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BGonline.org Forums
Simple Probability Problem
Posted By: Jake Jacobs In Response To: Simple Probability Problem (Jake Jacobs)
Date: Tuesday, 25 December 2012, at 4:12 a.m.
There are a couple of interesting wrinkles.
1) Match length does matter, in that if the matches are long enough X will have a probability of winning approaching 100%.
So let's suppose the matches are all to 25 points. Then X would win 90%. This isn't far from a 400-point rating difference. X would be a 76-24 favorite with a 200-point difference, so he would be roughly 200-points over "the field," while Y would be about a 200-point dog under the field, and X would be 400 points over Y. Actually, at 400 points X would be 90.9%, or 10-to-1.
Sam Pottle provides a neat shortcut, just multiply 3-to-1 by 3-to-1 to get 9-to-1. And this should work for other differences as well, e.g. X is 60% over the field, while Y is 10% under the field, so X is 8.5-to-1 over Y.
But now we come to another wrinkle. This came to me from a friend, Bill Benter, who has been involved in a contest to model chess results. He asks: What if the population of chess players is comprised by these 100 players with these ratings:
25 = 400 25 = 1200 25 = 2000 25 = 2800
Introduce Player X, rated 2400, and Player Y, rated 800, and let them play everyone a whole bunch of games. Assuming everyone's ratings were accurate to begin with, and no one's game improves or worsens, Player X will win 75%, Player Y will win 25%, but Player X is 10,000-to-1 over Player Y.
2) Distribution matters.
I did ask Sam if he knew a way to incorporate this into his neat shortcut, but Santa brought him better things to do this Christmas.
Which reminds me: Merry Christmas everyone! I'll see you all next week at the Thai Open. :-)
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