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off the gammon bearin theorem

Posted By: Bob Koca
Date: Monday, 4 June 2007, at 7:16 p.m.

Corollary: Suppose that the following conditions are true: 1) Cubeless game 2) Only goal is to run off the gammon 3) Each of your homeboard points has a checker on it 4) There is no contact 5) You are allowed the option of changing the position by advancing an outfield checker forward and rearranging the homeboard checkers in a way that keeps at least one on every homeboard point

Then is optimal to accept the option.

Proof. Result seems obvious enough doesnt it?

Theorem: Suppose the following conditions are true:

1) cubeless game 2) only goal is to run off the gammon 3) each of your homeboard points has a checker on it 4) there is no contact

Then if you have a single die to play and it is possible to move exactly to the 6 point it is optimal to do so.

Proof. Let position A be the position resulting from moving to the 6 point. Let position B be the position that results from moving otherwise. By the corollary we know moving to the 6 point is at least as good as moving inside the homeboard or from moving a checker inside beyond the 6 point, so it suffices to consider the case where B results from moving an outfield checker forward to a more advanced point still in the outfield. The outfield positions are the same except for two checkers. Call * the checker in position Bs outfield that was moved to the 6 point in position A. Note that there is no corresponding checker in As outfield. Call # the checker in position Bs outfield that was moved forward and identify it with its unmoved partner in As positions.

Suppose that position B is played optimally from then on. Note that from the corollary we can assume there will be no moves inside the homeboard until a checker is borne off. It is possible to copy the optimal moves from position B to a strategy for playing position A as follows: 1) move exactly the same if possible. 2) If * is moved, move # instead.

Stop when * or # has reached the homeboard in position B. If the 6 point was reached exactly then the positions are exactly the same. If * of # has advanced beyond the 6 point then we have reached a position which by the corollary is at least as good for A.

Therefore position A is preferable to position B and the theorem follows.

Bob Koca

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