

= (1/2)(1 – 1/10,000) / (1 – 1/10,000) = 1/2.

We have P(H) = (1/2)/(1= 1/10,000) = P(T) and P(edge) = 1/10,000 and P(H)+P(T)+P(E) = P(H or T or E) = 1 . When we discard the edge landings, we simply say let the new value P(E) = 0 (i.e. the probability of the coin landing on it’s edge, given that it didn’t land on it’s edge is necessarily 0) and uniformly rescale all the remaining probabilities to make them add up to 1. The rescale factor is 1/(1-P(E)) (the original value of P(E) that is).

That’s all we do with the girls problem. We in effect say, let P(BBB) = 0, and rescale so that P(BBG+BGB+BGG+GBB+GBG+GGB+GGGG) = 1 in this case the rescale factor is 1/(1-P(BBB)) and so P(HHH given not BBB) = (1/8)/(7/8) = 1/7.

I usually take that natural/intuitive aproach, but the pesky formal approach gets the scaling factor from the other direction and it uses

P(BBG+BGB+BGG+GBB+GBG+GGB+GGGG) rather than 1 – P(BBB).

OK, so I quote some formula to “prove” the answer, what’s the bl***dy use of that – does anyone learn anything from it. To me it’s no better than saying “my teacher says the answer is [insert an incorrect answer here]“, and adding “and he’s smarter than you” [probably wrong again ]. It doesn’t help your understanding of anything – it has no merit in my books. That’s also the reason that I’m reluctant to suggest writing a computer program to simulate it. Maybe I’d suggest it as a last resort.

January 11th, 2011 at 3:25 am

Chris,

I have this coin that I’m going to toss. It’s a very special coin that I had made at the National Burou of Standards. It’s perfectly balanced and it is 1 picometer thick, so the probability of its landing on it’s edge is only about

1*10^-12, so I feel safe in ignoring it. I’m going to toss it in a vacuum, using a device I had made that guarantees a vertical throw. I believe that you will agree with me that the probability of the coin landing heads is ½ . But wait, I just remembered. Back in 1957, I tossed that very same coin under those very same conditions. And it landed heads. Wait, now I remember that I also tossed it in 1976, and amazingly it landed heads again. Well I guess I had better revise my estimate of ½ . What was it you said it was, oh yes P(H) =1/4 , P(T)=3/4. Uh oh, son of a gun, I just remembered I tossed it again in 1994 and got a head. Now where does that leave us, I guess that now P(H) = 1/8 and P(T) is 7/8 ?

The point I’m trying to make is that each coin toss is independent of any coin toss, either before or after. By what mechanism does one toss influence another. Is it Newtonian action at a distance, or quantum entanglement? There is no way. The probability of each toss is ½. Yes it is that simple.

And Wiz, how does conditional probability apply to the three girl problem? Are you suggesting that the fact that there is an older sister will influence whether the next child is a boy or girl? And anyways, where is it stated in the problem that the girl was born first. How can she influence the first two births if she is born last? Are we back to quantum entanglement?

January 11th, 2011 at 5:10 am

Hi Al. Please give me your solutions to the following questions.

I flip three coins:

1. What is the probability of flipping HHH heads?

2. What is the probability of flipping HHT in that order?

3. What is the probability of flipping HTH in that order?

4. What is the probability of flipping THH in that order?

Now using the combined state notation (where possible):

5. What is the probability of flipping hhh (i.e. in any order)?

6. What is the probability of flipping hht (i.e. in any order)?

7. What is the probability of flipping htt (i.e. in any order)?

8. What is the probability of flipping ttt (i.e. in any order)?

If you could give an explanation for the second and sixth say. That would be greatly appreciated.

January 11th, 2011 at 5:36 am

Hi Al. I’ll answer those questions for you. Then all you need to do is confirm or correct my answers.

First I note that the probability of flipping three coins in a specified order is 1/8.

1. What is the probability of flipping HHH heads? 1/8

2. What is the probability of flipping HHT in that order? 1/8

3. What is the probability of flipping HTH in that order? 1/8

4. What is the probability of flipping THH in that order? 1/8

Now using the combined state notation (where possible):

I note that e.g. htt is equivalent to HTT+THT+TTH, so

5. What is the probability of flipping hhh (i.e. in any order)? 1/8

6. What is the probability of flipping hht (i.e. in any order)? 3/8

7. What is the probability of flipping htt (i.e. in any order)? 3/8

8. What is the probability of flipping ttt (i.e. in any order)? 1/8

Do you disagree with any of those answers? If yes, please indicate what you think they should have been (and why)? Or if you think I have misinterpreted the meaning of the states that you’ve introduced, then try to explain what you mean in clearer terms.

January 11th, 2011 at 6:48 am

You don’t need to keep repeating that coins don’t have memories etc. I’m sure that the vast majority of people are aware of that.

I have tried to avoid repetition, but in view of the nature of the problem and in order to not require too much scrolling up and down the page, some repetition is inevitable.

Assume that we are talking at cross-purposes. You can also safely assume that I know exactly what the point is that you have been making (and the point that Wiz is making).

Your attack on Wiz was partcularly amusing. He is on your side. He was hoping to show that using conditional probability (and he was doing so in an entirely reasonable way) that he’d establish that I was wrong. Whereas it actually shows that I am right and that he is wrong. I have known Wiz on this site for well over a year and have a very high respect for him (as I occasionally brag, I suggested his nom deplume because I thought it very applicable and I still do). I do believe that (if he has time to wade through my posts) that he will agree with me and that might be very soon.

But your failure to state the probabilities that you associate with bbg or hht etc., means that I’m not precisely sure what your combined states really are. I.e. you haven’t defined them well enough for me to confidently determine the probabilities that you associate with them. I had asked for some of those probabilities back at post 43 (at the end of first paragraph 1).

I’ll modify that a bit, I more or less expect that you’ll agree with questions 1,2,3 and 4 (post 54/55). But I really am unable to predict with any confidence how you’ll answer questions 5, 6, 7 and 8.

Your explanation in post 47 sounds like a quantum entanglement to me (I’m calling it Vulcan melding to emphasise the ridiculousness of it). I’m keeping the coins/girls as separate entities (I’ve tried this by appealling to their ages and marking them with coloured spots) – i.e. I’m enforcing the very opposite of entanglement and you are forcing entanglement.

I’ve got to end for a while. But I will be back. I think I will first explain to you what I was doing when I gave the coins a date stamp.

January 11th, 2011 at 10:53 am

I’m going to try to make the statement of the problem as clear as possible, and then I am going to analyse it. I’m doing it this way so that if anyone sees a fault they’ll have an easy time telling which step they think I’ve gooofed up on.

I have three children whose names are X, Y and Z (more on that soon). I shall make reasonable assumptions, such as the probability that any of any one of them being a girl is 1/2 and isn’t influenced in any way by the gender of its siblings.

The probability that X is a girl is 1/2, the probability that Y is a girl is 1/2, the probability that Z is a girl is 1/2. With this information, you can only conclude that my children’s gender could be (listed for brevity in the sequence XYZ): BBB, BBG,BGB,BGG,GBB,GBG,GGB or GGG. If you can’t follow what I’ve written so far, then stop reading; you aren’t yet ready to do probability problems. Each of these strings represents a definite possible family of mine; you can’t interchange their names without legal proceedings, you can’t interchange their gender without expensive medical bills. The kids aren’t 1/3 Boy 2/3 girl. There is no quantum entaglement with one of the other possible families. In short, treat each of those strings as if they represented a real and definite family. If you haven’t understood what I’ve written so far, then stop reading; you aren’t yet ready to do probability problems.

As there are 8 equally likely possible combinations, the probability of each of those combinations of kids is 1/8 (that’s because any particular sequence has a probability of (1/2)(1/2)(1/2) = 1/8 of occurring.) Although the problem didn’t specifically mention that my kids have a name, it is reaonable to assume that they do. And even if I hadn’t given them a name, the act of temporarily assigning them a name cannot in any way affect the underlying mechanics of the posted problem. I’d rather do that than call all the boys “Boy” and all the girls “Girl”; that would simply make my daily life and the description and analysis of this problem as difficult as possible for no useful purpose whatsoever, and to have a stab at unintentional quantum entangling them to boot. I’ll stay with this simple, powerful, everyday common-sense description because it is far less likely to allow the sorts of errors that Al and Wiz have let in (soz Wiz, believe it or not, I love your version of the solution and I can see why anyone could be taken in by it).

Right that’s the preliminaries out of the way. If I now let you know that I have at least one girl in my family, you’d know that that means that my family cannot be the BBB type – and that’s it, you simply now know that my family is equally likely to be BBG,BGB,BGG,GBB,GBG,GGB or GGB. No laws of physics have been broken, no kids were harmed or modified in any way by my revelation that at least one was a girl. If you have a sibling of mixed gender, and you tell someone that, no harm will come to anyone (unles the person you impart that information is a vicious thug who doesn’t like the look of you etc.) Anyone who thinks I’ve invoked quantum entanglement, used magic or in some way violated the laws of nature, stop reading; you are wasting your time, you aren’t yet ready to understand probability problems (I am not joking).

So you are now left with 7 possible and equally likely families that I could have had. Therefore the probability that I have any particular one of these families is 1/7. In particular, there is a unique family in that set of possible families that has three girls. Therefore the probability that I could have had a family with 3 girls (given that I have at least one girl) is 1/7.

Now anyone who doesn’t accept that logic, should have some idea where the error lies in it. I challenge you to point out the error that you claim that I have made. I do not want to be told about the no memory effect, the independence effects etc., they were essential parts of constructing my argument. My argument is only true because of those very things.

—- guys ‘n gals. I’m sorry if I appear to be (or actually am being) irritable and/or condescending or overly offensive, it’s only because I couldn’t sleep last night and because I’ve put a lot of effort into these posts and I don’t think the intended receipients are reading them with anything like the care that I’ve written them with.

