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Calculating the best/worst scenario in a race

Posted By: Rich Munitz
Date: Friday, 8 May 2009, at 8:26 p.m.

In Response To: Calculating the best/worst scenario in a race (Matt Reklaitis)

I see your point.

Of course, your proposed tweak does not quite do the job either.

Say you have 6 on the ace and 3 on the 4 point. After 3 rolls, it becomes your counterexample position of 1(3) 2(3)

What I was thinking of before proposing my simpler but wrong formula was the following:

a = # checkers on acepoint

d = # checkers on 2 pt + 2*#checkers on 3&4 pts + 3*#checkers on 5&6 pts

if a <= d then r = roundup( pipcount/3 )

if a > d then r = d + roundup( (a-d) / 2 )

The idea here is that we play an ace off the acepoint and move the deuce elsewhere. If we run out of deuces before aces, then we need to add a roll for every 2 checkers that remain on the acepoint.

Note: this only applies once the bearoff has started. Obviously we can't bear off a checker from the ace if all the checkers are not yet home.

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