## Girls, girls

Posted by Chris on February 8, 2014 – 6:46 pm

(Sorry, I couldn’t resist)

I was walking through the park when I bumped into an old acquaintance.

He told me that he had two children. What is the probability that he has two girls?

A girl joined us. He said it was one of his children. What is the probability that he has two girls?

He said she was his eldest child. What is the probability that he has two girls?

February 10th, 2014 at 6:59 am

probability that both are girls = 1/2 & 1/2 = 1/4

given that one is a girl it does not matter whether elder or younger probability that both are girls = probability that 2nd one girl = 1/2

February 10th, 2014 at 8:01 am

Oh no! You fell for it Kali… That was Chris’ trap.

What you should have done is start with the possible families. Let’s sort them in age order (oldest first) just for fun.

You can have: BB, BG, GB, or GG.

So answer #1 is 1 out of 4, like you said.

For the second part, he only says one of the children is a girl (though we don’t yet know if she is older or younger).

So answer #2 cannot include ‘BB’, but can still be BG, GB, or GG. So the answer is 1 out of the remaining 3.

For the final question, we now know that the girl we met is oldest, so we are left with two options GB and GG.

So Answer #3 is 1/2 as expected (the same probability of child #2 being male or female).

February 10th, 2014 at 9:25 am

Good man DP. 1/4, 1/3, 1/2 is what I think too. In fact I goofed this when I was commenting on the Girls, girls, girls problem in post 74.

February 10th, 2014 at 3:10 pm

I’ll admit to having had initial doubts about the 1/4, 1/3, 1/2 results. This was due to me thinking about the probability of the events described actually happening. That was silly of me, the problem states that they definitely happpened. Among other irrelevancies I thought that the fact that I met a girl increased the probability of there being two girls. That’s a back to front way of thinking (on this occassion at least). The correct way of viewing it is that it is more likely that I’d meet a girl first if there was more than one girl. With only two children, if there is only one girl, then the probability of me meeting a girl is 1/2. If there were two girls, then I definitely meet a girl. In either case, the child I met would have a probability of 1/2 of being the oldest child. But as I say, that’s all irrelevant.

February 11th, 2014 at 11:11 pm

I think that there is some flaw with BG,GB,and GG combination

The reason is that out of BG combination one girl can be chosen in 1 way , from GB one girl can be chosen in one way but from from GG in 2 ways so to give BG, GG, GB same weightage is wrong. GG shall be given weightage 2 so perobability is from BG,GB, GG,GG ( GG is twice as weightage 2 ) so both girls probability is ½.

Looking at another way

The probability that the elder child is selected is ½

Probability that the yonger child is girl = ½

So elder child is selected (and is girl given) and both are girls = ¼

Simlilarly yonger child is elected (and is girl given) and both are girls = ¼

So one child selected to be girl and both are girls = ½

I have given 2 additional methods

If any flaw is there in my method it may be pointed out to me

February 12th, 2014 at 3:44 am

The flaw is that you are considering the probability of selection. For example, if the friend had two boys, then the described events could not have happened.

The information given can only be used to reduce the sample space.

I will confess that my confidence isn’t high at the moment. I’d be more confident if the friend had simply said that he didn’t have two boys, rather than having actually met a girl.

I’ve only just woken up and so am not thinking clearly. I’ll re-read your post and comment further later on. I’m not going to insist that I’m right yet.

February 12th, 2014 at 8:10 am

To kali’s post:

With that logic, you are saying that if the oldest child is born a girl, then you have 3 possible families (GG, GG, GB). So what you are stating is that regardless of the first child, the second child has a 2/3 chance of being a girl. But we know (assume) that each child has a 1/2 chance of being either gender.

Your first answer is correct (1/4), so you must not have said that the possible families are BB, BB, BG, GB, GG, GG. Otherwise you would have said there is a 1/3 chance of having 2 girls.

For the second question, age is irrelevant. That is just how I chose to sort them in my answer since it would help for the third part. You could have sorted by height. We may have met the tallest or shortest child, who happens to be a girl.

Sorted by height, possible families are SB,TB; SB,TG; SG,TB; SG,TG (S=short, T=tall, B=boy, G=girl).

