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BGonline.org Forums
7n+1
Posted By: Max Stockslager
Date: Friday, 8 November 2024, at 2:54 p.m.
It wasn’t clear to me why the EPC of an n-roll position was 7n+1 (why 7? why +1? why is this integer such a good approximation but not perfect?) so I tried to derive it to see if I got any insight out of the process.
If E(n) is the average number of rolls to bear off n pairs of checkers from the ace point, then E(n) = 1 + (1/6)*E(n-2) + (5/6)*E(n-1), which along with E(0) = 0 and E(1) = 1 gives E(n) = (6/49)*(7n + 1 - (-1/6)^n), as expected (I think).
But it still seems like a coincidence/miracle that this is almost-an-integer times a factor of 6/49 to perfectly cancel the 49/6 average pip value of a roll, giving an almost-an-integer EPC. The 6/49 seems to just appear out of nowhere solving the recurrence.
I understand the argument that the each checker pair should be worth about 7 effective pips: non-doubles have average pip count 7 and take off 2 checkers while doubles have average pip count 14 and take off 4 checkers, so checkers get borne off at a rate of 7 effective pips per checker-pair. So I may just have to live with that explanation. But the way the factor of 6/49 just appears when solving for E(n) makes me think there is something else going on.
More generally for k-sided dice I get E(n) = k/(k+1)^2 * ((k+1)*n + 1 - (-1/k)^n)), so again the factor of k/(k+1)^2 appears out of nowhere to perfectly cancel the average pip total of two k-sided dice, which is (k+1)^2/k if I did the math right.
Question: is it just a coincidence that the EPC of an n-roll position is so close to an integer, or is some reason this should be obvious/expected?
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