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(2): True Solution (Flaw in common answer.) by Joshua)
The three-doors situation is different. It's Monte's choice as to which door to show you that actually provides information. In the craps case there is no such information-giving.
The dice do not change from one game to the next; they are physically the same, unlike the situation in the card game 21, where the deck gets depleted of certain cards. Each craps game is independent, and consists of two fair dice with numbers 1-6 on each face.
Saying that it has been "proven mathematically" does not make it so. It is proven physically that the dice remain the same from game to game.
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Posted by Charlie
on 2011-09-04 10:50:11 |