Wednesday, July 7, 2010

Math Surprises

Some of the best math problems are the simplest to state. But they can be fiendishly difficult to understand, precisely because their domains are so commonplace. A good example is fairly well known: Suppose I have two coins. One of them is showing heads. What is the probability the other one is also showing heads?

The answer most people jump to is 1/2, since we've had it drilled into us that coins are independent, so the outcome of one shouldn't affect the other. But the phrasing of the question contains a subtle trick. By stating that one of the coins is showing heads, I have eliminated the possibility that both coins are showing tails. But there are actually three remaining possibilities: heads/tails, tails/heads, or heads/heads. In only one of these three are both coins heads, so the answer is in fact 1/3.

The trickiness of the phrasing comes from the fact that when we say that one of the coins is heads, we deliberately don't say which one. We could say equivalently that at least one is showing heads. The probability is the expected 1/2 once we specify ahead of time which coin it is.

Given this preliminary, consider the following problem, which has been making its way around the Web of late:

I have two children, one of whom is a boy born on a Tuesday. What is the probability the other one is a boy? (You can make all the expected simplifying assumptions: no twins, boys and girls born in equal proportions, each day of the week is equally likely, etc. It's a math problem, not an obstetrics exercise.)

I will post the answer in a few days.


  1. I have to respectfully disagree with you, James. The odds of the coin being heads is 50%. You have two coins -- one is showing heads. That means that coin is already flipped, the outcome is determined, the randomness of that coin is taken out of the equation. It's all up to the other one now....and as we know, the odds of a coin coming up heads on a toss is 50%, not 33.3333%

    For the problem to have the probability you state of 1/3, I think you would have to word it: "I am about to toss two coins. At least one of them will come up heads...what are the odds of the other one coming up heads?" And, of course, you'd have to have the gift of precognition to state it that way... :)

    For the kids, it's also 50%...and I don't care which day of the week the other is born (and you didn't ask, so it's not relevant). The odds of two children both being boys of course is 1/4, but we've already tossed the coin for one. The other will be either a boy or a girl...50/50 chance

  2. @Gary: If you don't believe me, you can easily do the experiment yourself. Do it like this: Take a sheet of paper and draw a line down the middle. Then repeatedly flip two coins. If they both come up tails, then obviously one was not heads, so you can ignore that run; make no marks on the paper. If either one does come up heads, make a mark on the left side of the paper. If both come up heads, also make a mark on the right side of the paper.

    After doing this 100 times, see if the number of marks on the right side is closer to 1/2 or 1/3 of the number on the left side. (Unfortunately, a mere 100 trials leaves some room for doubt; you could easily have 30 right/75 left, and what does that tell you? It takes closer to 1000 trials to really separate the distributions.)

    Or, if you're computer-savvy, write a program to generate 1 million trials.

    Your logical error is that you assume the FIRST coin is showing heads. But I didn't say that; I said AT LEAST one coin is showing heads. That is where the counterintuition derives from.

    And for the kids, obviously the answer is not 1/2, or it wouldn't be surprising, would it? You can apply the same logic to get the correct answer, though.

  3. But James, we're not talking about a 100 hypothetical trials, or a million. We are talking about one trial of two coin tosses, where one already came up heads. I'm not saying it was the first coin, YOU are saying it was the first coin, because you told me it IS showing heads. We no longer need to consider the four combinations (or permutations, whatever) of the two random coin tosses, because one is already determined. We only need to consider the randomness of one coin toss -- 50/50.

    And with the children -- it's not surprising at all. It's obviously 1/2. Because your child, unlike a player character in D&D while being generated, isn't a formless gray blob. It either IS a boy or IS a girl. You next child, regardless of the first child's gender, will either be a boy or girl, with a 50% probability of each.

    We're looking at this in two different ways. Yours is from a purely theoretical mathematical perspective of hypothetical events and outcomes, and I am looking at it from a perspective of events as they actually happen, with the outcome of the first event already determined prior to looking at the second.

  4. @Gary: You said, "I'm not saying it was the first coin..." But you also said, "I am looking at it from a perspective of events as they actually happen, with the outcome of the first event already determined prior to looking at the second." So you see, you ARE saying it was the first coin that came up heads. I am not. I am saying that one of them came up heads. Could have been the first or the second.

    You are certainly correct to point out that this is a critical distinction, since if I had said that the first coin came up heads, then the probability of the second coin coming up heads is 1/2. What you are missing is that there IS a distinction between knowing that a SPECIFIC coin came up heads versus knowing only that AT LEAST ONE coin came up heads. If you know the latter, the probability of the second coin coming up heads is 1/3.

    One more point: probability only has any meaning when considered in terms of millions of events, at least potential millions. So if you're unwilling to consider whose case - mine that it's 1/3, or yours that it's 1/2 - an empirical test would buttress, then you're unwilling to submit your belief to a test. I would be willing to put money on mine. Are you?

  5. Yes, James I am. Take two coins. Lay one down as heads. Flip the other coin 100 times. If the number of times it comes up as heads is closer to 33 than 50, you win.

  6. @Gary: Since you're changing the experiment, I assume this means you agree that in my experiment, it will come up 1/3. So the question is really one of semantics: did my original question correctly describe the situation that my experiment tests? If you still think it didn't, there's not much I can further say to explain it.