Gary Foshee, a collector and designer of puzzles from Issaquah near Seattle walked to the lectern to present his talk. It consisted of the following three sentences:

“I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?”The event was the Gathering for Gardner earlier this year, a convention held every two years in Atlanta, Georgia, uniting mathematicians, magicians and puzzle enthusiasts. The audience was silent as they pondered the question.

—“Mathemagical” by Alex Bellos

*New Scientist*, 29 May 2010

[This essay was revised after some helpful comments on John Baez’s blog about an earlier version.]

If you haven’t previously encountered the puzzle described above, you might like to spend a few moments pondering it yourself before reading the discussion that follows.

Your first thought might be: “*Tuesday?* How can that possibly change anything?”
To which I’d reply that sometimes it can and sometimes it can’t; it depends very much on precisely how you’ve obtained the information.

To deal first with a few potential red herrings, this is a puzzle about probabilities, not biology or obstetrics. So for the sake of the puzzle:

- we’re ruling out twins,
- we’re pretending that boys and girls are born with equal likelihood, and
- we’re assuming that there is no correlation between a child’s sex and the day of the week on which they’re born.

Now in Foshee’s formulation given above, it’s not really clear *why* he has chosen to disclose the particular facts he mentions.
If we don’t know exactly what parents in various circumstances would tell us, it’s hard to reason about the significance of these facts.
So we’re going to consider three variants of the puzzle, in which the same information is revealed in three different ways.

Puzzle 1: Suppose Eve knows that Anna has exactly two children. Eve asks her directly, “Is it true that at least one of your children is a boy born on a Tuesday?”Anna replies, “Yes!”

What is the probability that Anna has two sons?

Puzzle 2: Suppose Eve knows that Anna has exactly two children. Eve asks her, “Please pick a male child of yours at random and tell me what day of the week he was born. If you have no male child at all, say nothing. If you have two sons, choose at random between the first-born and second-born.”Anna replies, “Tuesday.”

What is the probability that Anna has two sons?

Puzzle 3: Suppose Eve knows that Anna has exactly two children. Eve asks her, “Please pick a child of yours at random and tell me their sex and what day of the week they were born.”Anna replies, “A boy, born on a Tuesday.”

What is the probability that Anna has two sons?

Before we go on to calculate any probabilities, let’s just look at the various combinations that might arise for Anna’s children. If we know only that Anna has two children, then any one of the four combinations shown below is possible:

If we know that Anna has at least one son, that leaves us with only three possibilities, indicated by the three remaining red-rimmed squares below:

And if we know that Anna has at least one son *born on a Tuesday*, we need to break down all the possibilities
for each child: boy or girl, and the day of the week on which they were born. That gives us 14 times 14, or 196 possibilities, but of those only the 27 squares marked in red below
include at least one son born on a Tuesday:

Now we’re in a position to answer **Puzzle 1** very easily. Of the 27 possibilities shown in the diagram above, exactly 13 lie within the deep blue square that corresponds to
Anna having two sons. So the probability for Anna having two sons is **13/27**.

What about **Puzzle 2**? We’ve ended up knowing the same facts, haven’t we, so why should the answer be any different?

But instead of a yes/no reply, in this version Anna might have said nothing, or she might have spoken any of the days of the week. Let’s look at the probability that she would have given the reply she did, under four different circumstances.

- If Anna has two girls, the probability that she’d say “Tuesday” is zero.
- If Anna’s first child was a boy and the second was a girl, the probability that she’d say “Tuesday” is 1/7 — the chance that her son was born on a Tuesday.
- If Anna’s first child was a girl and the second was a boy, the probability that she’d say “Tuesday” is 1/7 again.
- If Anna has two boys, it’s equally likely that she’ll pick the first-born or second-born son to talk about. In either case the probability that the son was born on a Tuesday is 1/7. So once again, the probability that she’d say “Tuesday” is 1/7.

Each of these four possibilities is equally likely, with probability 1/4, so:

P(Anna says “Tuesday”)

= (1/4) × 0 + (1/4) × (1/7) + (1/4) × (1/7) + (1/4) × (1/7) = 3/28

What we want is the probability that Anna has two sons, given that she has said “Tuesday”. This is just the proportionate contribution of the case where she *does* have two sons to
the overall probability that she says “Tuesday”:

P(Anna has two sons, given that Anna says “Tuesday”)

=P(Anna has two sonsandshe says “Tuesday”) /P(Anna says “Tuesday”)

= (1/4) × (1/7) / (3/28) =1/3

We can show this on a diagram similar to the previous one, by using *half squares* within the region where Anna has two boys. The left and right halves indicate the two equally likely
possibilities for
whatever it is that Anna uses to choose between telling Eve the birth day for her first-born son or her second-born son. The total marked area within this region is 7 squares, and the total marked area on
the whole diagram is 21 squares, agreeing with our result of 1/3.

