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: True Solution (Flaw in common answer.) by Charlie)

First, no I'm not assuming that you can quit.  You must play the full hand.

However, the percentage changes for each game.  In the long run, if you played every game, then yes, it does approach 50/50 (not 49.2929).

The problem here is in two assumptions that don't hold.  The first assumption is that each roll is independent.  The fact that the first roll imparts better odds to the shooter actually proves that this does not hold.  In order to compensate for this benefit, it actually takes 12 rolls to return the odds of any game back to 50/50.

The second assumption is that every game has equal odds of being any given length of runs.  This holds true if you set N (the number of games played) sufficiently large.  However, if you use a small enough sample of games, then the odds for the length of game N+1 are actually not equal to each other.

The problem is knowledge.  By observing the games that come before, you gain knowledge as to the deviation from normal.  Using this deviation, you can pick and choose a time to enter and leave the table such that your odds are actually improved at coming ahead, but only so long as N is sufficiently small.

A similar problem is the two lions, a car, and three doors, where the host asks you to pick a door.  Upon picking a door, the host reveals one of the other doors to hold a lion.  The host then asks you if you would like to switch doors.  First, the host can always pick a door, as he knows the position of each, and even if you pick a lion, he can reveal the other lion.

Now, if you say the odds are now 50/50, you would be incorrect, as the odds of the original door holding a lion have not changed.  You still have a 33% chance of having picked the correct door.  Actually, the odds are now 66% in favor of the other door.  This is due to the knowledge of the host affecting the probablistic outcome.  You would be wise to switch doors, and thanks for the extra 33% odds, host.

The same thing is happening here in the craps example.  If you observe a sample of games, you are gaining knowledge as to the probabilities that the next set of games is actually more in your favor, or whether it is in more favor of the house.  This is why gambling establishments hate counting.  Not just for cards, but any form of counting.  It changes the odds.  It is proven mathematically.

Posted by Joshua on 2011-09-03 19:55:30