The Two Child Problem

Conditions:

A child has a 50% chance of being a boy (and otherwise is a girl).

Given a family with two children, one of whom is a boy,
What is the probability that the other child is also a boy?

There are two children, so there are four equally-likely possibilities:

BoyGirl
BoyBoy
BoyGirl
Boy
GirlBoy
GirlGirl
Girl

We know that one child is a boy.
There are only three possibilities that have at least one boy:

BoyGirl
BoyBoy
BoyGirl
Boy
GirlBoy
GirlGirl
Girl

BoyGirl
BoyBoy
BoyGirl
Boy
GirlBoy
GirlGirl
Girl

Only one of out three cases has two boys.
This gives the answer: ⅓

But is it? One could argue that the second child has nothing to do with the first, so the probability of the second child being a boy is still ½.

Let's try it and see what happens!

(It randomly generates children and counts them.)

Results:

0 Families
0 Families with At Least One Boy
0 Families with Two Boys

As confirmed by the first four percentages calculated, a child has a 50% chance of being a boy, and the two children are indeed independent events with 50% chance each, as we would expect. ¼ (25%) have two boys, ¾ (75%) have at least one boy, and ½ (50%) have exactly one boy.

However, the answer is ⅓ (33%) because the question is only asking about families that have at least one boy, without saying which child is that boy. The choice of which child we are asking about depends on which child was the boy required by the question. There are two permutations: either the first is that boy or the second is that boy. The child chosen second is no longer independent of the first child chosen by the question.

Here is another way of explaining where the number 33% comes from:

The intuitive way to approach this question is to split the question into two parts. It is only asking about families that have at least one boy, but does not specify which one. There are two possibilities for which boy is used to satisfy this condition, so let's split the question.

If he is the first child, then we have two possibilites:

Boy,Girl
Boy,Boy

If he is the second child, then we have two possibilities:

Girl,Boy
Boy,Boy

There are 4 total possibilities, and two out of the four have two boys, so the answer seems like 50%. However, because we split the question in two, we have counted the symmetric case of "Boy,Boy" twice! This case has a 1 in 4 chance, just the the others, yet here it is 2 in 4 because we forgot to account for the fact that although it satisfies the question in two different ways, it is still a single case, so we must remove one copy of it before calculating the final probability. This leaves us with:

Boy,Girl
Boy,Boy
Girl,Boy

Now there are three possibilities and only one has two boys, so given that you have one boy, the probability of having two boys is ⅓.

The Two Child Problem on Tuesday

Conditions:

A child has a 50% chance of being a boy (and otherwise is a girl).
A child has a 1 in 7 (~14%) chance of being born on a particular day of the week.

Given a family with two children, one of whom is a boy born on Tuesday,
What is the probability that the other child is also a boy?

Because child is either a boy or a girl and can be born on any of the seven days of the week, there are 14 possibilities for each child. Since there are two children, this is squared, so there are 196 possibilities for each family:

The question is only asking about families with a boy born on Tuesday, so we only consider those 27 squares (highlighted above). Out of those cases, the probability of both children being a boy is equal to the 13 cases that have two boys (highlighted green) divided by 27.
This gives the answer: 1327 (48.148%)

However, your intuition probably says that Tuesday has nothing to do with it, and as with the simpler puzzle, the second child has nothing to do with the first, so the answer is 50%. Or perhaps 33% if you agree with the simpler puzzle's solution, but don't see how Tuesday can possibly make a difference here.

Which is right?

Let's try it and see what happens!

(It randomly generates children and counts them.)

Results:

0 Families
0 Boys
0 Children Born on Tuesday
0 Families with At Least One Boy Born on Tuesday
0 Families with Two Boys with One Born on Tuesday

If you let it run for long enough, you will see that it does indeed settle out to 48.148%, as predicted by the mathematical analysis of the problem, not 50% or 33%!

