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    feegs's Avatar
    feegs Posts: 7, Reputation: 1
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    #1

    Nov 7, 2011, 08:52 AM
    Odds in a spreadsheet
    My friend runs a poker league, the top 8 finishes get points (250, 200, 150, 110, 90, 70, 50, 30 & 10 points for playing) It runs over 13 weeks. What I want to do is create odds for each person on winning the league but I don't know how, any help?

    Steve
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    ebaines Posts: 12,131, Reputation: 1307
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    #2

    Nov 7, 2011, 09:43 AM
    If there are eight people in the league then prior to the first week's game the odds of any one of them winning is one out of 8. But I think maybe you want to ask a different question -namely, if you know each person's score after, say, 4 weeks can you determine each person's odds of having the most points after week 13? That's a more difficult question. It can be estimated by calculating the mean and standard deviation of how many points a person may gain or lose against the leading person, and calculating the 'z score' for the probability of each person overtaking the leader in 'n' games. If that's what you mean, post back and I can help you through the math. By the way - you said the top 8 finishers get points, but you listed points for 9 finishers.
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    #3

    Nov 7, 2011, 10:05 AM
    There's more than 8 people per week, that's just the final table. On average there's about 20 players per week. Each player gets 10 points just for playing but if they finish within the points they don't get the 10 points they get the points where they finish. I think your we're right with the 'z score' and 'n games' thing though. I've got a spreadsheet already for this series but there nearly finished so I wanted to set a new one up for when they start the new series.
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    ebaines Posts: 12,131, Reputation: 1307
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    #4

    Nov 7, 2011, 12:03 PM
    The odds are going to depend in part on how many players there are, and on an estimate of the typical number of players each week. Also - with the technique I proposed there are a couple of inherent assumptions that have to be made:

    1. That all players are equally skilled at the game, and thus no one player is more likely to win more points any week than any other.
    2. That the probability of a player shows up to play in any given week is independent of his current standing. For example, a player in first place going into the final week is just as likely to not show up as the player in last place.

    I think neither of these assumptions hold water, but you should at least be able to build a model that is a reasonable estimate of the odds. Good luck with it!
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    feegs Posts: 7, Reputation: 1
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    #5

    Nov 7, 2011, 12:13 PM
    Thanks for that. How would I go about building a model?
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    ebaines Posts: 12,131, Reputation: 1307
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    #6

    Nov 7, 2011, 01:09 PM
    I would start by looking at the expected number of points any player can expect to gain (or lose) to an opponent each week. The mean of this is 0 (no one player is expected to do better on average than any other). But the standard deviation is fairly substantial - about 114 by my calculations. Thereore the odds of gaining 114 points on an opposing player in a round of play is about 32%.

    Mathematically what you have is that the odds of gaining x points is equal to CDF, where CDF = cumulative distribution function. In an Excel spreadsheet this can be found using the "nomdist" function.

    If there are N weeks left to play the standard deviation is increased by square root of N. So the probability of the player who is in the lead by X points staying in the lead over N weeks is


    Let's try some numbers: if player A is currently ahead of Player B by 200 points, then the odds of Player A staying ahead of B is as follows: for one week: CDF(200/114) = 96%, for two weeks: CDF(200/(114*sqrt(2)) 80%, for 5 weeks CDF(200/(114*sqrt(5)2)78%, and for 19 weeks 65%. To find the odds of Player A beating out all other players you need to find these probabilities for A versus every other player, and multiply them all together. For example: if A is ahead of players B, C, and D by 200 points each with 5 weeks to play the odds of A staying ahead of all three is 0.78^3 = 48% (so chances are A will not win).

    Note - all this assumes that there are 9 players, and that the same 9 players show up each week. If there are more players, and/or some who don't show up, the odds for the guy in the lead get better (unless he doesn't show up to play).

    Hope this gives you some food for thought.

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