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BGonline.org Forums
OLM Sa 08/06/11
Posted By: Daniel Murphy In Response To: OLM Sa 08/06/11 (Jason Lee)
Date: Sunday, 7 August 2011, at 4:54 a.m.
I found this problem quite interesting for a day on which nothing is wrong on the internet that demands one's attention, despite the little equity involved, since we're so far behind.
(1) I thought it obvious that being so far behind we should cater to getting off in 3 rolls, which means catering to rolling 2 great doublets, and so a stacking play is called for, not a distributing play. But how to quantify this? I realized that XG's 21 raw pips means XG's EPC is somewhere between 3-roll and 4-roll equivalence. Call it 3.5, say. So XG's likely to be off in 3 or 4 rolls, highly unlikely to be off in 5, therefore we should aim to bear off in 3 rolls.
Suppose, however, that XG had 1 or 2 more checkers somewhere on the board. Then XG's EPC roll-equivalent might be over 4 rolls, or still under 4 rolls. Is that sufficient to determine whether we distribute or stack?
(2) Given that our main winning chance must, I would think, come from bearing off 9 checkers with 2 doublets, I decided to consider only the possibility of rolling such possible doublets in our next two rolls. For each play there were 3 such doublets, {6-6 5-5 4-4} or {6-6 5-5 3-3}. It's not possible here to make 2-2 or 1-1 bear off 4 checkers.
At first I assumed that we should want to maximize the number of permutations that bear off 8 checkers in 2 rolls. But I realized that's not the whole story. Bearing off 8 checkers in 2 rolls still leaves us with 2 checkers to bear off. Thus it seems very important to consider where will the two remaining checkers be?
If we play 4/3(2) to cater to {6-6 5-5 3-3} we get 7 permutations of doublets that leave 2 checkers, but the best of them leave 2 checkers on the 3 point (3 times) and the others on either the 4 and 5 points or on the 4 and 6 points.
If instead we cater to {6-6 5-5 4-4} with 2/off 5/4 and 6/5 to rebuild the 5 point -- 2/off 6/4 -- we get an interesting result: 4-4 twice leaves us with 6 checkers but the other 8 permutations all leave the same position: 2 checkers on the 3 point. So 2/off 6/4 seems to be slightly better than 2/off 4/3(2). And if anything is better than 2/off 6/4, it has to move a checker from the 3 point to the 2 point. So I look for a fourth ace to go with 3/off.
2/1? Of our possibly good permutations of doublets, 6-6 6-6, 6-6 5-5 and 5-5 6-6 give us 3 gin positions with 1 checker each on the 1 and 3 points; the other permutations leave us with 3 or more checkers.
But both 3/off 6/5 or 3/off 5/4 give us 4 permutations of doublets that leave us with 1 checker each on the 3 and 2 points -- a redoubling position as a 25:11 favorite -- and 4 permutations leaving us needing 2-2 or better as a 5:31 underdog. Great!
As far as I can tell these two moves are equal, and better than 2/off 6/4 (slightly underdog in 8 good doublet permutations) because (25-11)*2 = 28 less (5-31)*1 = -26 is positive.
At this point I came to doubt whether examining only the doublet-doublet sequences of 3 possibly good doublets was sure to find the most winning play, since some other play might create enough random wins by rolling three doublets in a row. That seems unlikely, but I can't think of a proof. However, because we have the cube, I think the doublet-doublet sequences taking 8 off and giving us an excellent redouble are crucial so I vote, if I may, twice: once each for 6/5 3/off and 5/4 3/off, whichever play needs a tiebreaker.
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