The Three Card Puzzle

You are approached by a gambler who shows you three cards, one that is black on each side, one that is white on each side, and one that is black on one side and white on the other. You are allowed to shuffle the cards yourself, without looking at them, and place one card on the table. Let’s say the face showing is black. The gambler then says, “We know this is not the card that is white on both sides, so there are only two possibilities: this is either the card that is black on the other side, or it’s the card that is white on the other side. I will bet you that the other side is black. You have a 50-50 chance.”

You should not take the bet. While it may seem that there are only two possibilities, thus giving you a 50-50 chance of winning, there are actually three possibilities. The card may be showing the black side of the black-white card, it may be showing one of the faces of the black-black card, or it may be showing the other face of the black-black card. Thus, with the gambler taking black, you have only a one-in-three chance of winning. If you don’t believe it, make some cards and try it out!

3 thoughts on “The Three Card Puzzle”

1. JeffJo says:

Say the gambler adds another card that is white and black. Upon showing a black side, the gambler now suggests that we know the card is one of three, a d gives you 2-to-1 odds that the other side is black. Which color do you bet on?

The original problem is equivalent to the Monty Hall Problem, where you gave the correct answer of 2/3. The four-card one is not only equivalent to your second Two Child Problem, it is identical except for the names of the values (black/white vs boy/girl). But you gave the gambler’s answer, which is incorrect.

And all are variations of Bertrand’s Box Problem/Paradox. In Bertrand’s time, the word “paradox” did not refer to the problem with confusing answers, but how you could tell one must be incorrect. If the mere observation of one value (a card face is black, a child is a boy, Door #3 is a loser, a coin is gold) changes the probability that both values are present (black and white, boy and girl, #2 is also a loser, gold and silver), it must change it the same way regardless of what is observed. And if it changes regardless of the observation, you don’t need to make the observation to say it changes. But it can’t change without an observation. Paradox; so it can’t change.

2. Thank you for pointing out those equivalencies. Comparing your four-card version to the Two Children Problem is a neat way to demonstrate that it matters how the cards or families are chosen. If you start with four cards (black-black; black-white; white-black; and white-white), this is, as you point out, the same as the possibilities for two-child families (boy-boy; boy-girl; girl-boy; and girl-girl). If you choose one card at random, one side of which turns out to be black, then you know that the white-white possibility has been eliminated, but the probability that the other side is black is 1/2. However, if you instead eliminate the white-white card at the start (equivalent to selecting a family at random from among the two-child families with at least one boy), your probability of drawing the black-black card is 1/3. That is unsurprising in the card context, but makes for an interesting puzzle in the Two Children Problem — the wording of which has been clarified, thanks to you!

3. JeffJo says:

In 1889, Joseph Bertrand formulated the Box Problem as a cautionary tale. You cannot interpret the observation of a state, as having chosen the sample from the population of all samples ever the state is true. And in fact, that false impression that you can is what the gambler is trying to exploit.

And rewording the Boy Girl problems, saying they were ambiguous, is justifying the false impression. It would be better to put the original, and the new, but side by side and explain the difference.