My Math Forum Predicting the Premier League Table

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 August 6th, 2015, 12:03 PM #2 Senior Member   Joined: Oct 2013 From: New York, USA Posts: 673 Thanks: 88 20! = 2,432,902,008,176,640,000, which is much great than 25,000,000, but not all outcomes are equally likely. Another way is estimating how many positions each club could realistically finish in. For example, if eight clubs have 4 realistic positions, eight clubs have 7 realistic positions, and four clubs have 10 realistic positions, there are 3,778,019,983,360,000 possibilities. This ignores the fact that the positions are not independent events because each position has exactly one club. Your "1/X^20 = 25000000" is wrong. If you had a 50% chance at getting each club in the correct position, the number is 2^20 = 1,048,576, which again makes the false (but maybe close to true) assumption of independent events. http://www.soccermetrics.net/league-...premier-league made predictions for the 2014-2015 season and here is how many spots off it was for each club: 1, 1, 0, 0, 1, 1, 4, 4, 1, 10, 4, 3, 6, 2, 1, 2, 0, 8, 6, and 1, for an average gap of 2.8 positions. If you assume that being 4 spots off on a club means that it had 9 realistic positions (it's actual position and 4 off in each direction), then take 2x + 1 for those 20 numbers and multiply them to get 16,031,714,926,500. This ignores the fact that there are boundaries on each end. For example, a club predicted to finish 18th that actually finished 12th was 6 spots off, but it would have been impossible for the club to finish more than 2 spots worse than the prediction because there are 20 clubs. For the club that the prediction was 10 off for, 2*10 + 1 = 21, but I used 20 instead because there are 20 positions. So what I did for each of the 20 numbers was min(20, 2x + 1). For any bet with ratio odds that high, I wouldn't work too hard calculating mathematical odds. You could bet a small amount hoping to get rich if you win, but I wouldn't recommend betting much. It's somewhat like winning the lottery.

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