A family has two children. Given that at least one of the children is a boy who was born on a Tuesday, what is the probability that both children are boys?

Assume that the probability of a child being born on a particular day of the week is 1/7.

I'm wondering if there's a better way to calculate P(At least 1 boy born on a Tuesday) than the explanation. [In the solution, this is P(B) in Bayes' Theorem.

Here is how they calculate:

To calculate we note that there are 14^2 = 196 possible ways to select the gender and the day of the week the child was born on. Of these, there are 13^2 = 169 ways which do not have a boy born on Tuesday, and 196 - 169 = 27 which do, so P(B) = 27/196.

I understand this intuitively, but my statistics classes shy away from this "naive" definition of probability (although we do assume equally likely boy vs girl and day of the week for this problem). So is there a way I can calculate this P(B) in a more stepwise fashion (for example: {1/7 chance of born on Tuesday • 1/2 chance boy} + {1/2 chance boy • 1/2 chance other was not a boy • 1/7 chance of Tuesday • 6/7 chance the other was not}. I know the example I just wrote isn't correct but can someone intuit for me a way I could go about getting P(B) in this problem in a similar method? Perhaps I just need it illustrated.

Note: I found another thread with this question but wasn't sure I understood the explanation. Perhaps someone has an enlightening comment to help.


I don't know why you'd shy away from counting... It's often the best method.

But, yes, we can do it in a different way. If we have a general Bayes setup $$ P(A|B) = \frac{P(B|A)P(A)}{P(B)}$$ the denominator can be split up via the law of total probability into $$ P(B) = \sum_i P(B|C_i)P(C_i)$$ where the $C_i$ partition the sample space. In your case you have $A = $ "Both children are boys" which you can take as $C_1$ and then perhaps $C_2 = $ "Both are girls", and $C_3 = $"exactly one child is a boy". Those are three disjoint events that span all possibilities.

As I'm sure you have already computed, we have $P(B|A) = 1-(6/7)^2$ and that's the hardest of the three. Then of course $P(B|C_2) = 0.$ And then, almost as straightforwardly, $P(B|C_3) = 1/7.$ Putting it all together we have $$ P(B) = P(B|A)P(A)+P(B|C_2)P(C_2) + P(B|C_3)P(C_3) \\=\frac{13}{49}\frac{1}{4} + 0\frac{1}{4} + \frac{1}{7}\frac{1}{2}\\ = \frac{27}{196}$$

  • $\begingroup$ Thank you, this is helpful. At least for me, looking at it this way helps me see all the different possibilities I'm counting which is good for my learning! Cheers! $\endgroup$ – Hanzy Jul 8 '17 at 11:29
  • 1
    $\begingroup$ Yes and it's always good to do the problem in multiple ways and make sure you get the same answer $\endgroup$ – spaceisdarkgreen Jul 8 '17 at 16:07

Case 1: You wander the streets, asking everybody that you meet "Do you have exactly two children?" Most people ignore you, or say "no." But occasionally, one will say "yes." You ask these "Is at least one of them a boy who was born on a Tuesday?" After about four say "no," one finally says "yes." What are the chances that this person has two boys?

Answer: There are 14 different kinds of children according to the general description "[gender] born on a [day of week]." Assuming each description is equally likely, and independent in siblings, that means there are 14*14=196 combinations. The elder child is a Tuesday boy in 14 of these, and the younger is also a Tuesday boy in 14. But one of those has two Tuesday boys, so there are 14+14-1=27 combinations with a Tuesday boy. Similarly, there are 7+7-1=13 combinations with two boys, and at least one Tuesday Boy. The answer is 13/27.

But can you realistically assume that was the scenario that produced your question?

Case 2: At a convention for math puzzles[1], a man starts his discussion session with these three sentences: "I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?"

Answer: We can't assume that there is anything special about Tuesday boys. All we know is that this man decided to ask a variation of Martin Gardner's famous Two Child Problem[2] by adding a fact about one of his own two children. If that fact doesn't apply to both, we can only assume that he chose it at random from the set of two similar facts that he could have mentioned. Which means that of the 27 combinations I mentioned above, he would have mentioned the Tuesday Boy in only 14 of them. In the 13 with two boys, he would have mentioned the Tuesday Boy in only 7. The answer is 7/14=1/2. [3]

In fact, Gary Foshee's answer at that convention pre-supposed case 1.

Case 3: I also have two children. I just wrote a gender on a notepad in front of me. At least one of my children has that gender. What are the chances that I have two of that gender? [4]

Answer: If I hadn't written a gender down, you'd say that the chances that my two children have the same gender are 1/2.

If I had said "at least one is a boy," the logic in case 1 would say the chances change to 1/3: of the four combinations BB, BG, GB, and GG, only three include at least one boy, and only one of those has two. But if I had said "at least one is a girl," the same logic would again say the chances change to 1/3.

But if it changes to 1/3 regardless of what I say, then the act of writing it down makes that same change occur. This isn't possible, so the logic in case 1 can't be right.

Conclusion: Being told that some information applies is not sufficient to deduce a conditional probability based on that information. That was Martin Gardner's point when he said his Two Child Problem was ambiguous, and Joseph Bertrand's point in 1889 with his Box Problem which is essentially the same problem. But Bertrand went further than Gardner did, and showed why you can't just assume you know the information because you asked for that exact information.


[1] Say, the 2010 Gathering for Gardner, named in honor of the famous math puzzler Martin Gardner. Where this actually happened.

[2] Apparently, without realizing that Gardner himself admitted that the problem statement was ambiguous, and that you can't answer without making assumptions about how you learned the information.

[3] Similarly, Gardner said that both 1/3, and 1/2, could be answers to his question.

[4] This is a variation of Bertrand's Box Problem (https://en.wikipedia.org/wiki/Bertrand%27s_box_paradox). It's sometimes called Bertrand's Box Paradox, but the paradox Bertrand referred to was not the problem itself. It was the argument I provided for why the answer can't be 13/27.

  • $\begingroup$ Thanks so much for taking the time to write such a detailed reply. While I hadn't thought it out in as detailed a manner as you outlined, my initial thoughts were similarly guided. However, seeing the answer provided to the question (which I found in the "brilliant" app for iPhone), I wanted to operate on the same assumptions as they did just to get better at calculations. But this reply is much more informative than their short description. 👍 $\endgroup$ – Hanzy Jul 9 '17 at 14:03

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