Question for NorrisAlan (or anyone else good at statistics

Discussion in 'Sports' started by kidbourbon, Jun 4, 2014.

  1. kidbourbon

    kidbourbon Well-Known Member

    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.
     
  2. droski

    droski Traffic Criminal

    Just add up the numbers

    60 percent equals .6
    50 percent equals .5

    1.1 wins
     
  3. Tenacious D

    Tenacious D The law is of supreme importance, or no importance

    Que TennTra, IMO.
     
  4. IP

    IP Super Moderator

    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"
     
  5. IP

    IP Super Moderator

    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).
     
  6. snoball5278

    snoball5278 Contributor

    fyp. lulz
     
  7. IP

    IP Super Moderator

    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.
     
  8. kidbourbon

    kidbourbon Well-Known Member

    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.
     
  9. kidbourbon

    kidbourbon Well-Known Member

    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.
     
  10. snoball5278

    snoball5278 Contributor

    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. fl0at_

    fl0at_ Humorless, asinine, joyless pr*ck

    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. kidbourbon

    kidbourbon Well-Known Member

    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.
     
  13. IP

    IP Super Moderator

    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.
     
  14. kidbourbon

    kidbourbon Well-Known Member

    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.
     
  15. droski

    droski Traffic Criminal

    Can't he just run the standard deviation formula on it?
     
  16. fl0at_

    fl0at_ Humorless, asinine, joyless pr*ck

    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. kidbourbon

    kidbourbon Well-Known Member

    Droski, are you just pulling terms from a wikipedia article and writing sentences that include them?

    [video=youtube;gId6nrMDmUU]http://www.youtube.com/watch?v=gId6nrMDmUU[/video]
     
  18. kidbourbon

    kidbourbon Well-Known Member

    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.
     
  19. droski

    droski Traffic Criminal

    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. fl0at_

    fl0at_ Humorless, asinine, joyless pr*ck

    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.
     

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