## Morris: How Many Boys? On a Tuesday? Mrs Smith has two children.  The eldest is a boy.  What is the chance that both are boys?

Mrs Jones has two children.  One is a boy.  What is the chance that both are boys?

Mrs Brown has two children.  One is a boy born on a Tuesday.  What is the chance that both are boys?

From Gary Foshee published on Ed Pegg’s http://www.mathpuzzle.com/

1. ### kitty said,

December 1, 2009 at 4:05 pm

50%, 50%, 25%

2. ### Bob said,

December 2, 2009 at 11:51 pm

Given that Mrs. Brown has two children, the probability of two boys, if you know the eldest is a boy, is 1/2 , and the probability of two boys, if you know one of them is a boy, is 1/3.
I was surprised how hard it was to think about the final problem and went from thinking that Tuesday was a red herring to finally believing that it does, in fact, make a difference.  The probability of a woman having two boys, given that she has two children, one of whom is a boy born on Tuesday is 13/27.
I’d be interested in hearing if anyone thought about it differently.

3. ### Jonathan Lundell said,

December 3, 2009 at 6:34 pm

1/2 and 1/3, sure. But 13/27? I conjecture that this is (14-1)/(28-1) (2 genders * 2 children * 7 days = 28?), and wonder whether you (Bob) aren’t reading an extra condition in the the problem, namely that only one boy is born on a Tuesday—which, if it were the case, would need to be more clearly stated.

4. ### Bob said,

December 3, 2009 at 7:33 pm

Thanks for the comment, JIonathan.  I cerrtainly tried not to fall into that pit.
I assumed that the sample space was the set of all combinations which  involved two children, one of whom was a boy born on Tuesday.  I think there are 5 of these, (BT,BT), (BT, BN), (BT, G), (BN, BT), and (G, BT).  The first, second and fourth of these are the ones where there are two boys. As I see them, the probabilities of the independent events are P(BT)=1/14,  P(BN)= 6/14, P(G)=1/2. The probabilities of the five compound events above can then be computed.  I got P(BT, BT)=1/196, P(BT, B)=6/196, P(BT,G)= 7/196, P(B,BT)=6/196, and P(G,BT)=7/196.  I summed the probabilities of the three combinations that have two boys, and divided by the sum of all five of the probabillities.  Does that help, or have I missed something?

5. ### Jonathan Lundell said,

December 3, 2009 at 9:43 pm

Yes, that sounds right. Weird, isn’t it?

6. ### Bob said,

December 4, 2009 at 9:52 pm

I certainly agree with you, there!  I just had a colleague, Stuart,  who sent me the following solution.
If BT is the firstborn, there are 14 possible second-born, boys born on the seven days of the week, and girls born on the seven days of the week.
If BT  is the second-born, any of the girls mon-thru-sun could have been firstborn.  Considering the boys, there are only six possible firstborns, since we have already included (BT,BT).
That makes a total of 27 possibilities, with 13 of them including two boys.  He felt that his analysis made it easier to see why the result cam out the way it diid.
His comment was , ‘I now see the connection between all three problems, so I now think I better understand why #1 and #3 have very similar answers (50% and 48%).

7. ### Mike Jarvis said,

December 9, 2009 at 6:22 pm

I always find these problems a bit irksome, because it depends on how the woman was selected.
1: Ask random women one at a time if they have two children, and then if so, do they have a son born on Tuesday. Select the first woman who meets the criteria. In this case, Bob’s analysis is correct, and the probability is 13/27.
2: Ask random boys one at a time if they have exactly one sibling, and then if so, were they born on a Tuesday.  Select the mother of the first boy who meets the criteria.  In this case, the probability is 1/2.
3: Ask random women if they have two children with at least one of them being a boy.  Ask for the day the boy (or either boy if 2) was born, and write the question using that day.  In this case, the probability is 1/3.
4: Ask random women if they have two children.  Then ask them to select a child at random and to tell you that child’s gender and birthday, and write the question using that gender and day.  In this case, the probability is 1/2.
5: Ask random boys if they have exactly one sibling.  Then ask him what day he was born on, and write the question using that day.  In this case, the probability is 1/2.
The problem as stated sort of implies that the first case is what is desired, but only because it is the most interesting mathematically.  It’s really the most contrived selection process of the five.  Which is why this kind of question irks me.  :)

8. ### Stephen Morris said,

December 9, 2009 at 6:40 pm

Mike,

I kind-of agree with you.  It is hard to imagine how you could come up with this information without the method biasing the results in some way.  I struggled with this a lot.

