## Mathematical Recreations

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#### The Memory Game (Concentration)

I have young kids, and they've gotten to the age where they can now play games. Well, little kids' games aren't always that exciting, so my mind sometimes wanders a little. One of the games that they like to play is a variation on the old Concentration theme. We'll just call it the memory game. I've been trying to introduce the concept of game strategy, but they don't quite get it.

Here's the problem for April. You lay out 2n cards face down. There are n matches to be found. On your turn you pick any card, turn it over, then pick another card and turn it over. If they match, then you keep the match and get another turn. If they don't match, then you turn the cards over and the next player takes their turn. When all matches have been claimed, the player with the most matches wins the game.

If the match to the first card has not been seen, then my kids always try to turn over a new card in hopes that the unseen card will complete the match. While this may maximize their chance of completing that match, it also gives the next player a chance of seeing a new card. This is often a bad decision.

The specific problem is to consider the case where there are m matches on the table in a two-player game; you have turned up a card and the matching card has not been seen; k of the face-down cards have been seen before. You have perfect memory and you assume that your opponent does as well. Should you turn up one of the (2m – 1 – k) cards which have never been seen, or should you turn up one of the k cards which have been seen?

Consider two cases. In the first case, you value each match independently, so that your payoff is proportional to the number of matches you get. In the second case, you currently have j matches more than your opponent, and you want to maximize your chance of winning. (Count a win as +1, a loss as –1 and a tie as zero.)

This is a problem that I've never solved before, although I've thought about it some. It's possible that a numeric approach might be required.

Send all responses to .

Thanks,
Steve