1. #1
    Justin7
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    Thinking about betting season wins for NFL? (75 points at stake)

    Oddsmakers and players underestimate the value of a half-game on season wins futures in the NFL. So, two questions to show this point.

    1. A) If a team has a 50% chance of winning each of 16 games, what is the probability that it wins exactly 8 games during the season?
    1. B) In the same example, what is your EV betting under 8.5 wins at even money?

    The first correct answer to 1A and 1B gets 25 points.

    2. A) If a team has a 60% chance of winning each of 8 games, and a 40% chance of winning each of 8 games, what is the probability that it wins exactly 8 games during the season?
    2. B) In the same example, what is your EV betting under 8.5 wins at even money?

    The first correct answer to 2A and 2B gets 50 points.

    The same person cannot win both prizes. You must show your valid methodology (with the correct answer) to win.

  2. #2
    uva3021
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    1a. 19.63%
    1b. 19.63%
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    Justin7 gave uva3021 25 SBR Point(s) for this post.


  3. #3
    TomG
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    There is a good sports betting book called Sharp Sports Betting by Stanford Wong that answers your exact question (Chapter 10). It is also discussed in another very good sports betting book Weighing the odds in Sports Betting by King Yao (Chapter 8).

  4. #4
    TomG
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    Of course, I'm sure you're too busy to bother reading the plethora of sports betting books on the market.

  5. #5
    uva3021
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    2a. 14.17%
    2b. -13.7%

  6. #6
    Justin7
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    Quote Originally Posted by uva3021 View Post
    1a. 19.63%
    1b. 19.63%
    And how...?

  7. #7
    uva3021
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    just saw the qualifier at the bottom

    this is a binomial distribution, as there are two possible outcomes for each of the 16 games

    for the first problem above, from game to game the probability doesn't change, however the second problem appears harder than it really is because half the games are of a different probability

    regardless, each set of two games, in any combination, will always have the same probability of .6 * .4 = .24, which are grouped together 8 times over 16 games, multiply by this by the number of possible combinations of winning 8 games in a 16 game season (combin(16,8)), and the result is the answer to P(x=8)

  8. #8
    Justin7
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    Quote Originally Posted by uva3021 View Post
    just saw the qualifier at the bottom

    this is a binomial distribution, as there are two possible outcomes for each of the 16 games

    for the first problem above, from game to game the probability doesn't change, however the second problem appears harder than it really is because half the games are of a different probability

    regardless, each set of two games, in any combination, will always have the same probability of .6 * .4 = .24, which are grouped together 8 times over 16 games, multiply by this by the number of possible combinations of winning 8 games in a 16 game season (combin(16,8)), and the result is the answer to P(x=8)
    Pairing games... you created a trinomial distribution. It's a clever way to simply things, but how did you get a -EV result for under 8.5?

  9. #9
    x2
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    2a)20%
    Figure how many combinations you can win K games with 60% chance (pascal's 8th row); multiply by .6^k*(.4)^(8-k); sum the squares of each of the results.

    2b)20%
    0-7 and 9-16 have the same P (40%); (1*(40+20))-(1*40)
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  10. #10
    Justin7
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    Maybe I am not explaining this clearly.

    In a 16 game season, a team plays at home 8 times, and on the road 8 times. Of those 8 home games, it expects to win 60% of the time. Of those 8 road games, it expects to win 40% of the time.

  11. #11
    uva3021
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    don't even remember answering this last night

    pairing games would only work if the team won 4 at home and 4 on the road, a grand assumption however

    to work it out you would take every possible combination of winning 8 games using the home and road probabilities, then add up all the probabilities

    how this is done without doing a simulation i don't know

  12. #12
    AlwaysDrawing
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    2A)

    Multiply the individual probabilities of each combination of 8 wins, and add them up:
    P(Road Wins, from 0-8) 1.68% 8.96% 20.90% 27.87% 23.22% 12.39% 4.13% 0.79% 0.07%
    P(Home Wins, from 8-0) 1.68% 8.96% 20.90% 27.87% 23.22% 12.39% 4.13% 0.79% 0.07%
    Probabilities of combinations 0.03% 0.80% 4.37% 7.77% 5.39% 1.53% 0.17% 0.01% 0.00%
    sum(probabilities of combinations)=20.07%

    20.07%

    2B) the odds of 1-7wins and 9-16 wins will be the same distributions, so the odds of winning the UNDER 8.5 games would be the cumulative probability of winning 0-8 games, or 20.07+(79.93/2)=60.035%

    (+$1*60.035%)+(-$1*39.965%)=20.07%
    Last edited by AlwaysDrawing; 08-22-11 at 10:58 AM. Reason: Excel tables don't copy well
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  13. #13
    Spektre
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    2a. 14.17%

    Is the EV given above for 1b correct?!

  14. #14
    AlwaysDrawing
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    Quote Originally Posted by Spektre View Post
    2a. 14.17%

    Is the EV given above for 1b correct?!

    Yes.

    19.63+((100-19.63)/2)=59.815% of the time the bet will win.

    EV=Prob(win)*amount if won + prob(loss)*amt. if lost

    EV*=59.815%*$1 + (1-59.815%)*-$1= $0.1963 or 19.63%

  15. #15
    TomG
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    Justin7 infracted me for being snarky. I try to be on good behavior but it's hard for me when my Asperger's is acting up (I do not easily grasp social nuances).

