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): My premise is valid, Correction on outcome.
But a random number generator that guaranteed a specific number of a given value or given values would not be a random number generator. In any event, tosses of dice do not guarantee a specific number. As the number of tosses increase, the ratio of given numbers approaches the theoretical value, but in fact the differences from the expected values are in fact expected to increase, not decrease.
I think you said previously that if you carried your method out many times over the long run it would no longer work. However, in a casino, while you may be working on your umpteenth iteration of trying your method, and therefor losing expectation of its working, someone else may have just entered the casino and sat down at your table. For that person, if he were following the recommended method, you'd expect the system to be working, as he hasn't used it before. But it's the same rolls of the dice that for you are the "long run" and for him is just the beginning.
Posted by Charlie
on 2011-09-07 01:45:02