**Puzzle 3** is similar, but this time Anna always has to make a choice between her first-born and second-born child, whatever their sexes. Again, let’s look at the probability that she would
have given the reply she did, under the four possibilities corresponding to the four large coloured squares on our diagrams.

- If Anna has two girls, the probability that she’d say “A boy, born on a Tuesday” is zero.
- If Anna’s first child was a boy and the second was a girl, the probability that she’d say “A boy, born on a Tuesday” is 1/14 — the chance that she picks her first-born child to describe, 1/2, times the chance that her son was born on a Tuesday, 1/7.
- If Anna’s first child was a girl and the second was a boy, the probability that she’d say “A boy, born on a Tuesday” is 1/14 again.
- If Anna has two boys, it’s equally likely that she’ll pick the first-born or second-born son to talk about. In either case the probability that the son was born on a Tuesday is 1/7. So the probability that she’d say “A boy, born on a Tuesday” is 1/7.

Each of these four possibilities is equally likely, with probability 1/4, so:

P(Anna says “A boy, born on a Tuesday”)

= (1/4) × 0 + (1/4) × (1/14) + (1/4) × (1/14) + (1/4) × (1/7) = 2/28

What we want is the probability that Anna has two sons, given that she has said “A boy, born on a Tuesday”. This is just the proportionate contribution of the case where she *does* have two sons to
the overall probability that she says “A boy, born on a Tuesday”:

P(Anna has two sons, given that Anna says “A boy, born on a Tuesday”)

=P(Anna has two sonsandshe says “A boy, born on a Tuesday”) /P(Anna says “A boy, born on a Tuesday”)

= (1/4) × (1/7) / (2/28) =1/2

Again, we can show this on a diagram using *half squares*, where the left and right halves indicate the two equally likely
possibilities for whatever Anna uses to choose between telling Eve the birth day for her first-born or her second-born child.
The total marked area within the two-sons region is 7 squares, and the total marked area on
the whole diagram is 14 squares, agreeing with our result of 1/2.

The answer of 1/3 for **Puzzle 2** is the same as the answer we get merely by knowing that Anna has at least one son.
The answer of 1/2 for **Puzzle 3** is the same as the answer we get merely by knowing that one *specific* child of Anna’s is a son.
In both these cases, knowing the day of the week tells us nothing extra — which probably accords with your intuition that it really *ought* to be irrelevant.

But why, then, is it *not* irrelevant in **Puzzle 1**? After all, if we changed **Puzzle 1** so that the
question was about Monday or Wednesday or any other day, we ought to get the same probability in each case. So doesn’t symmetry alone mean that the day ought to be irrelevant, always?

At this point, it might be worth taking a brief detour to a simpler example. Suppose we throw a dart at the oddly shaped dartboard below. Assume that we manage to hit the board itself, and not the wall around it.

If the dart is taken to be equally likely to have landed anywhere on the board, there will be about a 60% chance it landed in the green circle, and a 40% chance it landed in a blue region.

We can think of this dartboard as being made by superimposing six identical teardrop figures like the one below:

In this single teardrop, the relative amount of green and blue is much more in favour of the green: close to 90%. Because the green regions overlap completely when we combine the six versions to make the full dartboard, and the blue regions don’t overlap much at all, the proportion of green becomes diluted.

Now suppose I tell you that a dart landed within teardrop number 2 of the dartboard. If you decide to take that information and calculate the probability of the dart being on a green point or
a blue one by assuming that the dart is *equally likely to lie anywhere within teardrop 2*, you will get a much higher chance for green, 90%, than you would have if you’d treated the dart as
being *equally likely to be anywhere on the dartboard*, when it’s just 60%.
The difference arises because you’re changing your assumption from the dart being at a random location on the whole dartboard, to that of it being at a random location on one of the teardrops.
These are very different assumptions!

So which assumption is the most sensible one to make? That depends on the precise question we’re trying to answer. If we’ve simply thrown a dart at the board, with an equal chance
of hitting it anywhere, the additional piece of information that we’ve struck some particular teardrop is irrelevant; the probability that we hit green is still just 60%.
It’s only if we think of playing a different game, where we rule out all throws that don’t strike a particular teardrop that we’ve chosen in advance,
that we can say that a successful throw (successful by that new, narrower criterion) will yield a probability for green of 90%. But the fact that all the teardrops are *identical* doesn’t mean
there is no difference between an individual teardrop and the dartboard as a whole.

Returning to Tuesday’s child, in **Puzzle 1** we’re playing a game where only a particular day of the week that we’ve chosen in advance gets a “Yes” answer, which is similar to
ruling out all throws on the dartboard that don’t strike a particular teardrop.

But in **Puzzle 2** and **Puzzle 3**, Anna is simply telling us the day of the week on which a child of hers was born, which is similar to us merely learning the number of one of the teardrops within which our
dart has landed — while we’re still playing with the whole board.