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?

the dice probabilities should be as follows:
1- 13/60, 2- 6/60, 3- 13/60,4- 13/60, 5-- 13/60, 6-0/60.
that would be a probability of 6/60 (1/10) of rolling a winning combination of 7.
on the second roll there is a 13/60 chance of rolling the same number combined with the 13/60 chance of rolling that number in the first place which is a 13/120 chance. on the 3rd roll there is a 13/90 chance of rolling the same combination as your initial roll(with all combined chances of the three rolls).Unfortunately for you his chances of rolling that same number grow with each roll. But fortunately so will the chances of not only the othere numbers but also the small cance of rolling a 2 increases.
I understand this isnt a completely correct answer but with a little tweaking it could become one.

Posted by mike on 2003-10-24 23:13:41