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Thread: Question for NorrisAlan (or anyone else good at statistics

  1. #1

    Default Question for NorrisAlan (or anyone else good at statistics

    You're a math major right? Help a kid out if you can.

    What I did: go through the schedule of a given team (we'll go with the Vols), and assign to each game a probability of winning.
    What I want to do with that: using the numbers I just came up with, figure out the chances that the team ends up winning [X] number of games. If we're talking the Vols, I might be looking to see what their chances of winning 8 games are, and if they're more likely (again, given my numbers) to win 6 games or 7 games.

    That's it. I feel like this shouldn't be terribly hard, but I have absolutely no idea how to do it.
    Balls deeep...

  2. #2

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    Just add up the numbers

    60 percent equals .6
    50 percent equals .5

    1.1 wins

  3. #3

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    Que TennTra, IMO.

  4. #4

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    Quote Originally Posted by Tenacious D View Post
    Que TennTra, IMO.
    This post made me look up to confirm, but it is cue. Might have been thinking of queue. So far as I can tell, there is no English "que"
    Bajakian must be dealt with.

  5. #5

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    Quote Originally Posted by kidbourbon View Post
    You're a math major right? Help a kid out if you can.

    What I did: go through the schedule of a given team (we'll go with the Vols), and assign to each game a probability of winning.
    What I want to do with that: using the numbers I just came up with, figure out the chances that the team ends up winning [X] number of games. If we're talking the Vols, I might be looking to see what their chances of winning 8 games are, and if they're more likely (again, given my numbers) to win 6 games or 7 games.

    That's it. I feel like this shouldn't be terribly hard, but I have absolutely no idea how to do it.
    Not a math guy, but I think you multiply them together. The problem is you will get tiny, tiny numbers as you are not just calculating the number of wins, but rather the chances of winning all of the games. Which isn't what you are asking so this isn't really much help.

    What may be more help is to look into something called Bayesian probability. That is going to get you where you want to go, but the journey will be boring and the results, while what you want, will likely not have the rigor or value you would expect (I distrust many applications of Bayesian logic and probability).
    Bajakian must be dealt with.

  6. #6

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    Quote Originally Posted by IP View Post
    **** you tenny, there is no English "que"
    fyp. lulz

  7. #7

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    Quote Originally Posted by snoball5278 View Post
    fyp. lulz
    I feel like there is a decent chance Tenny is some sort of crime boss akin to those portrayed on Justified that sort of hide in plain sight, so I wouldn't say it like that.
    Bajakian must be dealt with.

  8. #8

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    Quote Originally Posted by droski View Post
    Just add up the numbers

    60 percent equals .6
    50 percent equals .5

    1.1 wins
    That doesn't answer my question. I can do that and come up with a number. But that doesn't tell me the likelihood of different numbers, which is what I want to know.
    Balls deeep...

  9. #9

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    Quote Originally Posted by IP View Post
    Not a math guy, but I think you multiply them together. The problem is you will get tiny, tiny numbers as you are not just calculating the number of wins, but rather the chances of winning all of the games. Which isn't what you are asking so this isn't really much help.

    What may be more help is to look into something called Bayesian probability. That is going to get you where you want to go, but the journey will be boring and the results, while what you want, will likely not have the rigor or value you would expect (I distrust many applications of Bayesian logic and probability).
    If you multiply them together you get the likelihood of every event taking place. I don't want that. I know the odds of UT going 12-0 are roughly the same as the odds of me having unprotected sex with Emma Watson live on national television.

    I want to find out the chances of 6 wins, 7 wins, 8 wins, etc.

    I have looked into Bayesian Probability, and a few other concepts I don't quite understand, and, well, I don't quite understand them. The closest I've come is Binomial distribution, but that requires that each event have the same probabilit, and so it does me no good.
    Balls deeep...

  10. #10

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    Quote Originally Posted by IP View Post
    I feel like there is a decent chance Tenny is some sort of crime boss akin to those portrayed on Justified that sort of hide in plain sight, so I wouldn't say it like that.
    then play it safe and don't correct him at all. there is no telling what all he's done to nyy with a cue.

  11. #11

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    The simplest way is the sum of the probabilities divided by the number of events.

    If you want to do something more, do a Z score and compare to the z value for a 95 or 99 Percent distribution.

    But you'll need to define a mean number of wins.

  12. #12

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    Quote Originally Posted by IP View Post
    Not a math guy, but I think you multiply them together. The problem is you will get tiny, tiny numbers as you are not just calculating the number of wins, but rather the chances of winning all of the games. Which isn't what you are asking so this isn't really much help.

    What may be more help is to look into something called Bayesian probability. That is going to get you where you want to go, but the journey will be boring and the results, while what you want, will likely not have the rigor or value you would expect (I distrust many applications of Bayesian logic and probability).
    Just talked to a friend who is pretty good with these such things, and he said Bayes Theorem is the way to do it. So, I guess I'm just gonna have to go and do some learning about this Bayes fella.
    Balls deeep...

  13. #13

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    Like I said, you won't like what you find. The results will be what you want, but you will realize that in the end it is just a reflection of your own assumptions. Which is fine for this sort of use, but disturbing how it gets applied in the natural sciences sometimes.
    Bajakian must be dealt with.

  14. #14

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    Quote Originally Posted by fl0at_ View Post
    The simplest way is the sum of the probabilities divided by the number of events.

    If you want to do something more, do a Z score and compare to the z value for a 95 or 99 Percent distribution.

