Author: Peter Fendrich
Date: 09:03:30 08/15/98
Go up one level in this thread
On August 15, 1998 at 03:43:02, Dan Homan wrote: >Bruce, > >For 2 programs with a draw chance D > Pa = chance of program A winning any one game > Pb = 1-D-Pa = chance of program B winning any one game > >For a 4-0 result in a 4 game match, either A or B has to >win all games, so the probability, P, that there is a 4-0 >result is... > >P = Pa^4 + Pb^4 > >So if the programs are equal, Pa = 0.4 and you get a 4-0 (or 0-4) >result 5% of the time. > >If there is a 4-0 result, what are the chances that the stronger >program won? Assume A is stronger. The chance that A wins any >4-0 (or 0-4) result is simply Pa^4/(Pa^4+Pb^4). > >So 4-0 results happen about 5% of the time between nearly That's true for sure but you still don't know the confidence of the result. Results of the type N-0 are very special because the guy with 0 is in the state of 'free falling' without knowing where the ground is. The 4-0 result could be the 4 first games of a 100-0 result. It could as well be the first 4 games of a 75-25 result. The difference between these two cases is huge because the 100-0 result is still a 'free falling' result. It could as well mean 1000-0... In the 75-25 result we have at least seen the ground... //Peter
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