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Estimating Cubeless Game Winning Chances

Posted By: Keene
Date: Friday, 29 January 2016, at 7:06 p.m.

In Response To: Estimating Cubeless Game Winning Chances (Chuck Bower)

Operating on the fly here.

White wins with any of the following scenarios:

1. Rolls a set 33 or better, everytime blue doesnt roll 66 - so 35/36 * 4/36 = 140/1296 2. Immediately Blue rolls something containing a 3 or a 4, not doubles = 18/36 = 648/1296 3. after one roll, blue has the same problem but with only one of the numbers (its more complex than that, but for the sake of estimation..), another 18/36 * ~8/36 = 144/1296

Add them all up and you get 648+144+140 = 932/1296 = ~70%. Now to add some compensation for the doublet based events not already mentioned, probably +3% in white's favor. So I estimate 73% cubeless winning chance here.

K

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