Here is a puzzle related to one we had earlier involving a young man who must pick the daughter with the largest dowry. However, it seems to be simpler on the surface – and maybe it is.

Be warned and remember the quote: “This branch of mathematics (probability) is the only one, I believe, in which good writers frequently get results entirely erroneous.” (Charles Sanders Pierce)

In the original version, a deck of 52 cards is shuffled and placed face down. You must place a bet in advance of any further action of where you think the first black ace will be found as the cards are turned over one at a time. Which location down the deck would you pick to maximize you chances of winning the bet?

This is the solution given by my hero, Martin Gardner, in the first edition of Wheels, Life, and Other Mathematical Amusements.

Contrary to most people’s intuition, your best bet is that the top card is a black ace.

The situation can be grasped easily by considering simpler cases. In a packet of three cards, including the two black aces, and, say, a king, there are three equally probable ordering: AAK, AKA, KAA. It is obvious that the probability of the first ace’s being on top is 2/3 as against 1/3 that it is the second card. For a full deck of 52 cards the probability of the top card’s being the first black ace is 51/1,326, the probability that the first black ace is second (on the stack) is 50/1,326, that it is third is 49/1,326, and so on down to a probability of 113.26 that it is the 51st card. (It cannot, of course be the last card.)

Is this analysis and answer by a uniformly recognized excellent writer correct?

Rather than answer directly, consider a deck of 26 cards in which one red and one black suit have been excluded. Now there is only one black ace to worry about. The probability that is it the top card is obviously 1 out of 26. The probability that it is the second card is 1 out of 25 times the probability that the first card was not the ace. This gives (25/26)/25 or the same probability that it was the first card. A few minute’s calculation will show what we knew already. The probability before we start turning cards over is uniform for all positions. This is different from the case where a bet is placed after N cards are shown out of a total deck of M cards.

Can you generalize the 26 card deck with one ace to the 52 card deck with two aces? How does the strategy change?

Once you have this puzzle down cold, let’s try a variation: You place a bet and you are allowed to let any number of cards go by without being turned up. When you say stop, you are betting the next one turned up will be a black ace. Your payoff is based on the number of cards left in the stack. For instance, if there are only 4 cards left, and two of them could be aces, then your payoff is even. If you chose the first card, then the payoff is 25 to 1 and so on for the other positions. Given this modification, what is the most favorable place for you to bet? You can take any card from the top to the third from the bottom.

Finally, play the same game, but this time the cards are turned face up one at a time until you say to stop. Then you are betting the next card will be a black ace. What is your best strategy now? How does the game change if you are allowed to keep going down to the last card?

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.