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Calculating the best/worst scenario in a race

Posted By: Bob Koca
Date: Saturday, 9 May 2009, at 3:10 a.m.

In Response To: Calculating the best/worst scenario in a race (Rich Munitz)

Here is my attempt:

Having the last several checkers on the ace point can lead to extra rolls. Define W = wastage due to last checkers being born off all being on the ace point.

If the number of checkers remaining on the ace point A is even then W is A/2 since every 21 from then on wastes a pip. If A is odd then the very last roll wastes 2 pips. W in that case is 1 + (A+1)/2.

How many will there be on the ace point after everything else has been borne off? For every checker on an even point we can advance it 2 and bearoff a checker from the ace. For every checker on the 5 point we can advance it forward 2 and use the ace to bear off from the ace point. (Note that a checker on the 3 creates an extra checker momentarily on the ace so the ace point stack is not reduced). It is possible of course that the ace point stack dissapears.

For positions with all checkers in the homeboard define

P = pipcount

a = number of checkers on the acepoint.

b = (pipcount of the checkers on even points)/2

c = number of checkers on the five point.

W = 0 IF a-(b+c) is <=0

W = [a-(b+c)]/2 IF a-(b+c) is >0 and even

W = 1 + [a+1-(b+c)]/2 IF a -(b+c) is > 0 and odd

Then max number of rolls to bearoff = Roundup[(P+W)/3]

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