> Your 2 2 2 3 3 3 distribution is good, given a 15 checker > position with a pip count of 48. (In fact, subject to these two > constraints, I believe it is optimal, but I'm going on memory.)
It's close, but 1 2 3 4 3 2 is the optimal distribution. As pointed out by several, we can't control the pipcount. And then our initial posting with minimal pip-waste for 1-15 checkers left is not of much value, except for the 000357 position which should be the aim in the bear-in. As a consequence Eirik made a table of optimal boards for all possible combinations of pips and checkers left. Interested can get a postscript-file by emailing him: th...@stud.unit.no
Here's the optimal board for each pipcount, no matter how many checkers left. Interestingly, fewest checkers left is often not optimal: The 48-pip board 000008 (8 men at 6point) wastes 1.49 pips more than 000144.
You can always transform pipwaste to expected rolls left: Just add the pipcount, and divide it with 8.1667. Enjoy.. 8-)
