## Girls only, part 2

Posted by Chris on October 11, 2011 – 6:20 am

I have two children, at least one of them is a girl who was born on a Friday. What is the probability that they are both girls?

Assume that each child is equally likely and independently to be a boy or girl.

Now let the battle commence.

October 11th, 2011 at 7:08 am

13 out of 28 [later: DP meant 13 out of 27].

October 11th, 2011 at 7:22 am

I repeat my argument from part 1, the known child can have any qualities you like, as long as the second child has a 1/2 chance to be girl, the permutations show a 50/50 that the other has the same gender.

October 11th, 2011 at 8:20 am

if you make a desicion tree, you can easyly see, that the total chance of having at least a girl being born on friday is 27/196 and the chances of having two girls with at least one being born on friday is 13/196. so the total chance of having two girls with knowing one being born on a friday(or any other specific day) is 13/27 (ca. 48.1%)

October 11th, 2011 at 8:30 am

Interesting that you fail to mention whether or not they were born on the same Friday. If they were, are they fraternal or identical twins? Obviously, if they are identical twins they are both girls. If they are fraternal, or were born on seperate Fridays the odss are 1 in 3. The latter, of course, does not take into account the genetic tendency of the mother to bear one gender or the other, or the time in the menstrua cycle that the mother became pregnant.

October 11th, 2011 at 9:35 am

OMG, Born on Friday, Blonde Hair, Green eyes or with two Hang Nails and in Podunk, Iowa!!!!! Day of Birth and Gender of Sibling are completely independent variables. One has ABSOLUTELY no effect on the other. And for the record, “genetic tendency of the mother to bear one gender …. “, My biology course suggested that genetic decision was “made” by the father!

October 11th, 2011 at 10:42 am

Whoops…I meant to also say 13/27, not 28. I agree with Carc….for now.

October 11th, 2011 at 10:43 am

OMG, Carc and DP (whose correction I’ve seen) got it. I’m gutted, I was hoping to melt a few brains – perhaps that will still happen

Hi BearSprite. If that was the answer, I wouldn’t have posted the problem. Of course I’d have been gutted if nobody said 1/2 or 1/3 and tried to prove it.

Hi JB. I tried to cover all such possibilities in the problem definition. I appreciate that you are being informative though.

Hi DanC. You’re right, day of birth and gender are independent. It’s because of that, that the answer to the problem is 13/27.

October 11th, 2011 at 10:57 am

Well, it is very close to 50%. To make it easy, let’s go ahead and throw out any case when there is not a girl born on Friday. I hope I can explain this correctly.

Because we are selecting a specific person (the one born on Friday), we can choose who that one is for each permutation. B & G will represent boy and girl in order of birth. m, t, w, th, f, sa, & su will represent the day they were born.

For the first being a girl born on Friday, we have:

GfBm, GfBt, GfBw, GfBth, GfBf, GfBsa, GfBsu &

GfGm, GfGt, GfGw, GfGth, GfGf, GfGsa, GfGsu

(14 cases in total, 7 with 2 girls)

And for the second being born a girl on Friday, we have:

BmGf, BtGf, BwGf, BthGf, BfGf, BsaGf, BsuGf &

GmGf, GtGf, GwGf, GthGf, GfGf, GsaGf, GsuGf

(14 more cases total, 7 with 2 girls)

But we can see that there are two cases of GfGf, where both are girls born on Friday.

So them we have [(7+7)-1]/[(14+14)-1] = 13/27

October 11th, 2011 at 11:06 am

Hi DP. I see your confidence has returned. Your analysis looks good to me. Thanks. Of course the “two” cases of GfGf are the same case, that’s why you only count it once altogether.

The paradox is that, in light of part 1, is that we may have expected an answer nearer to 1/3 than 1/2.

October 11th, 2011 at 11:48 am

Here’s a link to a Monte Carlo simulator:

http://trickofmind.com/wp-content/uploads/2011/10/girls-only-part-2.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 11th, 2011 at 1:47 pm

Hi again BearSprite. As the probability of each child being a boy or girl is 1/2, then there’s a 1/4 probability that a family with two children could have two boys. But the question states that there isn’t two boys. So we are left with examining 3/4 of the possible (in the wild) families. As BB,BG,GB,GG are equally likely (in the wild), then BG,GB,GG are equally likely in families which don’t have two boys. In such families, the probability is 1 that they have BG,GB or GG, and so each must have a probability of 1/3.

There is a little bleed over from the part 1. Never mind.

October 11th, 2011 at 4:04 pm

What would one being born or Friday have to do with this?

Let’s assume the sex was set about 9 months prior to birth. Then nine months later, give or take a week, the baby is born.

Do you suppose the day of the week has anything to do with the sex?

