Round Numbers - Futility Closet

Alan and Bob are playing a game of marbles. Alan has two marbles, Bob has one, and each rolls to try to come nearest to a fixed point. If the two have equal skill, what is the chance that Alan will win?

There seem to be two contradictory arguments. On the one hand, each of the three marbles has an equal chance of winning, and two of them belong to Alan, so it seems that there’s a 2/3 chance that Alan will win.

On the other hand, there are four possible outcomes: (a) both of Alan’s rolls are better than Bob’s, (b) Alan’s first roll is better than Bob’s, but his second is worse, (c) Alan’s first roll is worse than Bob’s, but his second is better, and (d) both of Alan’s rolls are worse than Bob’s. In 3 of the 4 cases, Alan wins, so it appears that his overall chance of winning is 3/4.

Which argument is correct?

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The first argument is correct. Alan’s chance of winning is 2/3. What’s wrong with the second argument? It’s true that there are four possible outcomes, but (a) subsumes two possibilities — Alan’s first roll might come closest to the mark, with his second roll in second place, or the reverse might be true. Similarly, case (d) covers two possible outcomes — in one, Alan’s first marble takes second place and his second comes in last, and in the other these positions are reversed. So in fact there are six possible outcomes, with (a) and (d) each twice as likely as (b) or (c), and 4 of the 6 outcomes favoring Alan. From J. Bertrand, Calcul des Probabilités, 1889, via Eugene Northrop, Riddles in Mathematics, 1975.

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