May 26, 2001
RAY: You're sitting at a table with a bunch of pennies on it. Some are facing heads up and some are facing tails up. You're wearing a blindfold, and you're wearing mittens so that you can't actually feel the coins and tell which side is facing up. If we could trust you, we'd let you do it without gloves.
TOM: Well, you know me. I always sit around with mittens and a mask on anyway.
RAY: I will tell you that a certain number of the pennies are facing heads up. Let's say 10 are facing heads up.
TOM: You're telling me there are a bunch of pennies in front of me and 10 of them are heads up?
RAY: Right. With your mittens on, you can move the pennies around, you can pick them up, you can put them down again, you can shake them, you can do whatever you want.
Here's the question: Is it possible to separate those pennies into two groups, so that each group has the same number of pennies facing heads up?
TOM: The answer is yes! Give me the $25 Shameless Commerce gift certificate, you bum!
RAY: Shut up. You can't win. Part two of my question is: How do you do it?
RAY: Now the reason that you're wearing mittens is that you can't feel the, you know, if we could trust you, we wouldn't have you --
TOM: You can trust me!
RAY: Well, I don't trust you. You're wearing the mittens. Well, here's how you do it --
TOM: Yeah. Go ahead. I can hardly wait.
RAY: I'm going to tell you. Like I said, you don't know how many coins are on the table. Well, I'm going to tell you there are ten coins facing heads up. From that big group, take ten coins. Slide them over --
TOM: I can do, I can do that.
RAY: -- to you, now you have in front of you a little group of ten coins --
TOM: Got them right here.
RAY: -- and the larger group, OK, is off somewhere on the rest of the table.
TOM: And some of the heads could be in my little group and some not.
RAY: That's right. Now, I always like using limit theory when solving this kind of a problem.
TOM: Yeah. Go ahead.
RAY: What if you had, by luck, chosen all tails?
RAY: OK? Where would all the heads be?
TOM: In the other pile.
RAY: OK. They would be in the other pile. So how would you make the two piles so the same number--
TOM: I would flip over every single one that I just had, except I can't do it because I've got the stupid mittens on.
RAY: For this part, once you've gotten to this point, we'll allow you to remove the mittens.
RAY: And it turns out that no matter which, no matter what number of heads you manage to get out of that bigger group, you're going to flip over all your coins because as it turns out that the number of tails that you pull out always equals the number of heads remaining in the larger group.
TOM: Say that again? The number of --
RAY: Well, let's say for example that you pull out your ten coins, nine of them are tails, and one of them is a head.
RAY: How many heads are left in the big group?
RAY: Nine. Let's say it's eight and two. How many heads are left in the big group?
RAY: How many tails do you have?
RAY: No, eight. And when you flip those eight tails over, you have eight heads and your group and eight heads --
TOM: So the way you do it is you take out ten coins --
RAY: Or whatever number of coins you are told are facing heads up.
TOM: Right, and in this case it's ten, and I, and I flip those over --
RAY: And you're done.
TOM: And I'm done.
RAY: Pretty good, huh?
TOM: Wow. That's very good.
RAY: Do we have a winner?
TOM: You bet we do. William Innes from Missoula, Montana. By the way, this only works if it's a two-sided coin. Three-sided coins, it wouldn't work.
RAY: No.TOM: The winner this week is William Innes from Missoula, Montana.