Again, we don’t know if we met the shortest or tallest child, but we know she is a girl, so it cannot be SB,TB. And so 1 of the remaining 3 families (SG,TG) has two girls.

February 12th, 2014 at 9:27 am

I think the discussion is due to some ambiguity in the statement of the problem. Kali interprets the problem as: “one of the children joined us. It was a girl. What is the probability the other one to be also a girl?” The answer to this question is indeed 1/2. DP and Chris interpret it as: “the guy has at least one daughter. What is the probability the other kid to be a girl, as well?” The answer here is 1/3.

February 12th, 2014 at 12:36 pm

I have found this one to be very difficult to interpret correctly.

The easiest way for me to argue it is to start with it being equally likely that the friend has BB,BG,GB or GG. Fortunately, no one disagrees with that, so I’ll assume that it’s a safe starting point.

If I were to have many friends who did indeed have that distribution of kids, then 1/2 of visits to BG or GB friends would result with me meeting a girl, and all visits to a GG friend would result in me meeting a girl. So that indicates 1/2 i.e. 1/(1/2 + 1/2 + 1), probability for GG (given that I saw a girl), where I’m assuming that it is equally likely that I visit a BG, GB or GG friend.

Similarly if the girl was the oldest child, that would happen on 1/2 the visits to the GB friend, and 1/2 the visits to the GG friend, so the GG probability is (1/2)/(1/2 + 1/2) = 1/2, again. So the answers to the original problem are 1/4, 1/2, 1/2 in agreement with Kali’s answer.

I’m still not quite fully 100% confident (but nearly am), hence my slightly woolly statements.

I’ll now try to understand where the flaw in DP’s (and my previous) belief is. Don’t hold your breath waiting.

February 12th, 2014 at 12:57 pm

Hi DP. I saw our flaw almost immediately after my last post. The sample space isn’t BB,BG,GB,GG. If I change notation, and use lower case to represent the child I met, given merely that I met a child, then we have the sample space bB, Bb, bG, Bg, gB, Gb, gG, Gg where each case is equally likely.

4 of those is where I met a girl, and of those 2 is where there were two girls => 2/4 = 1/2.

2 of those is where I met a girl who is the oldest child, and only 1 of those is 2 girls => 1/2 again.

February 12th, 2014 at 1:07 pm

It looks like I fell into my own trap on this one. Thanks for pushing it Kali – despite my previous posts my confidence wasn’t high. Thanks for your comments too Slavy, they confirmed that my change of perspective was correct.

Will DP or I learn anything from this?

Almost forgot. This doesn’t affect the result of the Girls, girls, girls problem. In that problem there is no definite girl, here there (definitely) is.

February 12th, 2014 at 1:28 pm

Agreed. Another way to point out our flaw is that we assumed that we knew no other comparable defining characteristic about the girl you had met. The other defining characteristic is that there IS a specific girl. It’s THE girl you met.

When trying to write up a defense to that point, I too quickly saw the flaw. I was going to try and say that slavy’s statement was false where he said “What is the probability the OTHER ONE to be also a girl?” My logic was skewed since in the previous similar(ish) problem there was no identified girl so there was no such thing as the “other” girl.

Just the thought of disagreeing with slavy should have been enough to scare me into changing my entire belief system =)

February 12th, 2014 at 1:34 pm

Excellent, DP. I was just about to say that in this case, it is quite sensible to talk about the other child(ren). I’m now going to return my post 74 statement on the Girls, girls, girls problem back to where it was.

Fortunately for me, I had already seen that the 1/4, 1/2, 1/2 was right before seeing Slavy’s post, so I was in no danger of having to contradict him.

February 12th, 2014 at 11:52 pm

To DP’s post number 7

I did not say that if the oldest girl is G then combination is GG,GG,GB. Nor my statement implies it.

From one of the lines below

Sorted by height, possible families are SB,TB; SB,TG; SG,TB; SG,TG (S=short, T=tall, B=boy, G=girl).

Out of which the combination of at least one girl is TGSB, SGTB,SGTG

Out of which there are 4 posibliities of a girl ( one from TGSB, one for TBSG, and 2 for SGTG)

There are 2 combinations TGSB or TBSG that is one boy and one girl.hence probability = ½

March 26th, 2014 at 4:22 pm

I say 50/50.