Suppose that I have a friend whose name is X. X has an elder brother who got married 10 years before. But X being a silly guy has the habit of forgetting things; you can say a “Short time memory loss”. This summer he is going to meet his brother after 10 years, and he has planned to purchase some clothes for his brother’s kids as a token of love. Each and every time he faces a problem he comes to me, so this time also, he came to seek help, and I have not left him empty handed. I told him that as far as I can recall, his brother has two children and one of them is a boy who was born on a Tuesday. Now, it’s your turn to help him. What is the Probability that X's brother has two boys? Assume an equal chance of giving birth to either sex and an equal chance to giving birth on any day.

  • 1
    $\begingroup$ Welcome to Puzzling! This is a well know puzzle which has been asked before in various ways. $\endgroup$
    – Glorfindel
    Jul 10, 2018 at 9:25
  • 2
    $\begingroup$ This variation that adds the day of the week is a rather interesting one though. I think it was first thought up only about 5 years ago. $\endgroup$ Jul 10, 2018 at 10:04
  • $\begingroup$ @Jaap, I was pretty sure that "One of the boy" was a hidden hint that the probability is 100%. $\endgroup$
    – xhienne
    Jul 10, 2018 at 10:05
  • $\begingroup$ Is lateral-thinking an applicable tag here? $\endgroup$ Jul 10, 2018 at 10:29
  • 4
    $\begingroup$ Fun fact: boys and girls can wear the same clothes. $\endgroup$ Jul 10, 2018 at 11:43

3 Answers 3


I'll use BoT to denote a boy born on a Tuesday, BnT for a boy born not on Tuesday, and G for a girl. The possible combinations for the two kids are

 BoT, BoT    1/14 * 1/14 = 1/196
 BoT, BnT    1/14 * 6/14 = 6/196
 BnT, BoT    6/14 * 1/14 = 6/196
 BoT, G      1/14 * 7/14 = 7/196
 G  , BoT    7/14 * 1/14 = 7/196
plus various combinations that do not involve any BoT, but those are not important.
The respective probabilities for each these 5 combinations is given. Note that I changed $1/2$ into $7/14$ in order to give them all the same denominator.

We want to know the probability of two boys, given that there is (at least) one BoT. We know that:

$P(2\ boys\ \&\ BoT) = \frac{1+6+6}{196} = \frac{13}{196}$

$P(BoT) = \frac{1+6+6+7+7}{196} = \frac{27}{196}$

Therefore $P(2\ boys\ |\ BoT) = \frac{P(2\ boys\ \&\ BoT)}{P(BoT)} = \frac{13}{27}$, or almost 50%.

Without that day of the week, and just knowing there was at least one boy, the answer would be $\frac{1}{3}$.

On the other hand, if you knew that the oldest child was a boy, then the answer would be $\frac{1}{2}$.

By saying the day of the week the boy you happen to know was born on, it is almost as good as specifying a particular child of the two, almost like saying that the oldest child is definitely a boy, and thereby raises the probability from $\frac{1}{3}$ to almost but not quite $\frac{1}{2}$.

  • 1
    $\begingroup$ Glad to see the math works out here, I just couldn't convince myself of this. I went and created a truth table for this, and came up with the same thing. $\endgroup$ Jul 10, 2018 at 16:05

The answer is

6 / 13
Because your friend's brother has one (not two) boys born on a Tuesday
He has either a boy not born on Tuesday (6 ways) or a Girl (7 ways)
So, 6 / (6 + 7) = 6 / 13

  • 4
    $\begingroup$ If we go by the wording of the puzzle, namely, that "one of them is a boy who was born on a Tuesday", the constraint against the other child born on a Tuesday has the same validity as that other child being a boy. So if you say that that other child wasn't born on a Tuesday (regardless of gender), then you also need to say that that other child is not a boy (regardless of the day of week they were born). By that logic, the probability of the other child being a boy is 0%. $\endgroup$
    – Lawrence
    Jul 10, 2018 at 12:48
  • $\begingroup$ I don't quite follow your comment. I'm saying that one Boy on Tuesday allows the other to be a Girl on Tuesday or a Boy on Wednesday but not another Boy on Tuesday. $\endgroup$
    – Jay
    Jul 10, 2018 at 21:43
  • $\begingroup$ Apologies for having pushed the thought too far. But having one boy born on a Tuesday doesn't preclude having two boys born on a Tuesday if it doesn't preclude having [one boy and one girl] born on a Tuesday. $\endgroup$
    – Lawrence
    Jul 11, 2018 at 0:22

The answer is



We are only concerned about one child; the other is already given as a boy (assuming your memory is correct). Since there is an equal chance of both on any given day, we are left with the standard probability for a single child.

The mention of a day is a red herring. The mention of the first child is a red herring. This is a reduction of the last stage of a Monty Hall problem, essentially. You have been given a bunch of information up front which does not at all influence the final question: Given that it is equally likely that a boy or a girl is born on any given day, what is the probability that this single child is a boy or a girl?

  • $\begingroup$ If all I knew was that X's brother had two kids, which are a boy and another child, would you then also say that the other child has a 50% probability of being a boy? $\endgroup$ Jul 10, 2018 at 15:53
  • $\begingroup$ Yes, of course. $\endgroup$ Jul 10, 2018 at 16:49
  • $\begingroup$ If I flipped two coins and I told you one of them is heads, the probability of the second being heads is still 50%. These are independent events. $\endgroup$ Jul 10, 2018 at 16:49
  • $\begingroup$ But the phrase "one of them" does not uniquely and consistently specify one of those two events. If you said "the first coin I flipped is heads", then you'd be right with 50% for the second coin. However, saying "one of them is heads" would also happen when the first coin is tails and the second heads. There are three equally probable outcomes, HH, HT, TH (since it is a given that TT did not happen), so the probability that both coins are heads given that at least one of them is heads, is 1/3. If you don't believe it, try it out for real. $\endgroup$ Jul 10, 2018 at 21:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.