Oh, wait, I forgot, only females are born on Fridays… I should have remembered that.

..

Just kidding, I think more boys are born on Fridays as I remember my kids.

Hey, I remember twins that were born a year apart… one before midnight on Dec 31, and one after midnight on Jan 1.

How could they be the same sex if they were born on different days and years?

October 11th, 2011 at 4:28 pm

Hi Ragknot. Apparently hospital staff like to induce births on a Friday, so that they can have a clear weekend.

I chose Friday because of “Girl Friday”. The original problem actually referred to a boy that was born on a Tuesday, but I didn’t want to make it too easy to Google for the answer. I chose Friday because “Girl Friday” would be likely to get millions of irrelevant hits.

October 11th, 2011 at 10:48 pm

I see what you were asking now, For some reason, I thought it meant a girl was MORE likely to be born on Friday.

October 12th, 2011 at 9:23 am

Must…..not…..answer….this….question…… it’s one in….t…w….

Aw heck, its 13/27….

October 12th, 2011 at 10:02 am

Hi Karl LOL. Was that a Kryten or a Homer impersonation?

October 12th, 2011 at 1:10 pm

possible cases are:

BB <- ruled out, at least one is a girl

BG <- 1/3

GB <- 1/3

GG <- 1/3

the GG-case is 1/3. how did you get to 13/27?

i don't see why the day of birth should matter at all.

October 12th, 2011 at 1:47 pm

@hamsterofdeath:

I have one child. What is the probability of it being a girl?

I have one child. What is the probability of it being a girl born on a Friday?

I have two children. What is the probability that they are both girls?

I have two children. What is the probability that one is a girl born on a Friday, and the other is a girl?

I have two children. What is the probability that one is a girl born on a Friday, and the other is a girl also born on a Friday?

October 12th, 2011 at 1:51 pm

Expansion of post #18 -

@hamsterofdeath:

I have two children. Neither is a boy. What is the probability that at least one of them is born on a Friday?

I have two children. Neither is a boy. What is the probability that at least one of them is a girl born on a Friday?

I have two children. Neither is a boy. What is the probability that one of them is a girl born on a Friday, and the other is a girl?

October 12th, 2011 at 2:09 pm

Also see DP’s explanation in post 8. Carc has given a small clue too.

Draw a table. Head the columns and rows with e.g. B-Mon B-Tue etc. That’s 14 columns and 14 rows. The intersections represent the 14*14 = 196 equally likely possible combinations of two children on the basis that either could be a boy or girl and that either could be born on any day of the week. We then use the given fact that at least one is a Friday girl. That means that the only possibilities left correspond to the column or row (or both) which is headed G-Fri. Each column/row is 14 units long and each share one cell (i.e. the G-Fri,G-Fri cell). That means that only 2*14-1 = 27 of the original possible combinations are admissible. Each column or row has 7 girls, and the shared intersection is G-Fri,G-Fri, so the 27 cells have 2*7-1 = 13 girls. So the probability of two girls given that at least one is a girl born on a Friday is 13/27. Easy peasy LOL.

I was quite amazed when I found the problem and saw the answer. You can be sure that I just had to know how that worked.

October 13th, 2011 at 2:30 am

Chris – it was more Homer than Kryten. I actually referred back to Girls Girls Girls to check my maths and reasoning!

Retirement is busy and taxing – hopefully back in full swing before too long!

October 14th, 2011 at 9:28 am

JB said something aboout a tentacle growing out of the girl’s head. By using the same sort of argument that DP used, and by letting the probability of a girl having a tentacle growing out of her head = p, then the probability of having two girls is(1/2) (1-p/2)/(1-p/8). So if p is very small, that ≈ 1/2. That isn’t surprising, if p is small, it is very unusual to have a daughter with a tentacle, and so it is likely that she is the only such daughter, and so she is quite well defined. If 50% of girls had tentacles growing out of their head, we’d get (1/2)(1-1/4)/(1-1/16) = (1/2)(6/7) = 3/7 = 0.42857…

As p varies from 0 to 1, the probabilty of you having two girls given that you have two children and at at least one girl with a tentacle growing out of her head, varies from 1/2 to 1/3.

October 14th, 2011 at 9:39 am

Karl, I shoulda removed tha Kryten reference once I’d recognised Homer. But, that you agreed with 13/27 is excellent. Don’t take the following Mark Twain quote personally. I’m only putting it up as I came across it a few days ago: “When I was a boy of 14, my father was so ignorant I could hardly stand to have the old man around. But when I got to be 21, I was astonished at how much the old man had learned in seven years.”

“That’s you that is” – The Mary Whitehouse Experience. http://www.youtube.com/watch?v=5nCKYEM8qRc