Fun Probability question (Craps) :)

Craps is a gambling game in which two dice are rolled and the numbers added. Meanwhile, players can place bets on the pass line or don't pass line.

First, the shooter makes a "come-out roll" with the intention of establishing a point. If the shooter's come-out rolle is a 2, 3 or 12 it is called "craps" (the shooter is said to "crap out"), and the round ends with players losing their pass lien bets. A come-out rolle of 7 or 11 is called a "natural," resulting ina win for pass line bets. If the point numbers 4, 5, 6, 8, 9, or 10 are rolled on teh come-out, this number becomes the "point" and the come-out roll isnow over. The shooter now continues rolling until the point number or a seven is rolled. If the shooter is successful in rolling the point number, the result is a win for the pass line. If the shooter rolls a seven (called a "seven-out"), the pass line loses.

Determine the probability of a win for the pass line bets.

Thanks all for your efforts.

The most obvious way to do this problem, is knowing the odds of rolling certain numbers. I will explain in terms of the game as opposed to the classic stats calculation.

How many different combinations are there to roll? 6^2 = 36 ie. 1&1, 1&2, 2&1, 1&3, 3&1 etc...

so, then ask yourself, what are the odds of rolling any given number? Say 7, there's 1&6, 6&1, 5&2, 2&5, 3&4, 4&3, so, there's 6 ways of rolling seven. How about eleven? only 2, 5&6 or 6&5. So, if there's 8 ways of winning on the come out roll and four ways of loosing, that means there's 36 - 12 ways to make a point. Then once you establish that point, what are the odds of crapping out? (rolling a seven)

do you see where I'm going with this?