It’s nowhere near as polished as I would like, but beautification takes a long time to achieve. I’m not getting paid for it either

—– Appendix

Notes. On counting how many children I have in each of the 8 possible families, I find that in 1 case I have 3 boys and 0 girls, in 3 cases I have 2 boys and 1 girl, in 3 cases I have 1 boy and 2 girls, and in 1 case I have 0 boys and 3 girls. I could now represent the possible families in another way. To avoid confusion (try to keep some continuity with previous posts) I’ll use lower case letters to represent the kids, but the sequence of letters in the string has no meaning, only the number of b’s and g’s has significance. I’ll call the objects represented by the strings such as bbg “collections”. I’ll always write the string in alphabetical order.

The collection bbb only contains the possible family BBB. The collection bbg includes BBG, BGB and GBB. The collection BGG contains BGG, GBG and GGB. The collection ggg only contains GGG. For convenience I’ll use P(x) to denote the probability of x occurring. If I consider the probabilities of my possible families as a collection, I immediately see that P(bbb) = P(ggg) = 1/8 and P(bbg) = P(bgg) = 3/8. Although I could use these collections (in fact I have a few posts back) they are not best suited to my purposes. They lead to less elegant solutions.

January 11th, 2011 at 3:59 pm

I will agree that it’s 1 in 7. The actual wording of the question appears to be unambiguous and I can’t find myself able to argue any other angle … woe is me, what have the holidays done to me ?

January 11th, 2011 at 4:21 pm

Consider the following. I have a chess board. On one square I place a grey counter. On another square I place a black counter. On the other 62 squares I place mauve counters (one per square).

What is the probability that if I uniformly randomly pick one of the squares that it will contain a grey? That’s easy, it’s 1/64.

Similarly, the probability that I could have picked the square with the black counter is 1/64.

And finally the probability that I could have picked a square with a mauve counter is 62/64.

I now remove the black counter, leaving 1 grey and 62 mauve counters. What is the probability that I could pick the square with the grey counter (given that I actually picked an occupied square)? That’s easy too, it’s 1/63.

I return the black counter to the board. What is the probability that I could pick the square with the grey counter (given that I don’t pick the square with the black counter)? That’s the same as last time, it’s 1/63.

I now consider a family of 6 children. Let them be named U,V,W,X,Y and Z. Each child has a 1/2 probability of being a boy or a girl. So there are 2^6 = 64 possible gender arrangements. Exactly one of those arrangements is GGGGGG and exactly one is BBBBBB. The rest are some mixture of boys and girls.

I’ll represent the BBBBBB combination with the black counter, the GGGGGG combination with the grey counter, and each of the mixed gender families with a mauve counters. A perfect one-one correspondence.

I know ask the same questions as above, but can’t help that notice that I could substitute the phrase “black counter” with “BBBBBB family”, “grey counter” with “GGGGGG family” and “a mauve counter” with “a mixed gender family”.

Conclusion. If I have 6 children, and at least one is a girl, the probability that they are all girls is the same as the probability of me picking the grey counter, given that I don’t pick the black counter. So the probability is 1/63.

So Al and Wiz. Whats’s wrong with that argument? I’ll give you a hint – nothing is wrong with it.

It happens that although you two seem to have a different philosophy, your actual calculated values are the same. You’d say that the probability of having 6 girls, given that you had at least one girl is (1/2)^5 = 1/32. How do you reconcile yourselves with that, that’s almost double the correct value.

Imagine I scaled my argument up to a beach, where each grain of sand represnted a possible family. But one particular grain represented the all girls family and another represented the all boys family.

By what physical mechanism does removing the boys only grain of sand cause a new grain of girls only sand to appear on the that beach?

I promise that I’m not laughing at you. I’m not really that sort of bloke. I only get a bit silly, and that’s not too often.

January 11th, 2011 at 4:29 pm

Thank you Eketahuna. Wow, that’s a turn up – I hadn’t expected that.

I’d like to say I’m done arguing this one. But the fact is that even if I get crotchety etc. (I am only human), I love coming up with new ways of understanding and explaining. In my heart, I enjoy helping people to understand things.

I enjoy the fact that these problems stump even the brightest people. Check out “Ask Marilyn” on the Monty Hall problem. University mathematics professors get that wrong and insist that the people who’ve got it right are nutters. It’s so simple.

January 11th, 2011 at 6:38 pm

With this (and the Monty Hall problem), it’s all in the wording, and removing all ambiguity, or allowing people to read things into it that aren’t actually mentioned, and so trapping them into a false answer. Or failing that (as with MH), specifying what criteria you have assumed, to remove any ambiguity – or argue the opposite for the sake ot it

Wow, that was rather rambling …

January 11th, 2011 at 10:36 pm

Hi Chris

i don’t be mind at all.if at any the point,you feeling to change my posts to be making it have the more sence,i will see the thankyouness in it.b)

but really-i’m glad you see what i was trying to say

how many more posts to go?

January 11th, 2011 at 10:40 pm

i don’t try to make a smile face-and one appears

i want one to show up-and nothing

aaaaaaaaggggggggggB)

January 11th, 2011 at 11:12 pm

Time for me to withdraw from the Three Sisters fray.

I can’t prove that 1 in 4 is the answer to everyone’s satisfaction (except mine) and I can’t refute the 1 in 7 answer. I don’t really have the time or the knowledge of stats and probabilities to work through all the arguments, so I shall just leave it at that.

It’s been good fun and we got to 63 posts between all of us.

Before I sign off I’d like to thank Karl for pointing me to the Straight Dope website way back in post # 5. I liked it so much that I’ve now subscribed to it.

January 12th, 2011 at 12:47 am

Hi Chris,

First of all, I did not “attack” the Wizard of OZ, in fact I did not attack anyone. The Wiz came up with the correct response, which I acknowledged way back in post 16. I agreed with his post #14, but I did not agree with his use of conditional probabilities in his post 49. I don’t understand about his being on my side. I didn’t realize that we were involved in some sort of team competition…

In post number 53 you posed a set of problems. First, why stop at 4 in the first set? Why aren’t we interested in TTT, TTH, THT, HTT ? Anyways, it doesn’t matter since the answers to these 4 are the same as the answers to the previous 4, namely 1/8 (as you wrote in post 54. Now I’m afraid I don’t understand problems 5-6. In problem 7 did you mean to flip the coin 9 times and get HTT THT TTH ? if so than the probability is .5^9, or if you prefer 8^-3. If you meant what is the sum of the probabilities of getting HTT, THT, TTH than the answer is 3/8 . If on the other hand you are asking for the probability of tossing a head and two tails , in any order , then that is the same as getting the probability of HTT (1/8) or the probability of THT (1/8) or the probability of TTH (1/8). So the probability of tossing a coin 3 times and getting ANY combination of heads and tails is (drum roll please) 1/8. My question to you is; what has all that got to do with the question you originally proposed? Had you asked; what is the probability that the man’s three children are all girls? Then the answer would be 1/8. But that’s not what you asked. If you had asked; what is the probability that one of the following scenarios occurred GGG, GGB, GBB, BBG, BGG, GBG, BGB? (Which I think is the question that you have been trying to answer?) . Then the answer is 1/7. But that is not the question that you posed either. Here is what you wrote, and I quote: “I have three children. If at least one of them is a girl, what is the probability that all three of them are girls?” One of the children is a girl. That’s an established fact, there is no probability associated with it, i.e., the probability is 1. So the question we have to answer is; what is the probability that the other two are girls? That answer is .5 * .5 = ¼ . Or if you insist on including the other child, known to be a girl, then we have .5 * .5 * 1 = ¼. Or if you prefer, look at the Wiz’s post # 10. He also has the correct answer, and his reasoning may be more to your liking.

Let’s face it Chris, you are wrong. You seem to have problems with the idea that someone can disagree with you, or worse, that you might even be (gasp) wrong! Yes I do indeed find your posts offensive.

January 12th, 2011 at 3:57 am

I’m pretty certain chris has the correct answer to the problem as worded (i.e 1/7)

for those who think order is of no consequence in this discussion – consider flippping coins and only two of them –

does anyone think there are only three outcomes? and the possiblity of and a head and tail is 1/3.

If so i’d like to meet you for some poker.

January 12th, 2011 at 5:52 am

This’ll put some of you to shame, my friend Seldon, aged 16, gave me this:

If we’re assuming one of the children is going to be a girl then isn’t it just a simple binomial distribution? each trial (the children) is independant of the others, theres just 2 possible outcomes (boy or girl) and im assuming the probability of it being a boy or girl is constant? or is that not true?? so for P(x=2) were success is a girl and failure is a boy? B(2 , 1/2) which gives us 1/4…. or am i way way off with that line of thinking?

I’m now rather confused, because I still want to believe it’s 1/7. :S

January 12th, 2011 at 7:20 am

Thank you everyone for helping to give this a high post count I hadn’t expected it to get more than 10 responses.

Hi Iain. Nice one.

Hi Krazeedude. Once the case of three boys has been removed, the family (of 3) is no longer described by the binomial distribution. So it can’t be used directly. But your friend also has assumed a particular child, and so has effectively forced a particular girl to be the one alluded to. I only introduced the binomial distribution because of the people who gave 1/3 as the answer, and for a more subtle reason to do with a part of Al Gelman’s analysis. I wish Iain had posted earlier, that would have saved me from exploring that avenue.

Hi Wiz. I really do understand why you believe in your analysis. The way you reason it is very persuasive, it cost me a lot of brain cells before I really saw the nature of the flaw. I still am not completely clear on how to explain it in terms of your analytical technique. The best I’ve done is in posts 28 and 29. [Update, I have now fully dealt with it, see posts 69 and 71].

A (very slightly) different way of looking at is: before finding that there was at least one girl, you would have said that the probability of three girls was (1/2)*(1/2)*(1/2) = 1/8. It is also fairly obvious that that’s because there were 8 different possible families. We’ve only eliminated 1/8 of the families, so we can only have increased the probabilty of them all being girls by a similar fraction. So you’d roughly expect the 1/8 to change by 1/8th. The exact calculation is (1/8)/(1 – 1/8).

