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?

Source: J. Bertrand, Calcul des Probabilités, 1889, via Eugene Northrop, Riddles in Mathematics, 1975.

Well, the first argument is correct. Bob wins 1/3 of the time.

The problem with the 2nd argument is that the 4 probabilities only have probability 1/4 if Bob's makes the median shot, in which case he will win with probability 1/4. If he makes better than the median shot, he will win with probability greater than 1/4, and if he makes less than the median shot he will win with probability less than 1/4.

If his shot is in the xth percentile of all possible shots (where x = 100% is a sure winner), then he wins with probability x^2. If you integrate x^2 from 0 to 1, you get 1/3.