    Part 1 is just binomial distribution. You have n, p, and q. Get the probability of 0, 1, ... 16 wins from the binomial formula. Add up each discrete probability to get the probability of 8 wins.

    Part 2 probably has a more elegant solution. But you can do the same approach as part 1, just divided into 2 parts. Use the binomial split for home/away. Get the Probability of 0, 1, ... 8 wins for each home/away split. Then combine them by summing together the possible ways of winning 8 games (e.g., win 0 away games, win all 8 home games; win 1 away game, win 7 home games, etc).

    I will donate any potential points awarded to the Foundation of Children with Asperger's.

  16. #16
    Spektre
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    2a) P(8 games won) = 14.17%
    2b) P(<=8 games won) = 28.39%

    So EV = $1 * 0.2839 - $1 * (1-0.2839) = -.4322 or -43.22%

    Spektre

    Method of calculation - Sum the probabilities of each number of win 0 to 8. These are easily found via binomial distribution, recognizing the constraint of only 8 possible games each at home and away.
    Last edited by Spektre; 08-22-11 at 12:09 PM.

  17. #17
    AlwaysDrawing
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    So here's my question for you Justin:

    How would you handicap a team's expected home/away winrate?

    That is, how would you handicap an individual team (e.g. the Ravens), given the following total season wins:
    Ravens to win on Pinnacle:
    OVER 11 +168
    UNDER 11 -199

    This implies a no vig theoretical hold of 64.7% for the UNDER (at +-183)

    How is the book handicapping that event, and how would you?

  18. #18
    Justin7
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    Always,

    A quick, not perfectly accurate way to attack this problem is to assume each 50 cents is worth 1/2 a game. At -183, it suggests that the Ravens are likely to win about 11 - (83/100) or 10.17 games.

    If 10.17 = 8p + 8(p+0.2) where p is the odds of winning on the road, and p+0.2 is the odds of winning at home.

    This assumes the odds of winning at home are always 20% higher than on the road, no matter how good the team. 18% is probably a better number, and I am not sure this is accurate as teams move away from the 50/50 point. But it is a starting point.

  19. #19
    Peregrine Stoop
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    these methods suck. There are simple ways to do it better.

  20. #20
    Justin7
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    Quote Originally Posted by Peregrine Stoop View Post
    these methods suck. There are simple ways to do it better.
    Care to share?

  21. #21
    Spektre
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    Quote Originally Posted by AlwaysDrawing View Post
    2A)

    Multiply the individual probabilities of each combination of 8 wins, and add them up:
    P(Road Wins, from 0-8) 1.68% 8.96% 20.90% 27.87% 23.22% 12.39% 4.13% 0.79% 0.07%
    P(Home Wins, from 8-0) 1.68% 8.96% 20.90% 27.87% 23.22% 12.39% 4.13% 0.79% 0.07%
    Probabilities of combinations 0.03% 0.80% 4.37% 7.77% 5.39% 1.53% 0.17% 0.01% 0.00%
    sum(probabilities of combinations)=20.07%

    20.07%

    2B) the odds of 1-7wins and 9-16 wins will be the same distributions, so the odds of winning the UNDER 8.5 games would be the cumulative probability of winning 0-8 games, or 20.07+(79.93/2)=60.035%

    (+$1*60.035%)+(-$1*39.965%)=20.07%

    I don't believe this to be the correct answer.

    Spektre

  22. #22
    AlwaysDrawing
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    Quote Originally Posted by Spektre View Post
    I don't believe this to be the correct answer.

    Spektre
    Why not? It's just a combination of two probabilities.

    (not to mention Justin gave me the points, though that shouldn't be taken of "proof" of anything)

  23. #23
    Justin7
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    Quote Originally Posted by Spektre View Post
    I don't believe this to be the correct answer.

    Spektre
    The method looked right, but I didn't add up the numbers. What do you get?

  24. #24
    Spektre
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    The methodology I used was similar but I don't find the probabilities are symmetric about 8 wins (which might show my method is wrong)

    I'll show for 2a)

    Probability of exactly 8 wins

    Phome(0-8)*Paway(8-0)+Phome(1-7)*Paway(7-1)+Phome(2-6)*Paway(6-2)+...+Phome(8-0)*Paway(0-8)

    I evaluated these binomials to:

    0.00065536*0.01679616+0.00786432*0.08957 952+0.04128768*0.20901888+0.12386304*0.2 7869184+0.2322432*0.2322432+0.27869184*0 .12386304+0.20901888*0.04128768+0.089579 52*0.00786432+0.01679616*0.00065536=

    1.10075E-05+0.000704482+0.008629905+0.034519619+0 .053936904+0.034519619+0.008629905+0.000 704482+1.10075E-05=

    0.141666929
    or 14.17%, which is in agreement with an earlier answer.

  25. #25
    x2
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    P of losing all of your 60% games is .4^8 same as winning all your 40% games. Etc.

  26. #26
    Justin7
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    Quote Originally Posted by Spektre View Post
    The methodology I used was similar but I don't find the probabilities are symmetric about 8 wins (which might show my method is wrong)
    I think these should be symmetric around 8.

  27. #27
    x2
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    14.17% of exactly 8 wins for a team with a 60% chance in all 16 games if it helps you see it better.

  28. #28
    Spektre
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    Quote Originally Posted by x2 View Post
    14.17% of exactly 8 wins for a team with a 60% chance in all 16 games if it helps you see it better.
    Aha, thanks for showing me my error.

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