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Thu, 26 Jun 2003 23:08:27 -0400
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Quoting Joseph Heled <address@hidden>:
> > Isn't that less than 1.5 standard deviations? Or do you mean cubeless money
> > games?
> Yes. those are cubeless money games.
In that case, the result is more statistically significant, and means a larger
difference in performance.
> I thought the STD is
> sqrt(1/(p*(1-p)*N), and assuming p ~= 0.5, this gives 0.0007
> Perhaps I am wrong.
That is roughly the standard deviation of the number of wins. However, the
standard deviation in the points per game is larger by more than a factor of 2,
since a single loss is -1 rather than 0, and there are gammons and backgammons.
> How many games would you deem necessary?
That depends on the purpose. If you want to conclude that 14 plays better than
13 on 0-ply with no cube, you probably have enough evidence to be very
confident of that. If you want to figure out how much stronger the 0-ply
evaluations are, the range of nonsurprising possibilities is still pretty large
compared with the advantage.
[Bug-gnubg], Hugh Sconyers, 2003/06/26