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BGonline.org Forums
Calculating the best/worst scenario in a race
Posted By: Alexander Hanhikoski
Date: Thursday, 7 May 2009, at 6:22 p.m.
Hi guys,
I am trying to come up with a way to calculate the minimum number of best (66) and worst rolls (21) needed to bearoff everything in a race.
Calculating how many double sixes is needed is trivial, since you can move the chequers unambiguously (it's chequer's slot / 6, rounded up, for every chequer on board). Adding all of these to total number of sixes needed and dividing this by 4 and rounding up again gives us the answer - the number of rolls needed.
For example, we have two chequers on 14 and one on the 7 point... (2*roundup(14/6) = 6 + roundup(7/6) = 2) = 8... roundup(8/4) = 2, so we need 2 rolls to bearoff everything.
Calculating the same with 21 isn't as simple I first thought it would be, and I'm not sure if there even is a way to calculate the optimal moves without heuristics. Basicly, I'm trying to calculate if there is any chance of winning left, if not, then try to figure out the resign value.
So, how should we play if we know all of our consequent rolls will be 21? I have developed some rules that seem to apply, but not necessary. Most of them are based on a greedy approach (ie. always bearoff from 1 and/or 2 point if possible, then from 3 point, etc.). In race scenario, we don't even need to take opponent chequers into account.
Is there perhaps some magic formula out there to calculate this? Thanks in advance, any idea is welcome and appreciated.
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