Probability of outcome of combined events

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Discussion Overview

The discussion revolves around the calculation of probabilities for winning a series of games based on individual game probabilities. Participants explore concepts of mutually exclusive and independent events, as well as methods for estimating expected wins over a season.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions the method of averaging game probabilities to estimate season wins, suggesting it may be more complex than simply adding probabilities.
  • Another participant proposes that if each game has an identical probability (e.g., 50%), it raises questions about the conclusions that can be drawn from such a small sample size.
  • A different participant asserts that the probabilities of winning multiple games should be calculated using the multiplication rule for independent events, expressing confusion over the application of this rule in the context of sports outcomes.
  • One participant agrees with the idea that win percentages can be summed for expected wins, but emphasizes that smaller sample sizes lead to less accurate predictions.
  • Another participant points out that total probabilities must remain between 0 and 1, questioning the validity of summing probabilities directly without considering their implications.
  • A participant illustrates the calculation of expected wins using a 50% win probability, noting that while the expected value may suggest a non-integer number of wins, actual outcomes must be whole numbers, leading to a range of possible outcomes.

Areas of Agreement / Disagreement

Participants express differing views on how to calculate expected wins from game probabilities, with some advocating for summing probabilities and others emphasizing the need for a more nuanced approach involving independence and the nature of the events. The discussion remains unresolved with multiple competing views present.

Contextual Notes

Participants highlight limitations in their reasoning, such as the dependence on sample size and the assumptions about independence of events. There is also a recognition that expected values do not directly translate to actual outcomes.

mayhawman
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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.
 
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Suppose each game showed an exactly identical probability (say, 50%). What would you conclude about season wins?
 
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.
 
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.
 
.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?
 
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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