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Probability of outcome of combined events

  1. Jul 3, 2010 #1
    I'm a bit lost on use of mutually exclusive and independent rule on this. I know probability of winning a game is a loose term, but....
    If I had a team's schedule of 12 games with probabilities listed of winning each game, would I add those probabilities, then divide sum by 12 for an average, and multiply by 12 for probability of season wins?
    I was thinking it would have to be much more difficult than this.
     
  2. jcsd
  3. Jul 4, 2010 #2

    EnumaElish

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    Suppose each game showed an exactly identical probability (say, 50%). What would you conclude about season wins?
     
  4. Jul 4, 2010 #3
    I'd conclude nothing froma sample this small, especially sporting events. Regardless of %, I thought it would still be the equation
    Pr(A+B+C)= Pr(A) X Pr(B) X Pr(C).
    It's just when I calculate all 12 "50%, I come up with the worls worst probability of winning even one game, much less 12. To accomplish a season so bad, they'd have to fumble every snap of every game and maybe forfeit some games from last year to make up for it.
    Are these not independent events? I have no evidence in a dispute with the individual who's saying just add the % and divide.
    Please show me where I err.
     
  5. Jul 4, 2010 #4
    You're correct, I think.

    If you have a certain percent chance to win a game, then you can (for seasons with lots of games) take that win percentage and add it to your season win total.

    For example:
    Game 1 - 40% chance to win
    Game 2 - 30%
    Game 3 - 70%
    Game 5 - 35%
    Game 6 - 25%

    You can take these percentages as parts of games won to find the expected wins for the season.

    For my example season:
    .4 + .3 + .7 + .35 + .25 = 2 games
    So you'll probably win two games. Just remember that the smaller the sample size, the less accurate this prediction is.
     
  6. Jul 5, 2010 #5
    I remember the probability must lie between 0 and 1, so it couldn't be 2 for total probability(1 would be an almost certain 5-0 season). Isn't that right?
    Since each is 1/5 of the total, maybe .4X.2+ .7X.2, etc
    =.08 + =.14 + etc?
     
  7. Jul 6, 2010 #6
    For example, if you're 100% sure you will win 5 out of 5 games, then you can calculate the expected number of games you will win using the method I described above (although it's obvious):

    1 * 100% = 1
    + 1 * 100% = 1
    + 1 * 100% = 1
    + 1 * 100% = 1
    + 1 * 100% = 1
    = 5 games

    If you play five games which you have a 50% chance to win:

    1 * 50% = .5
    + 1 * 50% = .5
    + 1 * 50% = .5
    + 1 * 50% = .5
    + 1 * 50% = .5
    =2.5 games

    Now since you obviously cannot win 2.5 games (unless you can tie?), you can infer that you have an equal chance to win either 2 or 3 games but you also have some chance to win 0, 1, 4, or 5 games.
     
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