    But you'll need to define a mean number of wins.
    Can you elaborate a little bit more.

    Simple example with 5 games.

    Game 1: .40
    Game 2: .15
    Game 3: .95
    Game 4: .35
    Game 5: .50

    Sum: 2.35

    So when you say divide by number of events, you can't possibly be referring to the number of events for which i'm trying to get an accompanying percentage, because that wouldn't work here.

    For example: 2.35/3 = .78. And clearly, given the listed percentages, there isn't a 78% chance of winning three games.

    So obviously you are saying divide by the total number of games. Here: 2.35/5 = .47 So from there can I go back and just assign a .47 to each game and then use the binomial theorem? Because that actually doesn't sound like a terrible idea.
    Balls deeep...

  15. #15

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    Can't he just run the standard deviation formula on it?

  16. #16

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    Quote Originally Posted by kidbourbon View Post
    Can you elaborate a little bit more.

    Simple example with 5 games.

    Game 1: .40
    Game 2: .15
    Game 3: .95
    Game 4: .35
    Game 5: .50

    Sum: 2.35

    So when you say divide by number of events, you can't possibly be referring to the number of events for which i'm trying to get an accompanying percentage, because that wouldn't work here.

    For example: 2.35/3 = .78. And clearly, given the listed percentages, there isn't a 78% chance of winning three games.

    So obviously you are saying divide by the total number of games. Here: 2.35/5 = .47 So from there can I go back and just assign a .47 to each game and then use the binomial theorem? Because that actually doesn't sound like a terrible idea.
    n is the number of events. Not the ones you want to occur, but the total, so the 2.35/5 = .47 is the way I meant.

  17. #17

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    Quote Originally Posted by droski View Post
    Can't he just run the standard deviation formula on it?
    Droski, are you just pulling terms from a wikipedia article and writing sentences that include them?

    Balls deeep...

  18. #18

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    Quote Originally Posted by fl0at_ View Post
    n is the number of events. Not the ones you want to occur, but the total, so the 2.35/5 = .47 is the way I meant.
    If I then use the binomial theorem to compute what I need, are my numbers going to be correct? In other words, for multiple independent events with differeing probabilities for each event, it's okay to just find the mean probability, assign it as the probability for each event, and proceed accordingly?

    That almost seems too easy.
    Balls deeep...

  19. #19

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    Quote Originally Posted by kidbourbon View Post
    Droski, are you just pulling terms from a wikipedia article and writing sentences that include them?

    Believe it or not I've taken both econometrics and statistics, and have a cfa charter (which involves a crapload of probability). I don't remember everything though.

  20. #20

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    Quote Originally Posted by kidbourbon View Post
    If I then use the binomial theorem to compute what I need, are my numbers going to be correct? In other words, for multiple independent events with differeing probabilities for each event, it's okay to just find the mean probability, assign it as the probability for each event, and proceed accordingly?

    That almost seems too easy.
    No, I would, if you had an average, use a z score.

    They are used frequently in writing a warranty, which is kinda similar to what you are looking for.

    For example, if a hair dryer has a mean life of 4 years, and a standard deviation of a year, a z score can be calculated to determine the probability of a dryer failing at 5 years. Or 6 years. And then based on that, the warranty period can be determined. If you find that 85% of your dryers fail by 5 years, and it would be detrimental to replace that many dryers, you set the warranty at 4 years. Or 3. Whatever you feel best, and whatever the market can bear.

  21. #21

  22. #22

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    Quote Originally Posted by fl0at_ View Post
    No, I would, if you had an average, use a z score.

    They are used frequently in writing a warranty, which is kinda similar to what you are looking for.

    For example, if a hair dryer has a mean life of 4 years, and a standard deviation of a year, a z score can be calculated to determine the probability of a dryer failing at 5 years. Or 6 years. And then based on that, the warranty period can be determined. If you find that 85% of your dryers fail by 5 years, and it would be detrimental to replace that many dryers, you set the warranty at 4 years. Or 3. Whatever you feel best, and whatever the market can bear.
    Cool.

    What I want to do with this little exercise is come up with my own numbers and likelihoods on win totals for CFB teams, and then see how it compares to those posted by Vegas. You know, for curiosity's sake.
    Balls deeep...

  23. #23

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    Quote Originally Posted by droski View Post
    Believe it or not I've taken both econometrics and statistics, and have a cfa charter (which involves a crapload of probability). I don't remember everything though.
    I thought your standard deviation post was a bit lacking in context, and it made me think of the scene. Just giving you a hard time.
    Balls deeep...

  24. #24

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    Quote Originally Posted by kidbourbon View Post
    Cool.

    What I want to do with this little exercise is come up with my own numbers and likelihoods on win totals for CFB teams, and then see how it compares to those posted by Vegas. You know, for curiosity's sake.
    Anything you do will likely be very limited. What you want is advanced statistics. But that has a high price (time) to pay.

    I wouldn't put much faith in any number you crunch.

    You can do it. But don't bet big.

  25. #25

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    Quote Originally Posted by fl0at_ View Post
    Anything you do will likely be very limited. What you want is advanced statistics. But that has a high price (time) to pay.

    I wouldn't put much faith in any number you crunch.

    You can do it. But don't bet big.
    I never bet big on preseason totals.

    Anything I do will be limited because my probabilities are just educated guesses. I realize that. It's just a rough sketch.
    Balls deeep...

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