But the fact is that 1/7 is the correct answer – I am certain of that, I never doubted that at all. The listing method of solving these problems is very powerful and should be deferred to; it is practically foolproof.

Hi Al. OK “attack” was too strong a word, “take issue with” would have been better. Being on the same side: that just meant you both were both proffering the same numerical answer and for essentially the same reason, you just use different words. I can understand Wiz’s version, but yours is ill-defined. But there is no basis for taking issue with Wiz’s suggested use of the conditional probability formula. I showed how it too led to the correct result in posts 50 and 51. I suppose that I’m a little disappointed that Wiz would have used it as a proof that he was right if it agreed with his result, but ignores it because it agrees with mine. But he at least freely admits that he doesn’t know how to apply the formula, and so I accept that he can’t tell if I’m cheating when I use it. So I have no real problem with that. I also conjecture that Wiz is having a bout of cognitive dissonance. I suspect that he does accept the logic that I’ve provided, but also is convinced that he is right at the same time. I had a run of those a few weeks ago, it really is a strange experience.

Post 53 problems. I didn’t want to ask you more questions than I thought were necessary; in fact I didn’t need to ask as many as I did. Based on your posts 36, 42 and 64, Although none of your statements individually unequivocally pin it down (as nearly every one is ambiguous – we both usee “any” with a different meaning, both reasonable), I have no real doubt that you believe that the order of flipping the coins (or choosing the children) doesn’t matter. i.e. you believe that flipping T then T then H is the same as flipping T then H then T and is also the same as flipping H then T then T (which I write as TTH, THT and HTT) are indistinguishable and presumably, there’s really is no need to have three different ways of writing that combination. That immediately implies that (again using my notation) that P(hhh) = P(hht) = P(htt) = P(ttt) = 1/8 (as that is a figure you have given). The fact is that P(hhh) = P(ttt) = 1/8 and P(hht) = P(htt) = 3/8. This is easiest to appreciate if it is written as a table of the possible time sequence of flips:

123 => flip number 1st, 2nd, 3rd

HHH

HHT

HTH

HTT

THH

THT

TTH

TTT

Each of those eight possibilities is equally likely.

Although I used time sequence. I could just as easily as written “1″, “2″ and “3″ on the coins with a marker pen. If I count how many H’s and T’s there are in each of those, I get (in the same order as the table):

3H0T,2H1T,2HT,1H2T,2H1T,1H2T,1H2T,3T or using my lower case notation

hhh,hht,hht,htt,hht,htt,htt,ttt and I find that I have 1 hhh, 3 hht, 3 htt and 1 ttt.

So P(hhh) = 1/8, P(hht) = 3/8, P(htt) = 3/8 and P(ttt) = 3/8.

Bear in mind those sequences could be the combination of an electronic lock – only one will work.

Writing your statements in your post (42) in my notation, you correctly say that we are left with three possibilities ggg, ggb,gbb. I guess that the 1/3 people are just seeing a list of three things, one of which is the one sought so the just go, abracadabra, it’s 1/3.

What they should have calculated is P(ggg) / (P(ggg) + P(ggb) + P(gbb)) = (1/8)/(1/8 + 3/8 + 3/8) = 1/7.

See post 86 for an analysis of your misinterpretation of my question.

I can’t believe that after all the very clear posts that I have made that you can believe that you are correct. You accuse me of having difficulties in admitting that I am wrong LOL, you have that back to front. You are wrong, the phrase “at least one of my children is a girl” is logically identical to” they are not all boys”, so how did you manage to choose the one that is definitely a girl?

—-

Despite my invitations to do so, you have failed to point out the error(s) in any of my logic, you have continued to fail to notice that you haven’t answered the question that was posted (you keep on ignoring the “at least” bit of it, you’d rather resort to insulting me – so, to use your own words (as clear logical arguments have failed), it’s my turn: let’s face it Al, you are wrong. You seem to have problems with the idea that someone can disagree with you, or worse, that you might even be (gasp) wrong! Yes I do indeed find your posts offensive.

January 12th, 2011 at 7:57 am

Hi Wiz. Another point (I hope I haven’t made it before), in 3 out of 7 cases, it cannot be the first girl that was being alluded to. The same applies in each case. I promise that your logic is still causing my brain cells to die at a high rate. Thank you for giving me a really hard to correct perspective.

Hi Al. Your phrase “So the probability of tossing a coin 3 times and getting ANY combination of heads and tails is (drum roll please) 1/8″. “ANY” should be e.g. “any particular (or given or specific)”, otherwise the answer is 1 – but I’m nit-picking.

The statement “I have three children, and at least one of them is a girl”, is exactly equivalent to the statement “my family is one of the following combinations of children: BBG,BGB,BGG,GBB,GBG,GGB or GGG” (using the shorthand way of describing). Obviously I’m assuming (as are you, that each child is equally likely to be a boy or a girl, and that each child’s gender is independent of their siblings gender). Note that in each possible family (in the lists) at least one of the children is a girl (and so the probability of at least one of them being a girls is 1 is satisfied). So that list is completely consistent with the posted problem.

If the question had been “I have three children. If at least two are girls, what is the probability that all three are girls?”, then on inspecting the corresponding list: BGG,GBG,GGB,GGG, I’d deduce that 1/4 is the answer. I used that logic when I answered Knightmare’s first question (post 44) to get P(H) = 1/4, P(T) = 3/4, as I stated without proof in post 51.

The people who previously said 1/3 will probably say we’ve only got bgg and ggg so 1/2. The Wiz-Al type logic would give 1/2 also, but that is essentially a coincidence.

January 12th, 2011 at 12:38 pm

Hi yet again Wiz. I think this is what you’re looking for.

First a slightly strange way of examing the normal (8 possible families case)

Here’s the applicable list for reference:

BBB,BBG,BGB,BGG,GBB,GBG,GGB,GGG.

Examing the list shows that each child has a 1/2 probability of being a girl. If the first child happens to be a girl, then the possible families are GBB,GBG,GGB,GGG and it is apparent that the other two children obey the usual rules and so if the first child is a girl, then the probability of the other two children being girls is 1/4. Exactly the same applies if we choose to examine the second or third child.

Putting that together, the probability that a particular child is a girl is 1/2, and then if that child is in fact a girl, then the probability that other two children are girls is 1/4. So altogether the combined probability of having 3 girls is (1/2)(1/4) = 1/8.

It doesn’t matter which we girl we pick on.

Now for the at least one is a girl family. I copied and pasted the previous argument and edited only where absolutely necessary.

Here’s the applicable list for reference:

BBG,BGB,BGG,GBB,GBG,GGB,GGG.

Examing the list shows that each child has a 4/7 probability of being a girl. If the first child happens to be a girl, then the possible families are GBB,GBG,GGB,GGG and it is apparent that the other two children obey the usual rules and so if the first child is a girl, then the probability of the other two children being girls is 1/4. Exactly the same applies if we choose to examine the second or third child.

Putting that together, the probability that a particular child is a girl is 4/7, and then if that child is in fact a girl, then the probability that other two children are girls is 1/4. So altogether the combined probability of having 3 girls is (4/7)(1/4) = 1/7.

It doesn’t matter which we girl we pick on.

I accept that in the first (normal family) argument that the fact that the first was a girl wasn’t a requirement for the other two to be girls with probability 1/2 each. Obviously if the first child happpened to be a boy, the other two both being girls would have a probability of 1/4. But in the at least one girl family, the condition that the first was a girl was crucial. If the first had been a boy then the probability for each of the other two children to be girls would have risen to 2/3 each, and the probability that they were both girls would have been 4/9.

Does that do it for you? I doubt that I’ll come up with a better argument of that form.

January 12th, 2011 at 12:55 pm

I’d like to say a special thank you to Karl and an extra special big thank you to Knightmare for their support (and patronage )

I nearly forgot, thank you Dual Aspect, you went above and beyond…

Also a special mention to Wiz, he introduced an excellent paradox – I think I’ve now cracked it.

Knightmare, I usually use : then ) for a

or ; and ) for a

I’ll have to watch out for the 8 ) – what a pain that could be.

January 12th, 2011 at 3:20 pm

Hi again Wiz. Breakthrough. At the risk of sounding patronizing, thank you again for giving me such a hard time.

About two minutes before staring to write this, I had a flash of insight. I’ve never before so consciously appreciated what one’s really doing when calculating the probability of flipping two coins and getting two heads. So far I blindly just write down 1/2 * 1/2 and think nothing of it; I take it for granted.

When we are calculating the probability of flipping two heads, what we are implictly doing as we move along the terms in the equation is saying that when we flip the first coin, we note that the probability of it coming up heads is 1/2. But the next term is only meaningful if the first coin had actually came up heads (if it didn’t, we’re no longer examining a two heads problem). In turn this means that we must consider afresh the state of the system before we calculate each term. Of course we don’t normally concern ourselves with this level of thought as frequently the actual realisation of a particular event has no effect on the subsequent events. i.e. the coin probabilities don’t change because of previous coin flips (where have I heard that before, LOL).

Now try this with the girls problem. On examing the situation we find that the probability of picking a girl is 4/7. So we write down “4/7″. We then re-examine the siutuation using the assumption that we have already really picked a girl. We then find that the probability of picking a girl is now 1/2. So we write down “*1/2″ (i.e. we now altogether have written down 4/7*1/2). We then re-examine the system and find that (after having allowed the we have actually picked the previous two girls) that probability of the remaining child being a girl is 1/2. So write down “*1/2 “to get “4/7*1/2*1/2″.

It is now absolutely obvious, to me, why the probability is 4/7 * 1/2 * 1/2 = 1/7, using Wiz’s style of analysis. Of course I have used the list of possible families to examine the situation at all stages of the analysis.

