Like many puzzles, today’s seems to have a variety of origins. I first read it in a book by Martin Gardner. So I will give him credit for it, but one can find variations of it in many sites.

This puzzle gets right to the heart of statistics and probability, and has the charm of being rather simple, but with at least one trap for the unwary. For that reason, it could be the basis of a pleasant way to make a bit of extra money on the side.

Assume you have four cards taken from a deck. Two of them are black and two are red. The face values are not relevant. Shuffle the cards. Your sucker - I mean bettor - is allowed to choose two cards by indicating which ones he wants. You might have him place a coin on the back of each of the two cards he wants to choose. The cards are left face-down while you agree on the terms of the wager.

The bet is that the selected cards match in color. You are going to bet on the outcome when you turn the selected cards face up, but what are the fair odds? You point out that there are only three cases possible: (1) both cards are black, (2) both cards are red, or (3) the cards are mixed. Since the odds are obviously 2:1 that you will lose, you should put one dollar in the pot against his two. That is, is the cards are mixed, he gets his money back plus a dollar. If the cards match, you get your money back plus his two dollars. Sounds fair.

But unfortunately you have picked a victim who knows you have been reading this column and so he figures there is a trick somewhere. He thinks about it for a minute and realizes that he had come very close to being scammed. “Wait a minute. You almost got that one by me. You counted wrong. There are four possible outcomes, not three: (1) both red, (2) both black, (3) the left one red with the right one black, and (4) the left one black with the right one red. Two of these outcomes are winners for me and two are losers, so the odds are even. You put in a dollar and I put in a dollar. Winner gets both.

Well, you tried. Your presentation was good and you almost had him, but he was too smart for you, so you grudgingly agree to continue with the game anyway under his terms - even odds.

After a few minutes of rapid play, you leave with all his money while he scratches his head. What happened? He caught your attempt to conceal the true state of the outcomes, yet you ended up winning more than him. Why?

There are several ways to look at this puzzle correctly, and like all good puzzles, many more ways to look at it incorrectly. Both of the analyses given by the players are incorrect. Like a good magician, the scammer has misled the sucker into the wrong way of thinking.

The actual probability that the two cards are the same color is only 1/3. At even money, you have a tremendous advantage. One way to see this is brute force. (Unfortunately, brute force is often the only way.) Construct all 24 permutations possible of the four cards. Select all possible pairs and you will see that eight of them (or 1/3) have matching colors. Another way is to consider that you have drawn one card. That leaves three on the heap. Of those three, only one will match the card in your hand. Therefore you have a 1/3 chance of winning, regardless of the color of your card.

Note that if you had let the mark pick up the first card and look at it, the odds would not have changed, but instead of looking at four cards on the table, he would have been looking at three, and probably had a better chance of realizing his actual chance of winning. Putting the coins on the chosen cards was part of the misdirection.

For those who wish to delve further into decision theory without wading through a lot of equations, I have posted a tutorial on elementary decision theory. It shows examples of faulty physicians’ diagnoses (important for those considering surgery) and how to evaluate anti-terrorist activities (important for everyone). That tutorial can be found here.