Author: Ricardo Gibert
Date: 22:30:20 06/11/01
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On June 11, 2001 at 15:42:36, Leen Ammeraal wrote: >On June 11, 2001 at 13:55:31, Gian-Carlo Pascutto wrote: > >>On June 11, 2001 at 13:36:21, Leen Ammeraal wrote: >> >>>Although Peter's program can in many ways be better >>>than mine, I don't see how it can be more accurate, >>>that is, as long as we regard, for example, >>>10-5-0 as equivalent to 8-3-4. As you see, I simply >>>divide the number of draws by 2 and add the result >>>to either side. >> >>It is more accurate simply because it does not have >>to do that simplification at all! >> >>10 - 5 - 0 -> 89,4% chance that A is better >>8 - 3 - 4 -> 92,7% chance >> > >I had not expected this, and I wonder if someone >can explain this difference > >>>Would that be discutable, you think? >> >>Appearently it is, as it gives different results. >> >>>If you do, which of these two results would you >>>prefer? > >Yes, my question was unclear. I meant: which would >you prefer if you were player A (the first one). >If Peter's program is right, you would prefer >8 - 3 - 4 >to >10 - 5 - 0, >while I would not have any preference. I will think >about this, but any simple explanation is welcome. > >Leen Perhaps if you considered +5 -0 =10, you would see more clearly that the inferior side virtually no chance of ever winning a match, since his ability to win any game is minimal. As you should be able to discern, +5 -0 =10 is much more convincing than +10 -5 =0, where the greater volatility is obvious.
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