Thanks to reader Matt Lewis who sent me this link.
Martin Gardner, may he rest in peace, gave us a delicious probability problem:
Suppose that Mrs Smith has two children, at least one of whom is a son. What is the probability both children are boys?
Isn’t that lovely? Everybody immediately says, “One half!” But that answer is wrong because the counting is wrong. Discrete probability—which means all real-life probability—is all about counting, and counting isn’t easy. Counting is so difficult that once we depart from our standard ten-finger rules, our intuitions abandon us.
The first—and most important—rule of counting is this: What is everything that can happen? In the “Mrs Smith” problem, given the information provided, everything that can happen is this:
This is every situation that gives old lady Smith at least one son out of two kids, as stipulated by the problem. Now, there is only one case which produces two boys, and three different possibilities. That means the probability—given only the knowledge provided to you in the problem; Mrs Smith obviously knows whether she has two boys or one—of having two boys is 1/3. Simple!
Given the information, the situation “Girl, Girl” is impossible (no jokes, please).
Since this kind of problem is always fun, a conference inspired by Gardner regularly convenes in which participants offer simple-sounding, but difficult, probability problems. Science News covered the most recent convocation, in which Gary Foshee offered this puzzler:
I have two children, one of whom is a son born on a Tuesday. What is the probability that I have two boys?
Intuition—the base of all our thoughts, but a sometime shaky base—practically screams at us, “Tuesday? What’s the day of the week have to do with it? It’s irrelevant! The probability must be one-half.”
In fact, I thought so too, upon my first reading. But knowing it was meant to be a puzzler, and noting that it had a probability smell to it—this faculty is one developed through time, exercised by being burned regularly by simple-sounding problems—caused me to pause.
The Science News article provides an explanation of why the probability is not 1/2. But that explanation is difficult to follow. So let’s do it ourselves slowly and see what happens.
The second rule in counting is this: count a simpler, but similar, situation first. Do not try to jump to the right answer; that is a sure way of making mistakes.
OK, what’s a simpler, similar situation? How about this: births can only happen on Tuesdays. Well, if that’s true, then we are back to Gardner’s original problem, right? Prove that to yourself before moving on.
So far, so good. Now let’s suppose that births can occur on Tuesdays or Wednesdays. What is everything that can happen given the information that a boy was born on a Tuesday and that there are two children? Just this:
(1) Tuesday: Boy, Girl; Wednesday: none
(2) Tuesday: Girl, Boy; Wednesday: none
(3) Tuesday: Boy, Boy; Wednesday: none
(4) Tuesday: none, Boy; Wednesday: Boy
(5) Tuesday: none, Boy; Wednesday: Girl
(6) Tuesday: Boy, none; Wednesday: Boy
(7) Tuesday: Boy, none; Wednesday: Girl.
(Incidentally, and this has nothing to do with nothing, the word “Girl” is one of the words the spelling of which looks weirder the longer you stare at it.)
There are seven possibilities—again, and I cannot emphasize this enough, given the premises—only three of which lead to two boys. The probability—conditional on your information—is 3/7, which is just under one half. (And to all those who have been following the past two weeks, this is a logical probability; no “randomness” needed.)
The way this reads is this: At least one boy must be born on a Tuesday. He may be the first born or second. If two kids are born on a Tuesday, then none can be born on Wednesday (1-3). The first born might have popped out on a Tuesday—which must be a boy—but the second on a Wednesday (6-7); or the stork might have shown up on Wednesday first, leaving a boy to be born on Tuesday (4-5).
Solving the general problem is now easy: (1-3) stay the same, even if we add in the remaining days of the week. For example, here’s the full (1):
(1) Tue: Boy, Girl; Wed: none; Thu: none; Fri: none; Sat: non; Sun: none; Mon: none.
Situations (4-7) just repeat, once for each other day of the week. For example, here’s (4′):
(4′) Tue: none, Girl; Wed: none; Thu: Boy; Fri: none; Sat: non; Sun: none; Mon: none.
The number of situations is 4*6 = 24 for the original (4-7) for each day of the week, plus 3 for (1-3), which stay the same. That’s 27 possibilities. The number of those with two boys is 2*6 = 12 for the original (4-7) for each day of the week, plus 1 for (1-3), which stays the same.
Thus, the probability is 13/27, which is 0.48, or nearly one-half.
Update: Challenge! If you’ve figured out the original, then you’ll surely get this:
I have two children, one of whom is a son born on 29 June. What is the probability that I have two boys?
Ignore leap years.