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(3): My premise is valid, Correction on outcome. by Charlie)

You are correct in observing that a person entering at a different time would observe a different set and possibly come to different conclusions.  The thing though, is that in order to make reasonable assumptions about the game, the length must be sufficiently long so as to be governed by the same distribution as N->infinity, yet small enough such that there is a deviation from the normal.  Then you can make predictions on the games until the deviation is again minimized.  From the time you start making predictions to the time the deviation is minimized, your predictions (or expectations) should have a better chance of being right, however, knowing that each individual game does have equal probability within the whole of the distribution, it cannot guarantee you will come out ahead.  In fact, there is another level even above this one.  Assuming you make predictions consistently with this algorithm, and you play a number of sets L, then as L->infinity, your probability of being right versus being wrong goes back to 50/50. :)

Posted by Joshua on 2011-09-07 16:52:31