Note that we went from BBG,BGB,BGG,GBB,GBG,GGB,GGG to GBB,GBG,GGB,GGG, then when we consider the next child (prob 1/2) the assert that that child actually is a girl we are left with GGB,GGG, ready for the third term.

All that reminds me of an anecdote Slavy posted about a professor who was interrupted by a student during a lecture in which he was proving a mathematical theorem. The student suggested that it was a waste of time doing the proof, as it would probably turn out that the theorem was obvious. Apparently the professor went into deep thought for half an hour. Then he said, “Thank you dear colleague, you’re right, this theorem is obvious”.

—–

Addendum (24 hours after first posting this): I now realise how to write all of this in terms of conditional probability. I also realise that we do it all the time (in principle) but don’t realise it. Unfortunately, the full unambiguous conditional probability expressions are really quite a mouthful, and I only expect that Slavy would appreciate them. So I won’t bother to write it up.

The most relevant and compressed form of the calculation is:

P(XYZ=GGG | XYZ != BBB) =

P(X=G |XYZ != BBB) * P(Y=G | X=G & XYZ != BBB) * P(Z=G | XY=GG & XYZ != BBB)

That’s about the shortest equation I’d have to write. It’s no wonder that I couldn’t get that right much earlier.

If we didn’t specify that at least one was a boy, then for the ordinary case of three girls, we’d have:

P(XYZ=GGG) = P(X=G) * P(Y=G | X=G) * P(Z=G | XY=GG)

which readily simplifies to

P(XYZ=GGG) = P(X=G) * P(Y=G ) * P(Z=G)

January 13th, 2011 at 6:38 am

The contents of this post have been deleted by me, Chris.I can’t delete it, without making all the post reference numbers go out of kilter.January 13th, 2011 at 4:43 pm

Hi Chris, I’m going to sound surprisingly thick here, and I’ll understand if you rage-face but what does order have to do with it? Does arranging the 3 children BGG GBG and GGB really have an effect on the probability, it is still the same 3 kids? Long story short, removing the surplus, it leaves 4 results, one of which is BBB, removing it leaves 1/3. This hurts my head, so I’ve given it to my Maths teacher, who’s taken it away and will contemplate it tonight. With any luck she’ll have solved it by tomorrow. I’ll also present it to my physics teacher, who apparently has degrees coming out of his ears. I’m trying to drag as many people into this as possible, going by the theory of ‘wisdom in numbers’, it should wield good results :]

January 13th, 2011 at 5:38 pm

Hi Wiz and Al. Just another take. This time I’m considering the semantics of the problem.

In order to make it meaningful to refer to a particular child as being a girl we need some way of identifying that child. If in the question I had added a remark such as “one of my children went to the cinema last night. She said that the film was rubbish”. Then we’d have a definite identification of one of the children who was a girl and a means of (collectively) identifying the other two children i.e. the two who didn’t go to the cinema last night. Then it is perfectly sensible to calculate that the probability of the other two being girls is 1/4.

[Feb 2014: I've posted a new problem "Girls, girls" to better address this issue. As you'll see there, nothing is quite straightforward, but my above example is correct (I think)].

But if I have no means whatsoever of unequivocally and uniquely identifying a particular child as being a girl, I also have no means of identifing the “other” children either. It is meaningless to talk about the other two children – there is no “other” two children. So it is also meaningless to calculate the probabilities of the other two children being girls.

Even if you haven’t literally said “the girl” or “the other children”, you both have used equivalent concepts.

On the other hand, it is perfectly reasonable to theorise about what would happen if you had a means of uniquely identifying a particular child. You can then calculate that the probabilty of that particular child being a girl is 4/7. etc. I started of by identifying the children with their ages, I then tried a paint spot. But nowhere (except possibly to make a theoretical point) have I ever said the I knew that a uniquely identified/particular child was definitely a girl. But I don’t even need to specify how I could identify or distinguish them, I just need to know that in principle I can do that. Ultimately, I did it so that I could take full advantage of the major labour saving device exemplified by BGB as that allowed me to keep track of every available bit of information, in what I regard as self-evidently sensible way.

It happens, that in theoretical physics, it is possible to have systems of particles (the bosons, which are named in honour of Bose), where under certain conditions, the individual particles are indistinguishable, even in principle. These particles obey Bose-Einstein statistics. I believe tha B-E statistics are needed to explain e.g. superconductivity and superfluidity.

I’ve just read up on Bose. Apparently he got lucky, he made an error which was equivalent to saying that the probability of flipping two heads was 1/3. This error caused his equations to correctly deal with the infamous ultraviolet catastrophe.

If coins were bosons, the if you flipped two of them you’d find that it’s just as likely that you”d flip hh,ht or tt, and the probability of each case is 1/3.

If I have understood it correctly, then if in the original problem, my three children were bosons, and it was equally likely that each was a boy or girl, then it would be equally likely that I’d have 0, 1, 2 or 3 boys and the same for the girls. i.e. P(bbb) = P(bbg) = P(bgg) = P(ggg) = 1/4. Then the answer to the original question would be P(ggg)/(P(bbg) + P(bgg) + P(ggg)) = 1/3.

This difference in behaviour of bosons and fermions (everything known is one or the other) is so profound that, roughly speaking, you can say that matter is made from fermions and that the forces that glue the matter together (and let it interact) is made from bosons. The fundamental bosons are the photon, the gluon and the W and Z particle. They are the mediators of the electromagnetic force, the strong nuclear force and the weak nuclear force. If they turn out to exist then the graviton (the mediator of the gravitational force) will be a boson and the Higg’s particle (the particle that give arise to mass) will also be a boson.

January 13th, 2011 at 6:06 pm

Hi Krazeedude. Good question. I use “order” to define the sequence of letters in a string like BBG to mean that I associate the first letter with one particular child, the second with another particular child and the third with (you probably guessed) the remaining particular child. It could be in the order eldest, middle youngest. Or it could be the in the order of height. But whatever the letter is in each position, it is always referring to a particular child. It is a positional notation.

You’ve probably seen a table that shows the possinle outcomes of flipping two coins (A and B) it probably looked something like (I’ve deliberately chosen some crazy looking probability values):

A B P(A and B)

—————–

H H 1/10

H T 2/10

T H 3/10

T T 4/10

When discussing probability problems based on that probability table, it would be a pain if you constantly had to write the P(both A being heads and B being a tails) = 2/10. But if you say I’ll always write the state of A followed by the state of coin B as an ordered string, you can shrink) the previous down to P(HT) = 2/10, and also P(TH) = 3/10. But if you don’t used the position /ordered idea, well you’ve got a lot of typing to do.

In fairness, you could write P(A=H and B=T) but P(HT) is nicer.

I also introduced the idea that that would be true only if I wrote the letters in uppercase. If I wrote them in lower case, I meant that the positions of the letters no longer implied a particular girl. Moreover that e.g. bbg meant the collection (taken together) of BBG,BGB and GBB. The only thing that mattered was the number of b’s and g’s. I also endeavoured to write only one of ggg,bgg,bbg or bbb (i.e. n alphabetical “order” – I snuck that one in ) If I did that for the coins example, immediately above, Then P(ht) = P(HT) + P(TH) = 2/10 + 3/10 = 1/2

If I write the number 135, the order of the digits is important. In this case the first digit refers to the hundreds etc. If we didn’t do that, then we’d be well and truly stuffed.

January 13th, 2011 at 6:25 pm

… nearly forgot. I tend to use order in a seemingly different way. Sometimes I mean it chronologically. e.g. If I flip a coin three times and the first flip is an H, the second flip is a T and the third flip is a T, the I would write that as HTT (i.e. in the same order as I flipped the coins). In fact I’m really saying that the first letter is asociated with a particular event, but I think it natural to make the be the first event. But I could equally well have used ABC to mean 2nd, 3rd and 1st events (in that order). But whatever you choose, you must stick with it. If you change the order willy-nilly, then it will make things more difficult.

Have no fear. Al had to push me quite hard before I got annoyed. You might notice that I’m haven’t got the strops with Wiz. Unlike Al, Wiz hasn’t got a bad attitude. You notice in his first post he knows that he’s likely to be the sucker I’m looking for. He’s saying that he knows he’s fully aware that he might be wrong. He also did it amusingly. He made me laugh, but with him, not at him.

I don’t have any problem whatsoever with people not understanding or knowing things. I don’t know 99.9+% of what’s out there myself – but at least I know that and behave accordingly. I think I have a fairly good assessment of my abilities; at least, roughly speaking, I know what I know, I know what I don’t know. I know that I might be wrong even when I think I’m right.

January 13th, 2011 at 6:50 pm

Krazeedude. Be warned. Problems like this (e.g. the Monty Hall problem) have been answered incorrectly by Nobel prize winners. They have taken a stance not completely disimilar to Al’s. They accuse the people (who have actually understood the problem correctly and have given the right answer) of being idiots and so forth. They have even threatened to publish, in writing, their belief in their (unfortunately wrong) answer.

I won’t be even slightly surprised if your teachers get it wrong. One of my brothers is a Head of Science. He absolutely refuses to accept the correct answer (to the Monty Hall problem). His logical refutation is farcical. He refuses to accept any answer that contradicts his beliefs. He now refuses to discuss the matter yet alone to ever actually think about it. He has completely closed his mind – such is the strength of his belief.

Wisdom in numbers! I saw the results of a survey regarding the Monty Hall problem. I believe that of the 90% of people who felt that they knew the answer, 82% believed in the wrong answer.

A similar problem regard flipping two coins almost caused open war at a particular school. The headmaster(who unfortunately believed in the wrong answer) announced in a morning ceremony that the right answer was (the wrong answer). One student even simulated the problem with a computer and found the prediction to be true. But he still was so convinced that it was wrong, that he said that it [i.e. the right answer] was only true if you did it thousands of times in a row. He obviously thought that the wrong answer was right if you only flipped the coins once i.e. reality denial boosted on humongous doses of steroids.

