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21. November 2011, 14:37:36

playBunny

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Subject: Re: As luck would have it.

grenv: 9 rolls

Yes, and it can be done with nine 6-6s, which wastes pips, or nine rolls which bear off exactly.

For example, if the opponent starts and makes obligingly awful moves (in particular, abandoning their midpoint), the player can roll 6-6, 6-6, 2-2, 6-6, 1-1, 6-5, 6-6, 6-6, 6-6 to finish having rolled exactly 167.

Avoiding errors on the part of the player would need an extra move and an opponent who obligingly misses any necessary blots caused by bearing off with doubles. It can be done as a blitz with 6-6, 5-5, 4-4, 6-6, 6-6, 1-1, 6-6, 5-5, 3-3, 1-1 if the opponent starts with 5-2 and splits the back men.

21. November 2011, 15:31:15

grenv

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Subject: Re: As luck would have it.

playBunny: Right, the back 2 need to land on the opponents midpoint. this is as likely as 2 move fools mate in chess, but still the answer to the theoretical question :)

22. November 2011, 03:31:37

Walter Montego

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Subject: Re: As luck would have it.

grenv: And the theoretical question is what I asked. You both say nine is the least amount of turns even if we give the opponent any roll we want and have him move as we please? I think I will work on this myself, but nine seems very fast. I hadn't thought of it as grenv did, just figuring out each piece from where it starts. This makes sense as you can't use the roll just by the count as the number of pips needed may not divide evenly, 7 and 13 being good examples of this. Is my fifteen move game fairly exceptional, especially considering my opponent wasn't trying to help me?

Which point is what you are calling the midpoint? Is that the point in the outer table that starts with five pieces?