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 This or that? (Posted on 2015-08-19)
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.

Comments: ( Back to comment list | You must be logged in to post comments.)
 re: My take on it Comment 4 of 4 |
(In reply to My take on it by Charlie)

Charlie:

Both your explanation and mine are correct, but I like yours better.  I think yours addresses a little more directly the trick in the question.

I still think it is interesting that if he makes the median shot (50th percentile) that his chances of winning are only 1/4.  In order to have a 1/3 chance of winning, Bob needs to shoot in the 57.735th percentile, i.e. sqrt(1/3).

Steve

 Posted by Steve Herman on 2015-08-19 11:00:01

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