The first few posts that give 1/7 as the answer are very simple to understand. Don’t simply take some experts advice, convince yourself of the true answer, right or wrong.

Although it’s not quite relevant here as we haven’t done any experimental verification, Richard Feynman (a true genius and a very wise and wonderful man) once said something along the lines:

It doesn’t matter how smart a man is. It doesn’t matter what his name is. It doesn’t matter how expert he is. If his theory isn’t confirmed by the evidence, then he is wrong.

January 13th, 2011 at 10:25 pm

Hi Wiz. Yet another point. First, please believe me, I’m not bashing you, I am genuinely enjoying the intellectual thrill of this. I’ve become aware of so many different aspects of probability in the last few days, that I’m on a major high. Remember, not much over a year or two at most, I was a complete novice at probability. I hadn’t had any excuse to use what I’d learnt about it for 35 or so years.

I’ve just re-read your post (20) where you say:

” …

There are then 4 cases with equal probability after you’ve got your girl, one being GG, so the chances are 1 in 4.

Putting it mathematically:

(pG1 + pG2 + pG3) * (pBB + pBG + pGB + pGG) = 1

where pG1, pG2, pG3 are the probabilities of *THE* girl coming first, second or third, adding up to 1, and pBB is the probability of two boys out of *THE OTHER* two children, pBG and pGB are the probabilities of a boy and a girl, and pGG is the probability of two girls. Each of these is equal to 1/4, so pBB + pBG + pGB + pGG =1.

More pertinent to our situation is that (pG1 + pG2 + pG3) * pGG = 1/4.”

The problem is that you are explicitly identifying a particular girl (I’ve emphasised that with *THE* in the above. I understand that by pG1 that you mean the probability of child1 being THE girl.

[later: you're also stuck on "one girl", and ignoring the "at least", the latter is crucial.]

I also realise that you are recognising that THE girl could be child1, child2 or child3. But how can you actually determine the probability pG1? You can’t just say it’s 1/3, that’s simply the probability of choosing THE girl in a one girl family. If there were two girls, how do you decide which ONE is THE girl. You could arbitrarily, uniquely and randomly identify the children by getting them to pick a numbered label from a hat. In fact that has now completely identified each child in a unique but random way. And for that reason, there’s no useful purpose in further randomising by randomly picking a number – it doesn’t make your pick any more random. So you may as well always pick child 1. But whether you always pick child 1 (or 2 or 3), or always pick randomly (1,2,3) and take the average, you’ll get pGx = 4/7. Of course you could cheat – you could simply ask for A girl to be selected and say that she is THE girl. But then you’ll find pGx = 1 (not really x anymore as you’ve not actually picked randomly), because every family that isn’t all boys has at least one girl, so you always get A (but not THE) girl.

The good news is that, once you’ve picked a girl, you will find that pBB = pBG = pGB = pGG = 1/4, where the two children are not the girl (as we now know her to be) that you picked.

So your equation (pG1 + pG2 + pG3) * (pBB + pBG + pGB + pGG) = 1 is false, because the LHS = (12/7)* (1) = 12/7 != 1 either. There really isn’t a way to fix the equation, except by re-defining what it means. It seems to me that there is no clear way to do that.

Your equation (pG1 + pG2 + pG3) * pGG = 1/4 is also false. The first factor on the LHS is supposedly a probability. But that probability = 12/7 which is impossible. However the LHS can be fixed, by redefining its meaning. Clearly the first factor is really pGx = 4/7. So we get the final result, pGx*pGG = 4/7 * 1/4 = 1/7.

— See post 79.

January 14th, 2011 at 6:34 am

Heavily edited a few hours after posting.

It could be that I’ve never met my children or known their genders (unlikely, but not impossible), until, one day, when my wife (for the sake of argument) might have informed me that I had a daughter (call her Sarah-Jane Smith for definiteness). When I posed the problem on this site, that’s all I knew. I can calculate the probabillity of my other two children (the ones who aren’t Sarah-Jane Smith) being girls is 1/4, you can’t – you don’t have a “Sarah-Jane Smith” identifier. Moreover, if I met my children later, and I only had 1 girl, I’d know 100% who that was. But f there were two or three girls, I myself wouldn’t know which one was Sarah-Jane Smith. So you definitely can’t identify THE girl.

I think this is at the very heart of the interpretation of probability. When we calculate the probability of something, we can only do that based on the information available to us.

Get out of that without moving If you think you can, then see post 84 for the final nails in the coffin.

January 14th, 2011 at 7:20 am

Just thought I’d add my 2p’s worth at the request of some year 12’s in the school I teach at.

If you think it’s 1/3, you’re wrong as (among other things) P(2 girls) is not the same as P(one boy and one girl) since you could have boy then girl or girl then boy.

If you think it’s 1/4, you’ve actually answered the question “what is the probability that all 3 are girls given that THE FIRST is a girl” or similarly THE SECOND or THE THIRD.

The correct answer is 1/7. You could use the arguments mentioned right at the top of this thread, or use the formula for conditional probability mentioned later on

{P(B|A)=P(A&B)/P(A)}.

January 14th, 2011 at 7:21 am

Dear Lord,

So much talk over such a simple problem, the answers obviously 1/4, just kidding. 1/7.

A lot of what appears to be the problem is that some people attribute the same probabilities to all outcomes. i.e. you Toss a coin twice and can either get Two heads, two tails, or one of each, therefore the probability of getting each outcome is obviously 1/3. This is a thoroughly erroneous statement.

To solve this problem I drew a simple family tree style diagram to illustrate all the possibilities and I ended up with the classic 8 outcomes;

BBB,BBG,BGB,BGG,GBB,GBG,GGB,GGG.

I then read your statement that “at least one of them is a girl”, this knocks out BBB, reducing the list to 7;

BBG,BGB,BGG,GBB,GBG,GGB,GGG.

So out of these 7 outcomes in only 1 instance are there 3 girls. The probability is therefore 1/7th.

Monty Hall problem is simple when you use the million doors with a car behind 1 and goats behind the rest approach.

I choose a door out of the 1 million. What are the chances I chose the right one? 1/1,000,000.

The host of the show then opens 999,998 of the remaining doors (this’ll take him about 12 days if he’s quick) Anyway, whilst the chaos of releasing nearly a million goats into a live studio audience ensues, I still have my door and now the host has his.

I’m not going to be any more right than I was before, the car isn’t going to have magically transported itself behind my door, unless pushed there by determined goats, or me being crazily lucky and beating odds of one in a million in choosing the correct door straight off. So it has to be behind the hosts door. Has this been put to your brother in these terms? Would he really stick with his door?

Chris, this is the first time I’ve read your Blog, Nice. Bookmarked.

January 14th, 2011 at 8:09 am

Hi Chris, I discussed it with my teachers, the physics guy said 1/4, however the maths teacher said 1/7 and gave good reasoning. I’m back believing it’s 1/7, as does most of my class.

Also, that Monty Hall problem sounds exciting, I’m going to look it up :]

January 14th, 2011 at 11:32 am

Chris,

only 11 more posts to become the most popular. how does that make you feel? I noticed that you just beat your own post “Black or White”, where it came down to simple logic being the reason people couldn’t come to the correct answer.

I did notice, however, that you have nearly half of the posts. I don’t know if that should affect anyone’s opinion one way or another, just thought I would mention it.

Congrats on such a popular post, and I hope you don’t catch too much flak on future posts by upsetting anyone on here.

It was enjoyable to read through (in several sessions) all of the back-and-forth on this one. Thanks again for keeping me thinking.

I still think the 1 in 4 people have a good argument in thinking ’since you told me one is a girl, you must know which one’, BUT the math speaks for itself that there are 8 possible families before you start, and you are only telling us it is not one of them.

Maybe next time it should read something more like “I have three children. They are not all boys. What is the probability that all three of them are girls?”, unless you are trying to catch people in the wording of the problem.

Anyway, just doing my part to get you above “Man in a room”. By the way, we never did find out how he got out, since the person to post the problem never answered it. I like for this sort of “thinker” problem to be on top anyway.

January 14th, 2011 at 12:02 pm

Wow. Thanks folks. I know that I’m the main reason the post count has gone high – so I maybe I cheated I’m disgusted that the “the man in a room” is more popular – it’s

~~peurile~~puerile [Thanks Knightmare and Karl or spotting that].Seconds after pressing shutdown, I knew how to finally lay the 1/4 spectre to rest. In post 79 I had only considered the case of just one particular girl, Sarah-Jane Smith. Of course, I could have been told that I had another daughter (Tara-Jane Smith) or even a third one (Una-Jane Smith). So for me, who has definite girls in mind I’d calculate, 1/4, 1/2 or 1 as the probability of having three girls. NB I have just heavily re-edited post 79.

Only I know how many particular girls I had in mind – you don’t. That should finally dispose of 1/4 as being the answer based on the information provided. If I’d been smart enough to have thought of that earlier, I doubt if this page would have got more than 25 responses, tops.

I hope that most of you realise that I’ve not been writing all these posts because I’m trying to prove that I’m right. If I’d been proven wrong, that would have been even more exciting. I’m genuinely fascinated and thrilled at all the different aspects of this exceedingly simple problem and with the new personal understanding that I’ve gained.

No doubt, much of the deeper(!) stuff is covered in chapter 1 of a decent probability text book.

—

In summary: I believe that the question quiet clearly implies that I could have 1, 2 or 3 girls. The probability of that is simply 1 – P(BBB) = 7/8. You could calculate that as P(1 girl) + P(2 girls) + P(3 girls) = 1/8 + 3/8 + 3/8 = 7/8. Ordinarily I would say that the probability that I have 3 girls is 1/8. So the relative (aka conditional) probability that I have 3 girls, given that I have at least 1 girl is (1/8)/(7/8) = 1/7

January 14th, 2011 at 12:50 pm

Hi Krazeedude. Google with “ask marilyn monty hall”.

Hi Captainsteggs. I’ve tried the 1,000,000 doors.

According to one survey, when the problem is increased to 7 or more doors, most people still stick to the wrong answer.

