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BGonline.org Forums
Probability of rolling at least 2 double 6s in 3 rolls
Posted By: leobueno
Date: Friday, 9 May 2014, at 2:51 p.m.
I am not sure of the method to use to calculate the subject probability (P), so I tried a decision tree and came up with:
P = 1+35+35+35/1+35+35+1225+35+1225+1225+42876 =
106/46657 = 0.00227 = 1/440.2
Please correct or confirm my calculation. Thanks.
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