Weighted odds in sports events

Discussion in 'Physics & Math' started by Jennifer Murphy, Feb 10, 2016.

  1. Jennifer Murphy Registered Senior Member

    Messages:
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    The Golden State Warriors (GSW) have a record to date of 47/4 or 92.2%. There are 82 games in the regular season, so if we assume that they will average the same winning percentage for the remaining 31 games, they should be expected to win 75.6 games: 47 + 31*.922. Is that correct?

    But that assumes that (a) their odds of winning each game is the same and/or (b) the average strength of the remaining 31 games is the same as the previous 51. Neither of these is true.

    It seems like I could make a more accurate prediction if I took into account the record to date of the teams that they still have to play. I can easily get that information. For example, their next three opponents and their records to date are:
    Code:
    
      Team        W   L    %
    Phoenix      14  39  .264
    Portland     26  27  .491
    Los Angeles  35  17  .673
    
    I would expect their odds against Phoenix, a weak team, to be greater than .922, but how much greater?

    I would expect their odds against Portland, an average team, to be close to their average of .922, since that number is against the average of all teams. It should be slightly higher, since Portland at .491 is slightly below .500.

    I would expect the odds against Los Angeles to be less than .922, but how much less?

    Finally, if there were another team in the league with the exact same record (.922), I would expect the odds against that team to be exactly .500.

    I need a function, f(P1, P2), where P1 is the current record of Team #1 and P2 is the current record of Team #2, such that f(P1,P2) gives an adjusted or weighted probability that Team 1 will beat Team 2.

    I am at a loss to come up with this function. Can anyone help?

    Thanks
     

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