Hi DP. My last few posts specifically point out that they are wrong when they say I had a particular girl in mind. I didn’t say “

oneof them is a girl”, I said “ifat least oneof them is a girl”, that’s a very substantial difference.I think that I have only posted one intentionally trick question. Whilst they can be amusing, they’re not really my cup of cake.

January 14th, 2011 at 7:49 pm

Hi Al, just when I thought that there couldn’t possibly be any more to be said!!! I re-read your post42. You start it with: ‘Here is the problem you posed, which I believe I have solved. “I have three children. If at least one of them is a girl, what is the probability that all three of them are girls?”’

- That is the question I posed. The answer to that question it is 1/7.

You end it with: ‘Your question was “ There are three children one of which is a girl. What is the probability that the other two are also girls.” That is the question that I answered, and the answer is ¼.’

- That is not the question that I posed. I concur, the answer to that question is 1/4.

See post 74 for an explanation of why your question is are very different from mine.

I took some care when writing the question. But I had to make a decision when doing it. I could have said “… none of them are boys …”. I decided that was just too much of a giveaway – and wouldn’t have pulled in the suckers. If I’d tried “… the other two children…” that wouldn’t have made sense in conjunction with the previous statements. I might not post trick questions, but I can be mischievious.

The most amazing thing is that “AT LEAST one” is being interpreted as “one”. “at least one” could be written “one, two or three” in this problem.

Al, I don’t know if you give a jot, but I’d be more than happy to see you back on this site. You lost your rag, you said things that you might regret, but who doesn’t? I’ve been pretty bad on this blog too. So can we call it quits and just get back to having some enjoyment around here? I have deleted my harsher comments. I shouldn’t have posted them in the first place. I’m sorry that I did.

January 14th, 2011 at 8:29 pm

I pulled out of this puzzle way back in post 63 as it was giving me a sore head and there were other things I had to do. I now see that this has continued on like a runaway train and we’re now up to 87.

Is it 1 in 7, 1 in 4, or 1 in 3? I still don’t know for sure. I now think of it as a paradox with the answer depending on your interpretation of the wording of the question, in other words a question of semantics.

Chris, I’m glad to see you pull back from some of your earlier comments which I did think were getting a little heated. I do understand your frustration that your perfect logic is sometimes trashed by others with a different mindset. But it’s their arguments that we should attack, not their their mental capabilities.

Your passionate enthusiasm for puzzles of all kinds has made this such a great site that I have enjoyed immensely, and obviously many others have as well. Everyone should feel welcome to join in whether their contributions are serious or frivolous, brilliant or just plain crazy (oops!).

January 14th, 2011 at 9:12 pm

Hi Wiz. Although I think I’ve written some quite good stuff (if I say so myself), you only need to read posts 79 and 84 to see the error of your ways They’re very short and almost sweet compared to my norm.

I only got nasty with Al after his closing remarks in his post 64. He is the pot that called the shiny kettle black.

Hi Al. Although I don’t require it, if you give me permission, I’ll delete the last 3 lines of your post 64. I believe that I have now cleared out most, but not all, of my nasty OTT responses to it.

January 15th, 2011 at 9:53 pm

OK unless anyone else asks a question, I’m going to use every bit of my willpower and move on. I was going to write another complex piece regarding semantics, but I’ll spare you that. Instead, I’m going to (at risk of repeating myself) say that, in my opinion, the “one” in the question is indefinite, it is merely the lower bound of the sequence “one, two or three”. I think that “at least one girl” can only be sensibly interpreted as a slightly twisted way of saying “not all boys”. If I had put it that way, I think that nobody would have fallen into my trap. It also supports my pontifications on the semantics of the question.

I have been through a lot of my posts and have edited them severely in the last 24 hours.

I hope that most of you have enjoyed this blog. I definitely did.

89 posts LOL.

January 16th, 2011 at 1:20 am

come on Chris.i just read “the man in the room” question and their’s (that was on purpose) no way it should have the record at 92 posts!i don’t know what ‘peurile’ means-but i agree.

and i seem to remember the same thing on the old ToM site-a lame question being the most popular.it had well over 100 posts when it should have been more like 3-tops.as i recall it was something like…

“if a chicken weighs 8 lbs and one half of itself,how much does a chicken weigh?”

doesn’t this drive you mad?

January 16th, 2011 at 7:08 am

LOL

~~Peurile~~Puerile [oops, see post 95] means excellent. Not really, it means childish. I like the sound of the word, I think it sounds scornful.I hope they found that the chicken weighs 16 lbs.

January 17th, 2011 at 8:15 am

I thought I would make this at least equal to the “Man in a room” post

January 17th, 2011 at 8:18 am

Or exceed it with post 93 – it’s puerile, from the latin Puer – boy or child – which is sort of odd, considering the title of this question is Girls Girls Girls…

January 17th, 2011 at 11:50 am

Thanks and thanks again, Karl. I hadn’t spotted that.

January 18th, 2011 at 1:26 am

YES!! thank you Karl

you even spelled ‘puerile’ right-something i was trying to poke Chris for (their’s).

and,yes,the 16 lbs. answer was giving early on,and still they wouldn’t quit after 130 some posts!

who turns the computer on for them?

January 18th, 2011 at 5:49 am

LOL. I’d even spell checked “puerile” before posting. Your comment went straight over my head.

I’m aware I sometimes type the wrong one of “your”, “you’re”; “there”, “their” and “they’re”; “to” and “too”. I’ve no real idea how I manage to do that, yet alone how to prevent it from happening. I don’t seem to have the same problem with “it’s” and “its” though.

130 posts!!!! I’m lost for words.

January 18th, 2011 at 3:42 pm

130?? Seriously?? It’s basic equation jiggling!!

x/2+8=x

x/2=x-8

x=2x-16

x+16=2x

16=2x-x

16=x

6 lines by a 17 year old, and people argued to 130 posts. Having said that though, I do enjoy a good discussion :]

January 18th, 2011 at 7:24 pm

1 in 8 or 12.5% chance

January 18th, 2011 at 8:40 pm

Hi Krazeedude. LOL, if this keeps up, this one will get 130 posts too.

But I reckon w/2 + 8 = w => w + 16 = 2w => w = 16

January 18th, 2011 at 8:52 pm

If the probability that he had three girls is 1/7 and he had three girls then the probability of him having three girls is 1.

100

January 19th, 2011 at 2:43 am

For those of us who have the dubious pleasure of english TV and have watched Room 101 with Paul Merton, I think that is where this question should now go. Please, for the love of George Orwell, don’t torture me further with this probability question, because even I understand why it is 1/7.

I just wanted to write post 101.

January 19th, 2011 at 6:57 pm

Well done #101, I am number #102 (If you aren’t familiar with trivial bits of Austin Powers, this is funny. If you are, and it still isn’t, what can I say, it’s 2:00am) I have watched room 101, and this question should definitely join the likes of Justin Bieber. I do want to take this over 130, but is it really worth it, and would it really count? We’ve established the answer is 1/7, so any more (such as this) is surely off topic.

January 19th, 2011 at 7:22 pm

A few extra ones will give a safety margin for the inevitable further posts to “man in a room”.

I don’t care really. I hadn’t expected more that around two dozen responses.

I am number 103. I think the Austin Powers films are very funny too.

January 21st, 2011 at 7:13 am

Why do you consider 7 or 8 instance, while there are only 3? Namely: 1 girsl out of 3, 2 girls out of three and 3 girls out of three. BBG, BGB, GBB are really all the same; BGG, GBG, GGB are all the same, GGG is the third and BBB is out of scope due to the problem’s description. So the answer is clearly 1/3.

One more argument. When you analyse families with 3 children, in which at least 1 girl, you will not consider ordering children like BBG or BGB… you will just use a statistic on number of children and number of girls there. That’s it.

January 21st, 2011 at 7:41 am

Hi Max. What you ask has been done to death on this page. Also see Firls, firls,firls. But in a nutshell BBG, BGB and GBB are not the same thing (whatever that phrase means).

On average, for families with three children 1/8 will have three boys, 3/8 will have 2 boys and 1 girls, 3/8 will have 1 boy and 2 girls and 1/8 will have 3 girls. Your logic would say each of those would be 1/4. Then you’d be saying that the probability of having 3 girls is 1/4 – I bet you already know that it’s got to be 1/2 * 1/2 * 1/2 = 1/8 though.

January 21st, 2011 at 1:29 pm

Even better answer to Max, there are three ways to make GBB (GBB BGB BBG) whereas only one to make GGG, making GBB three times as likely, to 1/3 is unarguable.

January 22nd, 2011 at 1:08 am

These are perfect explanations. Acknowledged!

February 24th, 2011 at 3:46 pm

Once one of the three options is determined as being a definite (positive)then probability only holds for the remaining two which justifies the proponents of 1/3 probability. (1/4 is logically erroneous as has been explained; GB = BG).

February 24th, 2011 at 3:52 pm

Tom. The answer is 1/7 as has been proven using hard logic and a proper understanding of probability. GB does not equal BG. If I’m a boy and I have a sister, then that’s not the same as me being a girl and having a brother (a quick glance between my legs confirms that beyond reasonable doubt).

While you are welcome to post your comments, you could save yourself a lot of embarrassment if you bothered to read what other people have said. You definitely could be wiser (and quieter) after that.

February 24th, 2011 at 4:10 pm

Chris, I agree with you on my need to read what others have posted and I stand corrected in the double trouble problem. However on this I hold ground that GB = BG because it is not a question in reference to probability of sequence but occurrence.

February 24th, 2011 at 4:16 pm

Hi Tom, what do you mean by BG = GB? I have demonstrated, several times in fact, that that statement is practically meaningless.

I believe that you use it to imply that in all the families with 2 children, that on average 1/3 will have 2 boys, 1/3 will have 2 girls and 1/3 will have 1 boy and 1 girl. Well that is simply wrong. 1/4 will have 2 boys, 1/4 will have 2 girls and 1/2 will have 1 boy and 1 girl.

