A partial solutions was Ray Kremer.  Additional complete solutions were submitted by Bradley Mathematics faculty member Ollie Nanyes, as well as from Yan Fridman, Al Zimmermann, Ivan Lisac, Robert T McQuaid, Philippe Fondanaiche, Burkart Venzke, Jeff Downin, Cyril Terakopiantz, Tim Kelley. An additional partial solution was received from Maxim Ovsjanikov.

Here is Ben Brown's solution:

The possible ways the game could end are the following.
 

The only way that Player 1 can win is if he rolls a winning number on the 1st, 3rd, 5th, ... roll.
The probability that Player 1 rolls a winning number is 2/N, and so the probability that Player 1 rolls a losing number is 1 - 2/N = (N-3)/N.  The probability that Player 2 rolls a losing number is 1 - 3/N = (N-3)/N.

Therefore the probability of Player 1 winning the game is

For Player 1 to have the upper hand (or at least an even chance) 2N/(5N-6) must be no larger than 1/2, which occurs only for N = 2,3,4,5,6.  As several solvers noticed, there is no need to restrict N as strictly as I did in the problem.  Indeed, if N = 2, Player 1 wins on the first roll of the die.

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