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Wastage in the one-checker model

Posted By: Maik Stiebler
Date: Thursday, 6 May 2010, at 11:06 a.m.

In Response To: Wastage in the one-checker model (Fabrice Liardet)

Ok, I'll try to explain brushing over the details: Actually I slightly misquoted the article. The more interesting question is: In the one-checker model, what is the probability that the last roll of the bear-off is a double? The answer is also 2/7, with the analogous argument that doubles make up for 2/7 of all pips, so the last pip will be due to doubles with that probability.

You can use the same argument for determining the probability of the last roll being, e.g., double sixes. Double sixes take care of (1/36)*24/(49/6) of pips (49/6=A for real backgammon), so the probability is 24/(6*49). Generally, the probability that the last roll is a N-pip roll is p(N-pip)*N/A, with p(N-pip) the probability of an N-pip roll.

Now the average wastage in case that the last roll is actually double sixes will be (24-1)/2, because the checker is equally likely to have come from points 1-24 and thus landed at points 0-(-23). Generally, the average wastage is (N-1)/2.

Thus, the average average wastage is sum_over_N(p(N-pip)*N/A*(N-1)/2)=1/(2A)*sum_over_N(p(N-pip)*N*N - p(N-pip)*N), which is already a handy formula for calculating the wastage.

For recasting in terms of V and A observe that sum_over_N(p(N-pip)*N) = A and sum_over_N(p(N-pip)*N*N) = V+A*A.

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