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 My premise is valid, Correction on outcome. by Joshua)

"But you must also eat some.Notice that I have proven that should you be able to know that a game should end before the 12th roll, that you do have better odds overall.  Therefore, if you were to use the counting system and chose when to play a round of rolls (to their completion), then you can improve your odds."

Yes, you have "proven that should you be able to know that a game should end before the 12th roll, that you do have better odds overall", which I never denied. I do continue to deny that you can know that a game is more likely than normal to end before the 12th roll.  Just as in coin tosses of heads-heads-heads, where the next toss is still 50-50, just as before any history happened, in the sequence of games as long-long-long, the next game is no more likely to be short than it was before any such previous history.

In a counting system there has to be something to count that changes.  In 21 (Blackjack), a single deck starts out with 16 tens, and if 3 have been used, there are now 13, or 13/16 of the normal value. In craps there's no changeable value to count: there are always two dice, with the numbers 1-6 appearing once each on the faces.

Edited on September 6, 2011, 10:46 am

Posted by Charlie on 2011-09-06 10:31:03