Following on from the teasers in the probability paradoxes post, here is a closer look at “Tuesday’s child”. While it may not strictly be a paradox, it still has the rich potential for generating controversy. In fact, I don’t agree with what could be called the “classical” analysis of the problem. Here I will look at this classical approach and save my own interpretation for a later post.

A warning: this will be the most mathematical post on the blog to date, so it is not for the faint-hearted!

All of these probability paradoxes hinge on the notation of *conditional probability*. Conditional probability is the probability of one event *given* *that* another event has occurred. As a simple example, imagine roll a dice and A denotes “rolling a six” and B denotes “rolling an even number”. Then the probability of rolling a six is 1/6, but the probability of rolling a 6 *given that* I roll an even number is 1/3.

Now, before getting onto Tuesday’s child, I will go back to the simpler paradox, which I introduced in the Martin Gardner post. Note that throughout this post I will assume that girls and boys are equally likely and I will ignore identical twins (no offence to identical twins, of course!). I’ll quote from Gardner’s “Mathematical Puzzles and Diversions”:

Mr Smith says, ‘I have two children and at least one of them is a boy.’ What is the probability that the other child is a boy? One is tempted to say 1/2 until he lists the three possible combinations of equally probable possibilities – BB, BG, GB. Only one is BB, hence the probability is 1/3. Had Smith said that his

oldest(or tallest, heaviest, etc.) child is a boy, then the situation is entirely different. Now the combinations are restricted to BB and BG, and the probability that the other child is male jumps to 1/2.

Without going into my reasons in this post, I don’t agree with Gardner’s solution to the problem as he posed it. But, with a little tweak, I would agree. If instead you ask Mr Smith whether he has at least one boy among his two children and he says ‘yes’ (as opposed to having him volunteer the details), then the probability that he has two boys is 1/3.

I’ll tweak the Tuesday’s child problem in the same way for now. Imagine the problem is now as follows:

Mr Smith has two children. You ask whether he has at least one boy born on a Tuesday. He says ‘yes’. A lucky guess perhaps, but now you wonder what the chances are that Mr Smith’s other child is also a boy.

At this point, we could enumerate all the possible combinations of children and weekdays of birth. All up there are possibilities. Looking through that list, we would then scratch all of those that do not have at least one boy born on a Tuesday. In the list that remains, look at the proportion made up by families with two boys. Try it and you’ll find your revised list has 27 combinations in it and 13 of them have two boys, so the probability we are after is 13/27.

For me, that looks a bit much like hard work, so instead I would call on a bit more of the mathematics of conditional probability. Feel free to stop reading now if you don’t have the stomach for even more mathematics!

Mathematically, conditional probability is defined as follows:

where the left hand term denotes the “probability of A given B” (and P denotes “probability of”). The top term on the fraction is “A intersection B”, which simply means that both A and B occur.

I will denote by *X* the “event” that Mr Smith said ‘yes’ to the at least one boy born on Tuesday question (the inverted commas are there because “event” is actually a technical term in probability). The probability we want to calculate is

Starting with the conditional probability on the top of this fraction, if we have a two-boy family, the answer to “do you have a boy” will certainly be “yes”, so we need to know the probability of at least one of the boys having a Tuesday birth date. There are 7 possible birthdates for each child, giving 49 possibilities. Of these, 7 have a Tuesday for the elder child and 7 for the younger, but this double-counts the case where both were born on a Tuesday, so we have:

One way to calculate the probability of *X* itself is to break it down into different possible gender combinations:

Here I am using the fact that probabilities of “disjoint” (non-overlapping) events add up, i.e. if *A* and *B* are disjoint. Using the conditional probability formula, this gives:

Now the probability of the boy in a mixed gender family being born on a Tuesday is 1/7 and the probability of having boy-girl or girl-boy are both genders is 1/4. Combining this with the fact that the probability of a boy born on Tuesday is zero in a two-girl family and what we already know about a boy-boy family know gives us

Putting this all together, we have

While you are waiting for the next post with an alternative interpretation, you might want to think about Gardner’s two boy problem a bit more. In order to get to the classical conclusion that there is a 1/3 chance Mr Smith has two boys then you effectively have to assume that if he had one boy and one girl, he would definitely say ‘I have two children and at least one of them is a boy’ and *not* ‘I have two children and at least one of them is a girl’. Does that really make sense?

UPDATE: Here is a spreadsheet which simulates the classic version of the Gardner problem (i.e. assuming that you ask Mr Smith the question), and here is an alternative analysis.

### Possibly Related Posts (automatically generated):

- Eliminating the irrelevant (10 June 2010)
- Probability Paradoxes (7 June 2010)
- The Monty Hall Problem (21 June 2010)
- Vale Martin Gardner (28 May 2010)

{ 4 comments… read them below or add one }

I also came up with the 13/27 (but the hard way) before I decided that it was probably bogus. Still not sure. I mean, does the weekday really add much into things? And where does it end? If I said I had two kids and one was a boy who likes soccer, would that information be useful in guessing the sex of my other child (given that 1 out of every 10 boys like soccer)?

AHhhhhh!

#327:so I’m sure you won’t be surprised by the amount of discussion this problem has generated! Stay tuned for the sequel…with any luck it will ease your pain.Here is the follow-up post.

There is now a post up about the Monty Hall problem.