Craps is a 1-player dice game that is played as follows: Roll two 6-sided dice; their sum becomes your "initial" roll. If this initial roll is 2, 3, or 12, you lose. If the initial roll is 7 or 11, you win. Otherwise, keep rolling the dice until you reroll you initial number (and win) or until you roll a 7 (and lose).

You're betting that your adversary is going to lose his game of craps, which should be a favorable bet for you. But you receive an anonymous tip that he's secretly loaded one of the dice, so that it will always come up 5. This increases his chances of winning to 2/3.

Having learned of his evil deed, you're going to secretly load his other die so as to minimize his chance of winning. With what probability should you load each of the six faces? And how does that change his probability of winning?

(In reply to

re: some thoughts by Charlie)

I guess I need to learn a bit about conditional probabilities -- they seem to show up around here a lot.

Assuming Charlie's formula is correct, differentiating it and setting it to 0 results in a quadratic, with x=1/3 as a solution, which Charlie gave. The other solution is x = -1. Obviously, this makes no sense as a probability. I'm just curious if anyone has any speculation on what if anything this other root might mean.

On another note -- why not just weight the die so that 2 ALWAYS comes up. Then after about the 3rd time he rolls a 7, someone will catch on and they'll pound him to a pulp for cheating....