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Calculating the best/worst scenario in a race

Posted By: Rich Munitz
Date: Friday, 8 May 2009, at 7:32 a.m.

In Response To: Calculating the best/worst scenario in a race (Alexander Hanhikoski)

Best scenario is obviously: #rolls = roundup( #crossovers / 4 )

I am thinking that the general answer to worst scenario is the following:

let

p = pipcount;

a = #checkers on the acepoint;

h = roundnearest( (p-a) / 1.5 );

Then # rolls required to bear off = roundup( (a + h) / 2 )

The idea is to compute the number of half-rolls required to bear off all of the checkers not on the ace point. For every 3 pips we have, we'll get 2 half-rolls. If there are 1 or 2 extra pips, it will round to 1 extra half-roll. Then each checker on the acepoint takes another half-roll.

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