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22. Luglio 2005, 15:00:30
Wil 
Argomento: Re: Inifinite backgammon
Chessmaster1000:
For every move it's 1/36.
For 2 moves to happen is (1/36)^2
For 3 moves to happen is (1/36)^3
For an infinite number of times it's zero.


When n -> infinite, p -> 0, but it never reaches 0

If p=0, that means there has to be some maximum game length, which is smaller than infinite. What might that be?
If the maximum game length is not infinite rounds, what is it then? Say any number, there is allways 1/36 probability that it goes one round further.

If we count all the possible ends when one player doesn't throw 5+5, we get an infinite amount of games.
The game is at point when endless double 5 results endless game. After that, for every double 5, there is at least 21 other possible games that are different from that particular double 5 game. Wasn't the purpose to count number of possible games? If it is possible to throw 5+5 infinite amount of rounds, there has to be at least 21 * infinite = infinite possible games.

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