This week’s post is a puzzle. It is not even a relatively new puzzle… but I think it’s a *good* one, because the problem, and more precisely one of its several solutions, have an interesting relationship to some slightly more practical situations that I have been asked about recently. (As a teaser, I recall being involved at least once in such a “practical” situation, that nearly escalated to physical confrontation. It is strange the things that people feel strongly about.)

Before going into more detail, though, it has been suggested– correctly, I think– that I tend to spoil possibly interesting problems by including solutions in the same post. To remedy that, following is just the statement of the problem. Discussion is welcome in the comments; I will continue next week with my view of the problem and how I think it applies elsewhere.

Consider a standard, shuffled poker deck of 52 playing cards, of which 26 are red and 26 are black. I place the deck face down on a table. I will draw a card, one at a time, from the top of the pile and place it face up on the table for you to see whether it is red or black. At some point before drawing the last card in the deck, you must say, “Stop.” If the next card that I turn over is red, you win one dollar. If it is black, you lose one dollar.

What is the optimal strategy for playing this game, and what is the probability of winning?

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The obvious strategy would be to wait until I’ve seen 26 of one color or the other, and in the event I run into the second-to-last card without this happening, I just take a guess. Unless I’ve got this wrong, the odds of the last two cards being different colors is 26/51, and the odds of me guessing incorrectly in this event is 1/2, so that strategy gives me 13/51 chance of losing. Because the losses and wins are both $1, this looks like a good bet.

That’s just the obvious strategy, not necessarily the optimal one.

I just realized I understood the problem slightly. There’s no guessing the color, it’s just a “stop,” so that changes the odds (and I probably calculated these wrong): 25/102 that the last two cards are black (I lose), 26/51 of them being different (1/2 odds I win), leaving 25/102 that at least last two cards are red (I win). So that strategy has a 1/2 chance of winning.

Right, and this is a detail of the rules that is worth emphasizing. In fact, if you were allowed to guess the color of the next card (vs. guessing when the next card will be

red), then an optimal strategy wins with probability 1. For example, let 51 cards go by, after which you know what the color of the remaining card must be.Your calculation of probabilities for this strategy is correct; the problem is to find an improvement, or to show that none exists.