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Variation

Posted By: Taper_Mike
Date: Wednesday, 22 January 2014, at 10:17 a.m.

In Response To: Variation (Paul Weaver)

In the original position, White needs only 7 crossovers to beat the gammon. Meanwhile, Blue will need either 11 or 12 to win, depending how many checkers he bears off on this turn. If Blue rips off two, he will leave a shot on the following turn when he rolls a non-doublet containing two high numbers: 65, 64, and 54. If, instead, Blue plays 6/3 5/off now, ripping only one checker off, then his blotting numbers will be: 61, 51, and 41.

On the surface, the risk seem to be the same either way. The critical difference, however, is the double jeopardy. If Blue leaves three checkers on the 5pt now, then he will often get down to two checkers there later on. At that point, he will have to risk 61, 51, and 41. Therefore, it is best to face those numbers now, and not risk the 65, 64, and 54 in addition.

Paul's variation makes the decision much closer. I can't say for sure which play is right, but with 9 crossovers vs. either 11 or 12, I would still rip only one checker off.

Mike

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