The reasoning is the same as for the simpler puzzle. The intuitive way to start is by splitting the question in two parts.

If the first child is a boy born on Tuesday, then you have these 14 possibilities:

Boy (Tuesday), Girl (Sunday)
Boy (Tuesday), Girl (Monday)
Boy (Tuesday), Girl (Tuesday)
Boy (Tuesday), Girl (Wednesday)
Boy (Tuesday), Girl (Thursday)
Boy (Tuesday), Girl (Friday)
Boy (Tuesday), Girl (Saturday)
Boy (Tuesday), Boy (Sunday)
Boy (Tuesday), Boy (Monday)
Boy (Tuesday), Boy (Tuesday)
Boy (Tuesday), Boy (Wednesday)
Boy (Tuesday), Boy (Thursday)
Boy (Tuesday), Boy (Friday)
Boy (Tuesday), Boy (Saturday)

If the second child is a boy born on Tuesday, then you have these 14 possibilities:

Girl (Sunday), Boy (Tuesday)
Girl (Monday), Boy (Tuesday)
Girl (Tuesday), Boy (Tuesday)
Girl (Wednesday), Boy (Tuesday)
Girl (Thursday), Boy (Tuesday)
Girl (Friday), Boy (Tuesday)
Girl (Saturday), Boy (Tuesday)
Boy (Sunday), Boy (Tuesday)
Boy (Monday), Boy (Tuesday)
Boy (Tuesday), Boy (Tuesday)
Boy (Wednesday), Boy (Tuesday)
Boy (Thursday), Boy (Tuesday)
Boy (Friday), Boy (Tuesday)
Boy (Saturday), Boy (Tuesday)

There are 28 total possibilities, and 14 of them have two boys, so the answer seems like 50%. However, as with the simpler puzzle, by splitting the question we have counted the symmetric case twice! The probability of having two boys born on Tuesday is 1 in 196, just like all the other cases, yet here it is 2 in 196. As before, we must remove the duplicate to get the right answer. The symmetric case still subtracts only one, but for this puzzle the addition of seven extra categories (days of the week) dilutes the effect, so the answer is closer to 50%. The reason why Tuesday changes the answer from the first question's 33% is because a boy born on Tuesday is much less likely than just a boy. For every boy born on Tuesday, there are 7 other possible boys, yet there is still only one case where both boys are equal.

As you can see on the 14x14 square, if you separate the problem and only consider the highlighted row, or only the highlighted column, you have 14 squares on each. However, when you combine them to get the final answer of 14/28, the square where they intersect will be counted twice, so you must subtract one from both the numerator and the denominator.

1428 (50%) becomes 14 - 128 - 1 = 1327 (48.148%) 24 (50%) becomes 2 - 14 - 1 = 13 (33.333%)
The correct answers pop right out of 50%! All you need to do is subtract 1 for the symmetric case that was counted twice.

It may be easier to understand if you remove the confusion of children and days of the week. The problems can be rephrased equivalently like this:

You flipped two coins. One of them is tails. What is the probability that the other one is also tails?
You rolled two 14-sided dice. One of them is 3. What is the probability that the other one is less than 8?

Now it is much clearer why Tuesday changes everything, and why the answers will differ! In both questions, we are giving a specific condition for one die/coin, yet there are 7 ways to satisfy the other die (1,2,3,4,5,6,7), as opposed to only one way to satisfy the other coin (tails). However, there is still only one symmetric case, in which both are equal (tails,tails and 3,3).

Suppose you further increase the number of unique kinds of boys and girls, perhaps by using the 365 days of the year. The answer will be even closer to 50% because there is always only one symmetric case to subtract. For example, given a 365-day year, if one child is a boy born on January 3, the probability that the other child is a boy is 729 / 1459 (49.9657%)

As the number of unique kinds of children approaches infinity, the answer approaches 50%:

limn → ∞ n - 12n - 1 = limn → ∞ n2n = 12 (50%)