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

some thoughts...and a question by Brian Wainscott)

Actually, I don't think Charlie is correct in that it is *conditional* probability. (Although I think his calculation is correct.)

The remainder of his calculation, *((1-x)/4)/(x+(1-x)/4)*, is not conditional... it is simply the likelihood of getting 1, 3, 4, or 5 *(1-x)/4)* divided by (the likelihood of getting a 2 *(x)* plus the likelihood of getting the same number again *(1-x)/4* ).

Charlie, you are welcome to correct me if I misunderstood what you wrote.

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As for what conditional probability is... it is the probability of an even happening given that another event has already occurred.

*(Let's not deal with the trivial case when the former event has no effect on the latter case, such as... what is the likelihood of heads coming up on a fair coin toss, given that it previously came up tails).*

But, what's the likelihood that a blue die and a red die add up to a number greater than 8, *given that the blue die has already been rolled and shows six*. Well... normally, the odds that two (normal) dice add up to a number greater than 8 is *5/18*.

But since we know the blue die is 6, we can calculate the likelihood of them adding up to greater than 8 as *2/3*.

More formally, this probability can be written as:

P(B|A) (which is read as "the probability of event B given event A")

P(B|A) = P(A and B) / P(A)

So, to apply this to our die problem:

Event B = P(red die and blue die totalling greater than 8)

Event A = P(blue die shows 6)

P(A and B) is 1/9 (out of 36 possible red/blue combinations, only {3,6} {4,6} {5,6} and {6,6} will work)

and of course P(A) = 1/6

so... P(B|A) = (1/9) / (1/6) = 2/3.

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And yes, in this relatively trivial example, we could have calculated the probability directly rather than using conditional probability.

But I hope this helps.

*Edited on ***October 22, 2003, 8:24 pm**