The game of craps is played by rolling a pair of dice. If the total comes out to 7 or 11, the shooter wins immediately. If it comes out to 2, 3, or 12, the shooter loses immediately. If any other total shows on the first roll, the player continues to roll until either his original total comes up again, in which case he wins, or a 7 comes up, in which case he loses.

What is the probability the shooter will win?

For two 6-sided dice, there are 36 permutations. The number of permutations that yield each number is as follows: 2 or 12: 1, 3 or 11: 2, 4 or 10: 3, 5 or 9: 4, 6 or 8: 5, and for 7: 6. Thus the odds are 8/36 that one would roll a 7 or 11 (wins immediately), and 4/36 that one would roll 2, 3, or 12 (loses immediately).

If any of the six other numbers come up, 2, 3, and 12 are harmless, and in fact any subsequent rolls other than 7 and the desired number can be ignored. For 4 or 10, the odds of winning are 3 to 6 against, or 3/9. Similarly, the odds for 5 or 9 are 4/10, and the odds for 6 or 8 are 5/11.

Now the total odds of winning can be calculated by multiplying the odds of rolling a number with the odds that it will win, and summing the lot. So for 2 through 12, the odds of winning are

1/36*0 + 2/36*0 +3/36*3/9 + 4/36*4/10 + 5/36*5/11 + 6/36*1 + 5/36*5/11 + 4/36*4/10 + 3/36*3/9 + 2/36*1 + 1/36*0
= 0.492929

These odds are very nearly 50%, which means that for any one series of rolls that ends in success or failure, one has a nearly even chance of coming out on top. However, a lengthy stay at the craps tables will wipe out all but the luckiest shooters. Why? Because if the cold streak comes first, the player has to have enough money to get through it to be there for the hot streak (this is true for all games against the house). If you want good odds in Vegas, buy some real estate there.

Posted by Bryan on 2003-04-18 17:59:32