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OLM 20141229A The Prime Factors

Posted By: Taper_Mike
Date: Monday, 29 December 2014, at 10:19 p.m.

In Response To: OLM 20141229A The Prime Factors (Jason Lee)

On the assumption that the Dilly Builders roll a doublet that rips four checkers off, let’s count the ways we can bear a checker off after three candidate plays.

Candidate 1 – 7/5 3/2
If we play 7/5 3/2 (or, equivalently, 7/6 4/2), then we will have a gap on the 1pt and a checker on the 11pt. 66, 65, 64, 63, 62, 55, 54, 53, 52, 44, 33, and 22 will bear off a checker on the following turn. That’s 19 rolls. Of the remaining rolls, all except 21 would bring the outside checker at least as far as the 7pt. From there, we would be guaranteed to beat the gammon on the subesequent turn. Were we to roll 21, however, any non-doublet ace thereafter would fail to bear off on the second turn.

Candidate 2 – 7/6 3/1
Can we eliminate the 21 sequence by putting a checker on the ace point now? Take a look at 7/6 3/1. After this move, 66, 65, 64, 63, 61, 55, 54, 53, 51, 44, and 33 rip off a checker on the following turn. That’s only 18 rolls compared the 19 rolls we had before. All the remaining rolls advance the rear checker at least as far as the 8pt. From there, we can always bear off on the second turn.

Is is better to sacrifice 1 immediate number in order to avoid the 21 sequence? The only time it matters is when the Dillies roll a doublet that rips off four checkers on either of the next two turns. They could do that with 66, 55, or 44 on the next turn. If they roll 65, 64, 54, or 22 on the first turn, they would add 33 to that list for the second turn. If they roll 33 on the first turn, they would add both 33 and 22 to the list.

Candidate 3 – 11/9 7/6
If, instead, we play O = 11/9 7/6 on this turn, then we will have gaps on the 1pt and 2pt, with our outside checker on the 9pt. 66, 65, 64, 63, 55, 54, 53, 44, 43, 33, and 22 will bear off a checker on the subsequent turn. That’s 17 rolls. Any other roll will bring the outside checker home, so we are guaranteed to bear a checker off in two rolls.

Dilly Builders’
Roll on First Turn 		Odds That Dilly Builders
Bear Off Four Checkers
on Second Turn
33 5/36
65, 64, 54, 22 4/36
Any other roll except
66, 55, 44 		3/36

P( Dillies win in 2 rolls ) = (3/36)*(36/36) + (1/36)*(5/36) + (7/36)*(4/36) + (25/36)*(3/36) = 216/1296 = 1/6

Candidate Move Odds of Bearing Off
on Next Roll 		Odds of Bearing Off
Our First Checker on Subsequent Roll 		Odds of Beating the Gammon
7/5 3/2 19/36 (15/36)*(36/36) + (2/36)*(26/36) = 592/1296 (1/6)*(19/36) + (5/6)*(19/36 + 592/1296) = 90.8%
7/6 3/1 18/36 (18/36) * (36/36) = 648/1296 (1/6)*(18/36) + (5/6)*(18/36 + 648/1296) = 91.7%
11/9 7/6 17/36 (19/36) * (36/36) = 684/1296 (1/6)*(17/36) + (5/6)*(17/36 + 684/1296) = 91.2%

Provided, then, that I have not made a math error (something that seems increasingly far-fetched the more I work on this), candidate 2 should be best. Since that dovetails with my intuition, hopefully it will be right even if my arithmetic is not.

7/6, 3/1

Mike

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