February 24th, 2011 at 4:24 pm

For the family with three children (assuming as is most commonly the case) that there is an eldest a miiddle an a youngest. The we can have (in order eldest, middl, youngest) BBB,BBG,BGB,BGG,GBB,GBG,GGB,GGG. That’s 8 equally likely possibilities. 1 is 3B 0G, 3 is 2B 1G, 3 is 1B 2G and 1 is 0B 3G. So you are 3 times as likely to have a mixed gender children as you are of having single gender family. I think your logic would say that you are equally likely to have any mixture.

In particular, I believe your logic would say that you are equally likely to have 0B3G, 1B2G, 2B1G and 3B0G. i.e. each has a probability of 1/4. That is easily seen to be false. For each child the probability of being a girls is 1/2. So the probability of having 3 girls is 1/2 * 1/2 * 1/2 = 1/8.

Hence your reasoning was, there was at least one girls so 1/3 of having 3 girls.

The correct probabilities are 1/8, 3/8, 3/8, 1/8 so you need to calculate (1/8)/(3/8 + 3/8 + 1/8) = 1/7

February 24th, 2011 at 4:31 pm

Chris, the brilliant mathematical mind, correct my rational mind if it is mistaken. There are three children. Of this three, one is already determined (from the phrase ‘at least one’) as being a girl and therefore (as per my logical mind) the concept of probability (what remains to chance) is on the remaining two to also be girls for all the children to be girls. Ideally the resultant probability would therefore be restricted to 2 boys – BB or 2 girls – GG, or 1 boy one girl – GB or 1 boy one girl – BG. From this (much as you may not agree with the reasoning,) I’m sure you appreciate what I mean when I say BG = GB.

February 24th, 2011 at 4:34 pm

Tom, our posts have crossed. I’ve just demolished your BG = GB claim. Your claim is not based on logic, it is based on not thinking and on a complete lack of understanding of probability.

February 24th, 2011 at 4:39 pm

. My claim from a none thinking mind will settle then that the probability from your argument is 1/4. And I agree that you’ve correctly discounted GB = BG. Now kindly consider that if one entity is predetermined it cannot be used as a factor in calculating probability (chance).

February 24th, 2011 at 4:45 pm

Hi Tom. Which “entity” do you believe is pre-determined?

Now that I’ve eliminated BG = GB, are you about to switch from 1/3 to the 1/4 that you previously dismissed as logically erroneous?

February 24th, 2011 at 4:55 pm

… better still, just read posts 1 and 2. The solution had been fully explained by then.

NB “at least one” does not mean “one” unless there was only one child in the family. It is a deliberately misleading phrase. Altogether, logically, it meant that the three children weren’t all boys. So of the “not all boys”, which is the definite particular one that is is a girl (entity)?

February 24th, 2011 at 5:19 pm

Sorry I was experiencing this error ‘Fatal error: Maximum execution time of 60 seconds exceeded in c:\Inetpub\vhosts\trickofmind.com\httpdocs\wp-includes\wp-db.php on line 411′.

Using your calculation in #112 (2nd paragraph) but the reasoning in #34 that I want to agree with… if at least one (which though not a maximum asserts the minimum) child is a girl then we discount one 1/2 because at least one child is a definite 1/1 girl resulting in 1 * 1/2 * 1/2 = 1/4.

Kindly correct it if the ‘thinking’ is still wrong (I’m a student will appreciate the lesson). I’m logging out to go to bed but I’ll definitely come back to check on your response.

February 24th, 2011 at 5:30 pm

Hi Tom. Read #117. #34 contains no reasoning, just nonsense from someone who hasn’t got a clue.

I’m pleased that you’re keeping your mind open. That earns you a lot of Brownie points. So I’ll overlook your previous sarcastic responses and I’ll stop being so forthright in my responses.

I know this page is very long, but there’s some pretty good stuff on it (if I say so myself) and it’s well worth a read. Although Wiz has given a fallacious argument, it proved quite difficult for me to break it. It is about the only incorrect answer (to almost any problem that’s been posted on this site, ever) that I really liked. It made my brain hurt for quite a long while. Better still, I learnt a lot in consequence.

February 24th, 2011 at 7:21 pm

Hi Tom. Please take this the right way, it is just to illustrate the dangers of not defining things clearly. I accept that you have already withdrawn the remark. In your post 113, you ended it with “I’m sure you appreciate what I mean when I say BG = GB.” The problem is that I don’t know what you mean (I can and have made a guess though). The reason I don’t know what you mean is because I don’t think you do either – that’s why you weren’t able to tell me

exactlywhat you meant, even though I had asked you to. In particular what is the probability of BBG when it is the same thing as BGB and GBB? My point is that woolly gut feelings are not the stuff of logical or rational thinking. You actually do have to use grey matter (pen and paper helps too).Take it as a rule, if you can’t explain something to someone else, it’s very likely to be because you don’t actually understand it yourself. It’s for that reason that I like posting a lot on this site. I find that trying to explain things as clearly as possible (time permitting) to other people is a great way of learning those things and also of realising when you don’t understand something as well as you might otherwise have thought you did.

You may not be sensitive enough to notice yet, but many of the things that Al Gelman posted on this page were so ambiguous that I couldn’t really work out what he was saying. I found it very exasperating – I couldn’t easily point out his errors, because he hadn’t given me any well defined statments to work with. A lot of his stuff was merely insultingly repetitious statments of well known basics and incredibly woolly arguments.

I tried to get him to reveal what probabilities he associated with the BBG when it is the same “thing” (what thing?) as BGB and GBB, but failed to get that information clearly. He obviously didn’t know what he meant by the phrase. As far as I’m concerned, the phrase no meaning. No one who has used it has said what they actually meant by it or examined the consequences of it.

To try and clarify the idea, I proposed (an admitttedly poor notation) e.g. bbg to imply 2 boys and 1 girl and to mean the collection bbg = BBG+BGB+GBB, for which the probability of occurrence is 3/8 (ignoring the “at least one is a girl” bit of the problem). But Al didn’t trouble himself to read that definition and so I got nowhere with him.

I posted another problem “Firls, firls, firls” which I had specifically designed to demonstrate the “BBG is the same as BGB and GBB” is utter nonsense – and is merely the product of woolly thinking.

——

To do probability questions it is crucial to understand what is actually being expected. I think that the only sensible way to do this problem is to realise that if all you knew was that there were three children then (assuming there is an eldest, a middle and a youngest – which is just a convenient way of identifying the children i.e. it hasn’t altered the problem one jot) because each is equally likely to be a boy or a girl, then in order of age, the eight equally likely families are BBB,BBG,BGB,BGG,GBB,GBG,GGB,GGG. So if that’s all you knew, you could calculate things like, the probability of them all being girls is 1/8 and that the probability that there is 1 boy and 2 girls is 3/8 (i.e. BGG,GBG or GGB). You could also calculate the probability that he had at least one girl, and that is 7/8. You could also calculate the probability that he has 3 boys is 1/8.

But you are told that at least one is a girl. So the only combination that he can’t have is BBB. That doesn’t change the fact that the other combinations are all possible and equally likely, it simply eliminates 1 of the 8 possible families. The trick now is, given that there is at least one girl, we need to twiddle all the probabilities so that the probability of him having at least one girl is 1 (rather than 7/8). Obviously we must have (probability 1) that he has 1 of the other 7 equally likely combinations, so each must have a probability of 1/7 – we are now dealing with relative probabilities. We then have, P(0 girls) = 0 (that’s the special case), P(1 girl) = 3/7, P(2 girls) = 3/7 and P(3 girls) = 1/7. Note that 3/7 + 3/7 + 1/7 = 1.

Aside: If the father had said “at least two are girls” then we’d have concluded that the probability that all three are girls is 1/4.

Those who said 1/3, have somehow come to the manifestly incorrect conclusion that you are just as likely to have 0 girls, 1 girl, 2 girls or 3 girls. They have used the “is the same as” to get that. They seem to have correctly realised that the question means we are left with 1, 2 or 3 girls and say only one of those three possibilities is the one sought so 1 in 3. I suspect that I’m crediting them with more understanding than they actually have. I don’t think that any of them are familiar with the true significance of the concept of “equal likelihood”. They certainly haven’t noticed that they are (almost certainly unwittingly) saying that the probability that a family with three children having three girls is 1/4 (but I bet they’d all, correctly, say that it is 1/8 if they were asked for that probability directly). That’s why I was trying to get Al Gelman to state the probability of the bbg family etc.

Those who said 1/4 have completely ignored the “at least” part of the question. Al very clearly demonstrated that at the end of his post 42 where he rewrote the question in order to make his answer right (and mine wrong).

I hope that does it for you. Looking at the whole picture

is the only correct way to analyse these sorts of problems.The mathematicians call it conditional probability, but it is frequently more natural to think of it as relative probability.March 11th, 2011 at 7:23 am

Well, I needed to add this to the posts. A maths teacher I evacuated has used this thread to explain probabilities – mainly from the point of view of how to get it wrong!

So his thanks go to all of you who got it wrong. An invaluable aid to the furtherance of basic maths in the far flung corners of the globe.

March 11th, 2011 at 8:20 am

Hi Karl, you “evacuated” the maths teacher; I didn’t know you gave such “personal services”.

March 16th, 2011 at 3:15 pm

All part of the service! In north Africa at the moment, so not keeping up with ToM at the moment. Work is so inconvenient!

March 16th, 2011 at 5:06 pm

Hi Karl. I went to Libya for about 6 weeks, umpteen years ago. Bloody big beach there.

