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 That's a Load of Craps! (Posted on 2003-10-22)
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?

 No Solution Yet Submitted by DJ Rating: 4.4167 (12 votes)

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 re: some thoughts | Comment 3 of 12 |
(In reply to some thoughts by SilverKnight)

If we make x be the probability that we assign to rolling a 2 on the second die, make the probability of a 6 zero, and apportion the probability of any of the other four possibilities equally (so as to minimize the likelihood of getting the same point again), then the probability of the shooter's winning, which is to be minimized, is:

x+(1-x)*((1-x)/4)/(x+(1-x)/4)

whose terms represent the probability, x, of winning on the first throw, and 1-x of not winning on the first throw, (then multiply by the conditional probability of getting whatever point comes up before getting a seven).

Putting this into Excel, as =B2+(1-B2)*((1-B2)/4)/(B2+(1-B2)/4), where an initial guess of .2 or .3 (or anything reasonable) is placed into cell B2, and then using Solver to minimize this formula by changing B2, we get a minimum when x is 1/3. At that point the probability of a win is 5/9 (that is, the evaluation of this formula).
 Posted by Charlie on 2003-10-22 13:40:12

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