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Math Help - NBA formula(e)-derived "power rating" against pointspread

  1. #1
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    NBA formula(e)-derived "power rating" against pointspread

    Here's my situation:

    My buddy is sending me formula(e)-derived "power rating" selections in NBA basketball which are wagered at 11/10 but could also be wagered at 21/20 offshore.

    I figure the "fair coin" test would be what is applicable...but you guys are the experts.


    ALL GAMES CONSIDERED
    So far, the Against The Spread, bet on "SIDES" is 84 wins -120 losses - 1 tie(41.18%)

    So far, the Against The Spread,BET ON "total points scored" IS 90 wins - 108 losses - 4 ties(45..45%).


    WHEN I PARSE THE LINE ON THAT SAME SET OF GAMES FOR A 6-POINT DIFFERENCE FROM THE SPREAD:

    The subset produces these figures:


    SIDES: 38 wins - 47 losses - 1 tie(44.71%)

    TOTALS: 44 wins - 57 losses - 2 ties(43.56%)


    Do we have an "unfair coin" here that I could be wagering against(colloquially referred to as "fading" ) those pointspread betting events during the second half of the NBA season.


    Have enough betting events been generated or could these figures be due to chance?


    He's also sent me hockey picks which are producing a 3.11 % return but the volatility has been incredible...these NBA picks have been steadily poor(almost no volatility) and getting poorer as the season progresses.

    Thanks
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  2. #2
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    Quote Originally Posted by kahlmy_ishmael_xxiii View Post
    Here's my situation:

    My buddy is sending me formula(e)-derived "power rating" selections in NBA basketball which are wagered at 11/10 but could also be wagered at 21/20 offshore.

    I figure the "fair coin" test would be what is applicable...but you guys are the experts.


    ALL GAMES CONSIDERED
    So far, the Against The Spread, bet on "SIDES" is 84 wins -120 losses - 1 tie(41.18%)

    So far, the Against The Spread,BET ON "total points scored" IS 90 wins - 108 losses - 4 ties(45..45%).


    WHEN I PARSE THE LINE ON THAT SAME SET OF GAMES FOR A 6-POINT DIFFERENCE FROM THE SPREAD:

    The subset produces these figures:


    SIDES: 38 wins - 47 losses - 1 tie(44.71%)

    TOTALS: 44 wins - 57 losses - 2 ties(43.56%)


    Do we have an "unfair coin" here that I could be wagering against(colloquially referred to as "fading" ) those pointspread betting events during the second half of the NBA season.


    Have enough betting events been generated or could these figures be due to chance?


    He's also sent me hockey picks which are producing a 3.11 % return but the volatility has been incredible...these NBA picks have been steadily poor(almost no volatility) and getting poorer as the season progresses.

    Thanks
    The 95% margin of error m for a proportion p is m = 2*sqrt(p(1-p)/n) when n is the number of events. For the sides bet, m = 2*sqrt(.41(1-.41)/204) = .07. What you can say is that 95% of the time, the true proportion will be in the interval generated by the formula [p-m,p+m]. For p = .41 and n = 204, that interval is [.34,.48]. How confident does that make you?

    I think the biggest problem with betting systems is not about the data, it is that the behavior of bettors, NBA players, coaches, etc, changes in response to what happened in the past. So a system that worked successfully against them in the past, very likely will not continue to work against them in the future.
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  3. #3
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    Quote Originally Posted by JakeD View Post
    The 95% margin of error m for a proportion p is m = 2*sqrt(p(1-p)/n) when n is the number of events. For the sides bet, m = 2*sqrt(.41(1-.41)/204) = .07. What you can say is that 95% of the time, the true proportion will be in the interval generated by the formula [p-m,p+m]. For p = .41 and n = 204, that interval is [.34,.48]. How confident does that make you?

    I think the biggest problem with betting systems is not about the data, it is that the behavior of bettors, NBA players, coaches, etc, changes in response to what happened in the past. So a system that worked successfully against them in the past, very likely will not continue to work against them in the future.
    Thanks for the reply, Jake....

    So if this were an unfair coin, based on the results, could I be 95% certain that I was working to a 2% advantage over the house at 11/10?

    2.5% maybe at a 95% confidence level?
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    Quote Originally Posted by kahlmy_ishmael_xxiii View Post
    Thanks for the reply, Jake....

    So if this were an unfair coin, based on the results, could I be 95% certain that I was working to a 2% advantage over the house at 11/10?

    2.5% maybe at a 95% confidence level?
    I don't know how you are calculating the 2-2.5% advantage.

    The 95% confidence interval includes .48. Then you would be at a .7% disadvantage to the house at 11/10. You can be 95% confident you will do better than that and maybe much better. To narrow the confidence interval, you have to reduce the confidence level. A 90% confidence interval uses m = 1.68*sqrt(p(1-p)/n) = .06. The 1.68 is out of a table of the normal distribution. The 90% confidence interval is [.35,.47]. At .47, you have a 1% advantage over the house at 11/10. You can be 90% confident you will do better than that (with the caveat I raised in my first post).

