A "friend" offers to play the following game: you throw a die, and he throws two dice. If both his dice are either higher or lower than yours, he wins; otherwise, you win.

First, you reason: out of three dice, one will always be the "middle" one, and only one out of three times it will be mine, so my odds are just 1/3 -- I shouldn't play.

After a while, you realize that you forgot about duplicate numbers. About 50% of the time, all three dice will be different, and then you have 1/3 chance of winning. But on the other 50%, you assuredly win, so the game stands 2/3 in your favor.

It's clear that BOTH lines of reasoning cannot be right, if any. Should you play, or shouldn't you?

Note: you can solve this mathemathically, or you can use "lateral thinking"; can you find both ways?

We count the ways that a,b,c in [1,6]^3 have (b>a and c>a) or (b<a and c<a). It is not hard to see that 566; 466,465,456,455;...;166,165,164,163,162,...,126,125,124,123,122 comprise 5x5+4x4+3x3+2x2+1x1=55 ways and there are another 55 for the <a cases.  Thus 110 of the 216 cases are favorable to the friend, or 50.9%. If the dice are n-sided, the probability of success for the friend is (n-1)*n*(2n-1)/3*n^3 and n=6 is the smallest n that favors the friend.

Posted by Richard on 2004-07-08 14:59:23