Spoiler Alert: Coins and Mittens Puzzler

The Show
1

Ok, this was a tough one, but I think I solved it. OK, I admit I’m not the sharpest tack in the drawer. It took me almost a full hour of focussed effort, penciling it out, trying this and trying that. At first it seemed like it wasn’t solvable, but each time I tried to prove it unsolvable, I couldn’t, and I’d then realize “well, here’s yet another case that actually can be solved”, so I kept going. The solution is sort of non-intuitive. To me at least. I’m thinking this problem is related to a mathematical sub-field called “group theory”, and in particular the concept of operators.

I’m not giving my sol’n here, but here’s a hint if you like. Don’t give up. You need to build up the general solution from simple parts. So start by example, with the most simple case. 2 coins. Find that solution. Then 3 coins. And go on from there. Look for a pattern.

2

I don’t see where you are going with your sol’n. The objective is to produce two piles that have the same number of “heads”. I can see, like you imply, that turning coins over may be part of the sol’n, that’s why I mentioned “operator”, turning the coins over is an “operator” in group-theory-speak. But I don’t see how you are dividing the coins to make the two piles.

Do you mean you’d arbitrarily choose some number of coins from the pile and turn those over? And that would be one pile? And the remaining would be the other pile?

If so, why would you choose to limit the choices to 10, 10/2, 10+1, … etc … Why does it have anything to do with 10? How did you determine this?

3

@asecular, no I haven’t heard the solution yet. In the SF Bay area, the new show isn’t broadcast until Sunday. I’ll be listening then to see what they say. Unless I fall asleep. Then I’ll have to listen to the podcast. For some reason whenever Car Talk is on I get very sleepy. I’m not complaining, it’s just a fact. I guess I find it relaxing. lol …

I guess I was overthinking like you say. I was looking for a general solution guarenteed to work, not just for 10. I don’t recall ever needing calculus or algebra on the puzzler. But sometimes they helped by providing an alternative path to the solution.

Sometimes fairly simple ideas can lead to more important developments. Like the professor who watched the bubbles in his beer glass, wondered why they did what they did, and developed the bubble chamber. And, what’s his name – hmmm … Bell … he developed Bell’s theorem based on a very simple idea, which disproved one of Einstein’s ideas about quantum mechanics. The so-called EPR paradox. Bell’s theorem is really – at its core – no more complicated than this puzzler.

4

You incorrectly stated both the problem and the solution. Here is the way the problem shoul’ve been written:

You’re sitting at a table with a bunch of pennies on it. Some are heads up and some are 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

I tell you that a certain number of the pennies are facing heads up. Let’s say 10 are facing heads up.

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 the number of heads up in one group will equal the number of tails up in the other group?

Here’s the answer. You don’t know how many coins are on the table, but I’m telling you that ten are 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.

What if you had, by luck, chosen all tails? Where would all the heads be? Answer: in the other pile, the big group. So how would you make the number of heads in one group equal the number of tails in the other group?

It turns out that no matter how many heads you manage to get out of the big group, the tails in the little group will always be a number equal to the remaining heads in the big group.

Let’s say, for example, that you pull out ten coins and nine of them are tails. Only one is a head. Question: How many heads are left in the big group?

Answer: nine. Thus there are nine tails in the little group and nine heads in the big group. The numbers are equal.

Now let’s say of your ten coins, you get eight tails and two heads. How many heads are left in the big group?

Answer: eight. Also, since you have eight tails in the little group, the numbers are equal.

Try again. Let’s say that of the ten coins, you get four tails and six heads. How many heads are left in the big group? Answer: four. You have four tails and the larger group has four heads. The numbers are equal.

A final example. You get one tail and nine heads. How many heads are left in the big group? Answer: one. The numbers are equal.