I think the only fair way is:

Take all women within a large area at some point in time.  Count how many have two children, one of whom is a boy born on a Tuesday.  Then count how many of these have two boys.

There are some logistical, moral and legal obstacles to this, but I guess that’s why we invented mind experiments.

But essentially I agree, you need to know how the information was come by.

9. ### Bob said,

December 9, 2009 at 9:56 pm

I’m with  Johnathan, when he says, “weird.”  I love it when I get a result that to my eye is mathematically sound but seems as terribly counter intuitive as 13/27 did.  When I get a result like that I know I am going to learn something.
I thought it was not the porpose of this problem to be either useful or practical to carry out, butwas rather to make us puzzle over the situation and perhaps, by that effort, come to understand something about counting.  I learned quite a bit.  My thanks to the author.

10. ### ron osmond said,

June 8, 2010 at 10:48 am

The answer is 1/3, not 13/27. In the Boy Boy combinations there are 12 instances where the probability is shared by two days of the week, and one instance where there are two Tuesdays. So the numerator is 1 + 12/2 = 7 (not 13) and the denominator is 14 (being the number of Boy/Girl combinations) + 6 + 1 = 21.
So 7/21 = 1/3.

11. ### Ed Jones said,

June 10, 2010 at 7:07 am

This is nonsense. If instead you were told that the boy was born in January, or born on the 12th of March, you would come up with completely different probabilites. Conclusion: the timing of the birth of one child has no effect on the probability of the other being a boy. That’s astrology not science.

12. ### Jonathan Lundell said,

June 14, 2010 at 8:30 pm

Ed is right that the answer varies depending on the number of buckets, but of course astrology doesn’t enter into it. Try it this way, to cut down on the number of combinations:
Mrs Brown has two children.  One is a boy born in an odd year.  What is the chance that both are boys?
There are 16 combinations of odd/even boy/girl. 7 of them have an odd boy. Of those 7, three are BB and four are BG or GB, so the answer is 3/7.
Try writing down all the combinations and counting them.
There’s something about the nature of a proof in all this (Monty Hall comes to mind, of course): a proof does not compel its own acceptance. Or something like that….

13. ### strauss said,

June 15, 2010 at 7:35 am

One really amusing variation is that the more specific the information about one (or the other) child is,  the closer the probability that the remaining child is a boy is to 1/2.  For example, if you know that a family has a 7 foot tall boy named Xanadu [presuming one or more such people actually exist], the probability that the other child is a boy is almost exactly 1/2.

To see why this is, analyze the above knowing that one child is a boy born at 12:01:00 pm on a Tuesday.

14. ### Jonathan Lundell said,

June 15, 2010 at 9:57 am

The same observation goes in the opposite direction. My example of a binary odd/even choice isn’t the limit. Consider:

Mrs Brown has two children.  One is a boy born in a non-leapyear.  What is the chance that both are boys?

15/39, if I have it right, or ~.38, compared to 3/7 (~.43) or 13/27 (~.48), or Chaim’s example approaching .50. The lower limit is 1/3, yes?

The key is the number of combinations in which both siblings meet the stated criterion.

15. ### Stephen Morris said,

June 16, 2010 at 2:52 pm

I’m still trying to get an intuitive understanding of this.  Is it reasonable to expect that there is one?  discuss…

We can generalise to:

‘Mrs X has two children.  One is a boy who satisfies condition A.  What is the chance that they are both boys?’

where condition A has a probability of p for each child.

This varies from 1/2 to 1/3 as p varies from 0 to 1.

We would like to understand all three cases using this formula.   Cases 2 and 3 are straightforward, we have p=1 and p=1/7 respectively.

For case 1, Mrs Smith, we need a condition with p=0.  At this point it breaks down.

I think we need to introduce a new concept which is related to ‘dependency’.  Specifically put q = probability that Condition A is true for one child given it is true for the other child.

The answer is (2-q)/(4-q).  Note that p does not appear.

Now we can solve case 1 for Mrs Smith.  Condition A is that the child is the eldest.  As only one child can be the eldest we have q=0 and so the answer is 1/2.  In cases 2 and 3 we have q=p and so we get the previous results.

