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15. Januaro 2006, 16:03:04
Chicago Bulls 
Modifita de Chicago Bulls (15. Januaro 2006, 16:06:56)
It depends on the length your games had (match length and game length).
For example for 50 1-point/single games of a short 20 moves per game(that means i've taken the best case so your case would have even less probability) the probability of not having a single double is:
(5/6)^(50·20) = 6.5 E-78 %
That means a 100% FLAW on dice generator!!!!!

But i guess you are overreacting and that "50" and "NO double at all" is not correct....


PS: To give a feeling of what 6.5 E-78 % is:
6.5 E-78 % = 0.000000000....another 65 zeros....00065 %

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