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BGonline.org Forums
How to count?
Posted By: Daniel Murphy In Response To: How to count? (Chuck Bower)
Date: Wednesday, 9 December 2009, at 9:44 p.m.
To what the other advisors have said, I would add that if you must count:
{6,2}
11 - 2 + 4 = 13
Eleven being the number of rolls that contain at least one 6, because 11 = 36 - 5², two the number of rolls that contain one 6 and one 1, and four the number of other working doublets.
{5,3}
16 - 2 = 14.
16 being equal to 4², since any combination of four different numbers (3s, 4s, 5s and 6s) bear off, except for 4-3, which explains the -2.
If you have this many good numbers then you have this many bad numbers and the probability that a good number appears on at least one of your dice is and the probability that a good number appears on both your dice is 1 5 36 - 5² = 11/36 1² =1/36 2 4 36 - 4² = 20/36 2² =4/36 3 3 36 - 3² = 27/36 3² =9/36 4 2 36 - 2² = 32/36 4² =16/36 5 1 36 - 1² = 35/36 5² =25/36 Note that when memory fails you can use the appropriate equation to quickly calculate the probability of entering one or two checkers, or of hitting or covering a blot (with a direct number), in addition to calculating bear-off probabilities. You can easily memorize, and frequently use, the sequence 11, 20, 27, 32, 35, 36.
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