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?

(In reply to re: My premise is valid, Correction on outcome. by Charlie)

Let me try and explain this in a different manner.

Say that you have a random number generator that guarantees that every number comes up an equal number of times if N numbers are drawn, where N is sufficiently large.  Let's say for simplicity that the numbers being generated are the numbers from 1 to 10, and that N=20 is sufficiently large.

This setup is similar to playing an infinite number of games of craps, as given an infinite number of games, the odds that each length of the game has an equal chance of coming up.

While the number generator problem, we will be enforcing the rule, in craps, the laws of probability govern the distribution.

Then in the number generator game, if you have seen 10 games and all the numbers have come up except for 2 (and 1 has come up twice). Then for the next game, while any number except 1 could come up, as only 1 has been doubled, the likelyhood of 2 coming up is better, and the odds of 1 coming up are worse.  In fact, while the numbers 3-10 have equal odds (1/10) of coming up, 1 has a 0/10 chance and 2 has a 2/10 chance.

Now, let the laws of probability govern the game, meaning that all numbers are possible on any given step, but that as N->infinity all numbers come up equally.

That being said, looking at an individual roll, you could not say what the likelyhood of getting a given number is, but looking at a sufficiently large set, you could say one event is more likely than another.

In order to verify this, you could take a sample and determine the deviation from the normal, and knowing that as N->infinity that the distribution should be normal, then the expected value for the next event is altered. Counting with a plus-minus system is a shorthand way of coming to the same result. Yes, looking at a single game, the probability is equal, but looking at a sufficiently large set, the probability is different.  The general math behind this deals with limits, power series, and the PGF (Probability Generating Function).  The difference here is that to get the overall probability for an infinite set of games, you sum from x=0 to infinity, where as we are sampling a subset x=0 to x=N with N being sufficiently large to governed by the same distribution.  This does not guarantee the next run will be any given length, but it does mean that it is more likely to be a given length.

BTW--I don't gamble with money, but I have used this with a roulette wheel at a youth group entertainment event with quite a bit of success.  While not every bet placed won, the majority of them did.  (I kept the game simple by chosing only odds and evens. There are too many probabilities in roulette to keep track of more without being trained, or a being a savant.)

Edited on September 6, 2011, 7:52 pm

Posted by Joshua on 2011-09-06 19:43:50