Probability of outcome of combined events

  • Thread starter mayhawman
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  • #1
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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.
 

Answers and Replies

  • #2
EnumaElish
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Suppose each game showed an exactly identical probability (say, 50%). What would you conclude about season wins?
 
  • #3
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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.
 
  • #4
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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.
 
  • #5
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.4 + .3 + .7 + .35 + .25
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?
 
  • #6
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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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