March 24th, 2011 at 6:30 am

hiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiiii

June 8th, 2011 at 7:52 pm

hmmmm this is an interesting problem.

the question stated that one of them is girl and that’s where it all started.

you have 3 children

1 2 3

B/G B/G B/G

since u know ONE OF THEM IS A GIRL for sure

G B/G B/G

or B/G G B/G

or B/G B/G G

I’m pretty sure everyone agree up to now…

since the question didn’t say anything about which one being the girl. by that, it’s also implying that GBB & BGB & BBG is the same thing. that’s is one of them being a girl.

therefore, it gotta be 1/4

June 8th, 2011 at 8:11 pm

Hi Rin. The question did not say “one of them is a girl”, it said “AT LEAST one of them is a girl”. So all you know for sure is that there aren’t three boys.

If the question had said “one of them is a girl”, then the other two would have to be boys (and the question would have been a grammatical and logical atrocity); then the probability of there being three girls would be 0.

July 10th, 2011 at 2:16 am

Didn’t go through all the posts, only the first 15 or so.. The answer would depend on the question. In this case, “at least one of them is a girl” means you don’t know which one, it could be any of the three. So you have to consider all the 7 possibilities where at least one child is a girl, and so the probability of all 3 being girls is 1/7. So, I go with SP and Chris.

If the question stated that it was known which child was surely a girl, then the answer would have been 1/4.

July 10th, 2011 at 4:47 am

Hi Ankit. That’s right

October 11th, 2011 at 12:55 pm

Here’s a link to a Monte Carlo simulator:

http://trickofmind.com/wp-content/uploads/2011/10/girls-girls-girls.txt

You can load it into Excel. On the main screen press Alt+F11, then in the VB IDE (that should have opened), choose Insert then Module. Paste the code into the new blank module.

October 17th, 2011 at 7:43 pm

Because I’ve made a comment about the interpretation of these “paradoxical” probability problems elsewhere, I’ll repeat the gist of it here. I’m sure that I’ll have more or less said it on this very long page, although perhaps not quite so clearly.

This type of problem only really works because of the vagaries of everyday language. We usually automatically and subconsciously interpret what’s been said to what we think was meant. In turn we often say things that are ambiguous (under inspection) in blissful ignorance that we do so. We are not Vulcans. In the case of the posted problem, the situation described is quite unnatural. Any real person making the statement about their children would almost certainly actually have a particular girl in mind, and not be issuing a strictly logical statement. The strictly logical interpretation is reserved for use on websites like this and for amusement.

If you interpret the question in a way consistent with normal behaviour, it would be reasonable to think that a particular child was being referenced, and so the probability of three girls is the expected (and easily the most uninteresting – and that’s the clue that you should be looking for another interpretation of the question) 1/4. But if you go into puzzle mode (as the problem expects you to do – as that’s the most likely reason that it’s been posted), then you’re expected to use the strict logical interpretation of the problem and deduce (the not quite so boring) 1/7.

That understanding should enable you to resolve many of these problems and to be able to deal with both (or more) possible answers. Of course, another requirement is that you can actually solve such problems, but that’s a very different matter.

October 18th, 2011 at 7:43 pm

Hi Wiz. Are you ready to play?

I just re-read your post 14. You said, “If we did a survey of all three-child families with at least one girl then a quarter of them would have all girls.”

You didn’t stipulate that to be included in the survey, that there had to be a particular child who was the definite girl. If you did, then how do you define the particular child who is the definite girl? I should stop there, but I won’t

I’m sure that you’d agree that if you did a survey of 8 million 3 chiild families that you’d say that about 1 million would have three girls and 1 million would have three boys. And by elimination, 6 million would have a mixture of boys and girls.

If, from the survey results, you were to then select the families that had at least one girl (which is exactly the same as saying those families that aren’t 3 boys), that would leave you with 7 million families. 1 million of those have 3 girls and 6 million have a mixture of boys and girls. So 1 family in 7 has 3 girls, not 1 in 4 as you claimed.

Does that convince you, or do you want to modify your claim in post 14 (and acknowledge defeat (LOL)? You’d be wisest to appeal to post 131.

November 5th, 2011 at 12:42 am

Hi Chris,

Sorry to be so long replying – been distracted by some real world problems lately (fortunately not involving girls!). I can’t really add to what I’ve already said, and I guess time’s up anyway, so I’ll concede.

I still have the nagging feeling that there’s a flaw in your reasoning somewhere, but since I haven’t found it after the best part of a year then that’s probably just wishful thinking on my part.

So chalk up another win for Chris and another loss for Wiz. Till we meet again . . .

November 5th, 2011 at 6:33 am

Hi Wiz. It’s been very quiet here lately. The “flaw” is covered in comment 131. i.e if you really met someone (who wasn’t a Vulcan) who put the statment and asked the question, you’d probably think that they had a particular child in mind. (But there again why would they put the “at least” clause in). Then 1/4 is quite reasonable.

Consider the logically equivalent problem: “If I flip three coins and they don’t all come up heads, what is the probability that they all come up tails?”. The experiment hasn’t even been performed, so I can’t possibly have a particular coin in mind. What would you now say?

February 4th, 2014 at 11:58 pm

I had failed to properly respond to a point that Al Gelman had made. He said that I hadn’t mentioned that the children had different ages. I had responded with something along the lines that of course the children had different ages, and that I was only using age as a convenient way of identifyiing the children.

What I didn’t say was that he presumably thought that assigning ages was somehow making the logic wrong and therefore my reasoning was wrong.

Several points come to mind: how many families have all the children with exactly the same age? You’d be hard pressed to find a family where the children were born within one microsecond of each other. The probability of that is so close to zero, that I can practically guarantee that no such family has ever existed and that to assume that the family in the problem is such a family is unreasonable. (Even if such a family existed, it wouldn’t affect the result anyway – you just need to use another labelling scheme).

I had said that instead of using age I could have used height or weight or any arbitrary labelling scheme that I wanted. Whatever sensible (unbiased) labelling I use, it cannot possibly cause an error in the subsequent reasoning – it simply makes the reasoning straightforward. It is crucial that the labelling is independent of the data i.e. you must not do the labelling using the gender of the children. I am of course assuming that there is no correlation between the labelling scheme used and gender. I would have to be more careful if I really did a survey (in fact much more careful about every aspect of this problem).

I’m also saying that it is fine to introduce an “irrelevant” idea, especially if it is useful, as long as it doesn’t materially affect any conclusion you make (by definition, if the choice affects the result, then it isn’t irrelevant after all). In this case the age is irrelevant, but as age or any other such arbitrary labelling produces the same (well known results) e.g. that the probability of having three girls = probability of having three boys = 1/8, probability of having two girls one boy = probability of having one girl two boys = 3/8. If a labelling scheme does not produce those results, then it is biased. The bias is manifest in Wiz’s and Al’s logic as they keep on making one girl definite. If there are two or three girls, how are they choosing the one who is the definite girl? Once you have the idea of the definite girl, then you have also defined the “other” children. There is no definite girl, and so there cannot be (definite) other children either.

February 5th, 2014 at 8:05 am

To be continued?

February 6th, 2014 at 4:45 pm

OK, I’ll continue (LOL).

Wiz and Al have actually said (or implied) that there is a definite girl, I’ll emphasise with her with g, so the possible families are gBB, gBG, gGB, gGG and so 1/4 for all girls (given that there is at least one girl). In fact they should only say there’s definitely a girl and get GBB,BGB,BBG,GGB,GBG,BGG,GGG. By definition, “a definite girl” requires that a particular girl is identified e.g. the eldest child is a girl.

What would they say if I simply asked what is the probability that if I have three children, that they are all girls. Perhaps they’d say, well there could be three boys: BBB, if not, then they might (incorrectly) say there’s a definite girl and that gives us four more possibilities: gBB, gBG, gGB and gGG. So that’s five possible families, and the probability of having three girls is then 1/5 as is the probability of having three boys – no problem there (ignoring that we know it’s 1/8 not 1/5). Further, the probability of one boy and two girls is 2/5 and the probability of two girls and one boy is 1/5. So a family is twice as likely to have two girls one boy as they are of having one boy and two girls.

Of course, we could repeat the analysis, but swapping the roles of boys and girls, and conclude that you are twice as likely to have two boys and one girl as having two girls and one boy.

OTOH. If Wiz and Al don’t mind that the definite girl could be the oldest middle or youngest (or whatever), then emphasising the definite girl with g, we have gGG, GgG, GGg, gGB, GgB, GBg, gBB, BgB, BBg, (but only BBB) and perhaps gBG, BgG, BgG so perhaps it’s 3/9 = 1/3 or 3/12 = 1/4 for three girls. Whichever, you are three times as likely to have three girls as three boys.

How about the following, “I have three children, at least two are girls, what is the probability that I have three girls?” I’m sure that Wiz and Al will have said, there are two definite girls, so the other child could be a boy or a girl, so the probability of three girls is 1/2. On the other hand, if they’d started with their previous conclusion, they could simply remove GBB from the list GBB,GGB,GBG,GGG and said 1/3 for three girls.

Neither 1/2 nor 1/3 is correct though. The correct answer is 1/4. The initial list is really: BBB,BBG,BGB,GBB,BGG,GBG,GGB,GGG. Knocking of the cases with less than two girls, we’re left with BGG,GBG,GGB,GGG and so 1/4 for 3 girls.

February 7th, 2014 at 7:22 pm

Change of direction. I found the following http://mathforum.org/library/drmath/view/69420.html

Amusingly, it seems that for twins, approximately we have that the probability of 2 girls = probability of 2 boys = probability of 1 boy and 1 girl. This deviation from the norm is due to identical twins (nearly) always having the same gender. However the author has assumed that fraternal twins obey the usual rules of independence (which I find mildly surprising).

August 25th, 2014 at 10:52 pm

This is so simple. I haven’t read all the posts, so I hope it hasn’t already been said.

If you looked at 8000 families with three children, it’s obvious that 1000 will have three boys and 1000 will have three girls. That leaves 3000 with two boys and one girl and 3000 with two girls and one boy.

The question says “at least one is a girl”, which is the same as “they aren’t all boys”, so that leaves 7000 families that match. 1000 of them have three girls, so the probability is 1000/7000 = 1/7.