Author: Peter Fendrich
Date: 03:08:03 06/12/01
Go up one level in this thread
On June 11, 2001 at 17:46:13, Martin Schubert 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 > >Why do you get different probabilities for the same score? It is really different probabilities. We are not talking about scores only here but the 3 different possible outcomes: Win, Draw and Loss. The second result, from a statistical point of view is more "homogenous" and the first one is more "spread out" or unstable. This is why the probabilities are different. One way of measuring how "stable" the results are, is the variance: (Sum(X^2) - (Sum(X)^2)/N)/N The first score: 0,22 The second score: 0,16 meaning that the first result is more unreliable and will get lower probability than the second one. //Peter
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