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BGonline.org Forums
A comment about the takepoint formula
Posted By: Timothy Chow In Response To: World's incorrect 14-away/8-away pass: Analysis (Timothy Chow)
Date: Wednesday, 14 April 2010, at 2:18 a.m.
I wrote:
Mathematically, you can take if
wins + (effective gammon wins)*(gammon price) – (effective gammon losses)*(gammon price)
equals or exceeds your raw takepoint. (Here wins include redouble/drop, and I'm assuming backgammons and re-recubes can be ignored.)
So far I've just been asserting this fact without justifying it. It occurred to me that it might help if I say a word or two about why the above formula is true.
Suppose that there is no cube. Then I assume people are comfortable with the claim that you can take if
wins + (gammon wins)*(price of gammon win) – (gammon losses)*(price of gammon loss)
exceeds your raw takepoint (ignoring backgammons). Now, let's introduce the cube. Assume that the cube gets turned at most once more, and that gammons after the cube is turned are negligible. Then the effect of the cube is just to give you a new way to win 1 point (redouble/drop), a new way to win 2 points (redouble/take/win single), and a new way to lose 2 points (redouble/take/lose single). As far as the math is concerned, it doesn't care how you won or lost 2 points (whether the old-fashioned way with a gammon or the new way with the cube); all it cares is how often you win or lose 2 points. So to take the cube into account, just lump all your 1-point and 2-point wins/losses together, and use the exact same formula.
One can derive the above formula more formally by manipulating equations, but I don't think that would be illuminating; the above verbal description is the real explanation of what's going on, in my opinion.
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