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Denver race 8 cube

Posted By: bob koca
Date: Monday, 29 September 2008, at 6:49 a.m.

In Response To: Denver race 8 cube (neilkaz)

This is small enough to just brute force a win% to within a %. Give yourself 3 wins for an initial double 4's or better. Now you need to see how often he takes 3+ rolls and you take exactly 2. Stepan taking 4 and you 3 is also possible but clearly negligible.

Stepan taking 3+ is easy enough. He needs to get 22 or better in either of the next 2 shakes to be off in 2. Two rolls to get 5 numbers gives 10 minus overcount of slightly less than 1 for the overlap. So off in two almost 9 times in 36 games. He will take 3 rolls just under 3/4th of the time.

Now for you go roll by roll and keep a running count. This one is surprisingly easy as it breaks into cases very simply and naturally. Anything with a 1 or 2 will require double 3's or better (except 22 which makes 22 work also) so approximate as 20 X 4 = 80 so far. The 3 big doubles give no chance of being off in exactly 2. Double 3's give 36 chances. So up to a total of 116. Any of the other 11 rolls gives 17. So add 11 x 17 = 187 to the 116 from before giving 303 total. I would use 3/4 of 300 and realize this slightly undercounts. 3/4 x (300) = 3 X 75. Dividing by 36 is same as 75/12 giving 6.25 games out of 36. So you win a little more than 6.25 + 3 = 9.25 games cubeless.

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