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
A start by Brian Smith)
Well, we can't make him lose on the first roll (because we can't generate 2, 3, or 12 by using one '5').
So, we wish to maximize the chances that he doesn't get a 7 or 11 on the first roll AND he gets a seven on a later roll.
So, it is necessary to allow for a '2' to show up (to make the 7 on a later roll).
I think we want to eliminate the probability of rolling a six (this way, rolling an 11 on the first throw can't occur).
So, at least at first glance, it seems that we have to identify x, the probability of rolling a 2. And then we can divide the remaining probabilities among the remaining numbers (1, 3, 4, and 5), and assign (1-x)/4 as the probability to each of those numbers.
If this is all true, then we need to solve for x such that it minimizes your adversary's chances of winning.
--- SK
some thoughts
