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August 5th, 2015, 01:35 PM   #1
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Predicting the Premier League Table

Hi everyone,

I made an account here just to ask this question...

Okay so, this morning my friend texted me telling me about a bet he had taken. William Hill are offering 25,000,000/1 for anyone who can predict the exact order of the Barclays Premier League table this year. He asked me if this was a good bet...

So I told him that because there are 20 teams, the are 20! (about 2.4 x 10^18) different possibilities so it's a pretty bad bet! But obviously a lot of these possibilities are pretty unrealistic... Chelsea are not gonna get relegated for example. So I tried to come up with a more realistic answer.

First I divided the table into 4 groups of 5, using the assumption that the top 5 will finish in some random order, but they will all finish in that same top group (obviously this isn't true, but I was simplifying as much as possible). I did the same for the remaining 3 groups. So then I was left with 5! possibilities for the top group, 5! for the second, 5! for the third, 5! for the fourth. Therefore using this system, there are 5! x 5! x 5! x 5! which is about 2 x 10^8 possibilities. Is the maths here correct? I know I've over simplified, but in principle is the reasoning sound?

And finally, I wanted to work out how sure you would have to be with each guess to make it a "fair" bet. So for example, if I was 50% sure Chelsea would finish top, 50% Arsenal would finish second and so on, would this be fair? Well for that ended up with the equation:

1/X^20 = 25000000

This gives X as around 42%. Is my reasoning sound here? I'm pretty sure it's not, but I'm not sure why...

Anyways, hope someone here is as sad as me and has some time to think about this

Any thoughts/advice would be great!

Thanks,
Matt

Last edited by skipjack; August 5th, 2015 at 05:04 PM.
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August 6th, 2015, 12:03 PM   #2
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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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