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Antigammon puzzle SOLUTION

Posted By: Bob Koca
Date: Tuesday, 2 July 2024, at 8:27 p.m.

In Response To: Antigammon puzzle SOLUTION (Timothy Chow)

If you want one that can literally be proved here is a solution Nack sent me (and said I could post).

Opponent: 4 on ace point. You: Two on 5 point and one on 3 point.

With a 51 roll best play for both regular backgammon is 5/0 3/2. For regular get 19 good rolls instead of 17 good rolls. For anti-gammon there are swings on 64 and 54 which can be played to not bear off and have 30 good rolls instead of 26.

I won't attempt to give further justification for the first position. For the second position though let's just look at the chance of getting hit. For the 6/1* 4/1 play it is at least 1/36 just from the very first shake. For the 6/3 4/off play the following would need to happen. Opponent does not run past (6 rolls out of 36), then white does not bear off both (leaving one shot or possibly two), 26/36, then even if we make them all be double shots from the bar it is about 4/36. It is going to be significantly less than 1/54. I don't see the antigammon win chance being better in the second case after the hit since you would have 1 or 2 more checkers off and the chance of a second blot getting hit in the fray is very unlikely. The variations you win from opponent getting big doubles twice favor 6/1 4/1 as well since you are more likely to last for a second roll.

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