# Pennies, Mittens and a Mask... Oh My!

RAY: Here’s the answer. Now 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 so that you have a little group of ten coins in front of you. And the larger group of coins is another pile somewhere else on the table.

Now, I always like using limit theory when solving this kind of a problem.

RAY: What if you had, by luck, chosen all tails? Where would all the heads be?

TOM: In the other pile.

RAY: 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.

And it turns out that 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?

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. How many heads are left in the big group?

TOM:: Nine.

RAY: Nine. Let's say it's eight and two. How many heads are left in the big group?

TOM: Eight.

RAY: How many tails do you have?

TOM: Two.

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. 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!

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