    Your best estimate of the true proportion is .41, which yields an advantage of about 13%. But the probability you will achieve exactly that is zero, so you have no confidence in that. You have to give it a range. The wider the range, the more confident you can be you will achieve something in that range. But since the ranges, the confidence intervals, are centered about .41, your best estimate of your advantage in a range will always be 13% no matter what range you select.
    Last edited by JakeD; March 1st 2007 at 11:21 AM.
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    Quote Originally Posted by JakeD View Post
    I don't know how you are calculating the 2-2.5% advantage.

    The 95% confidence interval includes .48. Then you would be at a .7% disadvantage to the house at 11/10. You can be 95% confident you will do better than that and maybe much better. To narrow the confidence interval, you have to reduce the confidence level. A 90% confidence interval uses m = 1.68*sqrt(p(1-p)/n) = .06. The 1.68 is out of a table of the normal distribution. The 90% confidence interval is [.35,.47]. At .47, you have a 1% advantage over the house at 11/10. You can be 90% confident you will do better than that (with the caveat I raised in my first post).

    Your best estimate of the true proportion is .41, which yields an advantage of about 13%. But the probability you will achieve exactly that is zero, so you have no confidence in that. You have to give it a range. The wider the range, the more confident you can be you will achieve something in that range. But since the ranges, the confidence intervals, are centered about .41, your best estimate of your advantage in a range will always be 13% no matter what range you select.

    Wow! This is getting interesting!

    I can be 95% confident that if this were an unfair coin @ 11/10... i would be no worse off than 0.7% against me.

    Well, I can wager @ 21/20 offshore.....

    ...and I can be 90% confident that I am at a 1% advantage as long as this performance stays BELOW THE previously demonstrated 41.18%


    SAFETY FACTORS IF UNFAIR COIN'S PERFORMANCE STAYS BELOW the previously demonstrated 41.18%

    1) I'd be beting @ 21/20 instead of 11/10
    2) I'd be generating more betting events without any better performance than 41.18%

    Thank you, JakeD... you're the greatest!
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  6. #6
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    Quote Originally Posted by kahlmy_ishmael_xxiii View Post
    Wow! This is getting interesting!

    I can be 95% confident that if this were an unfair coin @ 11/10... i would be no worse off than 0.7% against me.

    Well, I can wager @ 21/20 offshore.....

    ...and I can be 90% confident that I am at a 1% advantage as long as this performance stays BELOW THE previously demonstrated 41.18%


    SAFETY FACTORS IF UNFAIR COIN'S PERFORMANCE STAYS BELOW the previously demonstrated 41.18%

    1) I'd be beting @ 21/20 instead of 11/10
    2) I'd be generating more betting events without any better performance than 41.18%

    Thank you, JakeD... you're the greatest!
    You're welcome. It's my pleasure.

    I'm not sure what you mean by "this performance stays BELOW THE previously demonstrated 41.18%." But to be correct, you should say that the unknown true proportion p of the system is the same in the future as it was in the past.

    I did this as an exercise. Suppose the unknown p is really .47, the upper end of the 90% confidence interval. What is the probability that you will actually make a positive profit in 200 bets at 21/20?

    To break even at 21/20, your win proportion must be 21/41 = .512. So to make a profit, your realized p must be below 1 - .512 = .488 since you are betting against. The standard deviation of your realized p in 200 bets is sqrt(.47(1-.47)/200) = .0353. To use a normal approximation, calculate the Z score = (.488 - .47)/.0353 = .51. The proportion of the normal distribution with Z < .51 is .69. Thus you have a 69% chance of at least breaking even if the unknown true proportion is .47.

    To sum it all up: IF the unknown true proportion p of the system is the same in the future as it was in the past, you are 90% confident that you have a 69% chance of at least breaking even in 200 bets at 21/20.

    Go for it! Good luck!
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  7. #7
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    Quote Originally Posted by JakeD View Post
    You're welcome. It's my pleasure.

    I'm not sure what you mean by "this performance stays BELOW THE previously demonstrated 41.18%." But to be correct, you should say that the unknown true proportion p of the system is the same in the future as it was in the past.

    I did this as an exercise. Suppose the unknown p is really .47, the upper end of the 90% confidence interval. What is the probability that you will actually make a positive profit in 200 bets at 21/20?

    To break even at 21/20, your win proportion must be 21/41 = .512. So to make a profit, your realized p must be below 1 - .512 = .488 since you are betting against. The standard deviation of your realized p in 200 bets is sqrt(.47(1-.47)/200) = .0353. To use a normal approximation, calculate the Z score = (.488 - .47)/.0353 = .51. The proportion of the normal distribution with Z < .51 is .69. Thus you have a 69% chance of at least breaking even if the unknown true proportion is .47.

    To sum it all up: IF the unknown true proportion p of the system is the same in the future as it was in the past, you are 90% confident that you have a 69% chance of at least breaking even in 200 bets at 21/20.

    Go for it! Good luck!
    That is a little bit more sobering.

    I think I'll go to one of the "play for free" fictitious betting account contest sites on Yahoo! and use "play money" to prototype this before spending any REAL CASH in an offshore casino.

    Again, thanks JakeD
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