So the answer does not depend on the actual probability of the condition being true, it only depends on the probability of it being true for one child given it is true for the other.

16. ### Jonathan Lundell said,

June 16, 2010 at 4:50 pm

You need to restrict Condition A somewhat (independence of gender?) to get to the 1/3..1/2 range.

‘Mrs X has two children.  One is a boy who is an identical twin.  What is the chance that they are both boys?’

17. ### Dominic Tunks said,

June 24, 2010 at 7:36 am

I initially made a similar mistake to you Ron (ie forgetting that BN,BT and BT,BN are distinct). I therefore also had a denominator of 21 – although for some reason that I cannot now reconstruct I made it 10/21. But of course you shouldn’t divide by 2, and, Ron, your approach then gives the same answer as everyone else: 1 + 12 (not 12/2) = 13 over 14 + 12 (not 12/2) + 1 = 27.

Going back to Mike’s ‘irksome’ comment, isn’t this the precise point of the problem? The solution derives from the information you have. ‘I have a sibling’ is different information from ‘I have a sibling and my mother has a son’, etc. ‘There are two people in a room, at least one of them is a male’ is different from ‘There are two people in a room, one of them is the Pope’.

18. ### Bill Steigerwald said,

June 29, 2010 at 8:12 am

So… did anyone consider there are 1.059 boys born for every 1.000 girls. Just throwing in another variable to make this a little more fun.

19. ### Dan Asimov said,

July 2, 2010 at 1:27 am

This is the best discussion I’ve found so far on the subject.  Solving the Tuesday problem is easy for anyone with a background in probability, but it’s much trickier to get the feeling that the answer makes sense.  Especially when every child is born on some day of the week, and so what difference should specifying the day make?

But I find it unfortunate that some visible mathematicians have publicly stated that the problem isn’t well-defined, since how the information was obtained hasn’t been stated.

What an utter red herring!  The problem is interesting and tricky enough without muddying the waters with this kind of nonsense.  Of course, any statement in English is bound to be ambiguous in some way if you look hard enough.  But there is a long tradition of interpreting probability problems of the form “Given a random member M of a population P satisfying conditions C_1,…,C_n, what is the probability that M also satisfies additional condition C ?” in a completely standard way: What fraction of the members of P that satisfy C_1,…,C_n also satisfy C ?”  Or in other words, exactly what Stephen Morris wrote above.

I find it irksome that some people find it necessary to display their cleverness by pointing out an unintended interpretation of this problem.

20. ### Gary said,

April 26, 2011 at 5:29 pm

This is more a biological question than a math question. There are many biological variables that determine the sex of offspring like the age of each parent, the health of each parent, the ability of the parents to reproduce, just to name a few. It really is not just a matter of mathamatical probability.

21. ### JeffJo said,

October 10, 2011 at 3:55 pm

On one hand, the question is about math, not biology. Since the biological values are not constant, the question can’t be answered unless you assume a 50/50 boy/girl ratio, and independence.

On the other hand, once you make that assumption, the proportion of two-boy families, among families with at least one, is 1/3. The proportion of two-boy families, among families with at least one boy born on a Tuesday, is 13/27. It goes up because it is almost twice as likely that you will find a Tuesday boy in a two-boy family, than in a one-boy family.

On the gripping hand, probability is not the same thing as proportion. In the equally paradoxical game show problem, the proportion of games where (1) you choose Door X, (2) there is a goat behind Door Y, and (3) switching will win, is 1/2. But the probability you will win by switching after the host opens Door Y to show a goat is 2/3. The difference is because, if the prize is behind Door X, the host will not always open Door Y. Sometimes, he will open Door Z. The actual answer is 1/(1+Q), where Q is the probability he will open Door Y in that case. It varies from 1/2 when Q=1, to 1 when Q=0. The accepted answer is 2/3, when Q=1/2, because you can only assume the host is unbiased.

Similarly, the answer to the Jones/Brown questions depend on how you learned the facts. They are 1/(1+2Q) and (1+12Q)/(1+26Q), respectively, where Q is the probability you will learn the stated fact when it applies to only one child. Both are 1/2 is Q=1/2. And the reason the 1/3 and 13/27 answers seem unintuitive, is because Q=1/2 is the better assumption if you aren’t told how to determine it. Just like in the Game Show Problem: you can only assume the method is unbiased. And frequently it is not even an assumption, based on wordings like “If I tell you that…”