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Math Help - Optimal Kelly Bet Sizing for "simultaneous events"

  1. #1
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    Optimal Kelly Bet Sizing for "simultaneous events"

    This is a football betting question in which the hoped deliverable would be a formula where 3 variables are plugged in and a result is produced.

    If one is betting a football game and one assumes one's self to be 53.5% proficient at picking winners against the 11/10 vigorish (0.91) the bookie charges(52.38% winners needed to break even)....

    ...then the formula would be:

    53.5% - (46.5% /0.91) = 2.35% of one's bankroll should be bet on every play.


    ..but NOW sportsbooks offer better odds(some at 21/20 vig(0.95)...some at other ratios).







    My question is an extension of what Ken Uston discussed in his textbook on Blackjack Card Counting.

    Uston discussed the card-counting player that was enjoying a 1.5% edge over the house in Blackjack(and betting 1.5% on each hand).....

    ...could increase that amount by a small fraction if he played 2, 3, or 4 hands against the same dealer's upcard(simultaneous Kelly betting events)...





    ...and then he provided the reader with the allowable percentage increase over that 1.5% bet.....


    for example...spreading to two hands one would MULTIPLY the $150 bet for a $10,000 bank by 1.15

    $150 * 1.15 would be $172(2 bets of $86 each)




    FOR 3 BETS THE MULTIPLE would be 1.21

    $150 * 1.21 = $181.5..... 3 bets of $60.50 each



    TRANSLATING TO FOOTBALL WAGERING

    For simplicity's sake.... all picks bet simultaneously have the SAME probability of winning


    Variable "X" would then be the probality of winning.

    Variable "Y" would be the number of betting events being wagered simultaneously

    Variable "Z" would be the differing VIGORISH extracted by the bookie,e.g....

    -110(0.91)
    -107(0.93)
    -105(0.95)

    The result of the equation would then presumably provide how much to wager on each bet OR provide the multiple.

    Thank you....

    If you'd like attribution by your username or real name, please mention it... because this is going to be posted on about 6 sports forums.

    Thank you,
    kahlmy_ishmael_xxiii
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  2. #2
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    Quote Originally Posted by kahlmy_ishmael_xxiii View Post
    This is a football betting question in which the hoped deliverable would be a formula where 3 variables are plugged in and a result is produced.

    If one is betting a football game and one assumes one's self to be 53.5% proficient at picking winners against the 11/10 vigorish (0.91) the bookie charges(52.38% winners needed to break even)....

    ...then the formula would be:

    53.5% - (46.5% /0.91) = 2.35% of one's bankroll should be bet on every play.


    ..but NOW sportsbooks offer better odds(some at 21/20 vig(0.95)...some at other ratios).







    My question is an extension of what Ken Uston discussed in his textbook on Blackjack Card Counting.

    Uston discussed the card-counting player that was enjoying a 1.5% edge over the house in Blackjack(and betting 1.5% on each hand).....

    ...could increase that amount by a small fraction if he played 2, 3, or 4 hands against the same dealer's upcard(simultaneous Kelly betting events)...





    ...and then he provided the reader with the allowable percentage increase over that 1.5% bet.....


    for example...spreading to two hands one would MULTIPLY the $150 bet for a $10,000 bank by 1.15

    $150 * 1.15 would be $172(2 bets of $86 each)




    FOR 3 BETS THE MULTIPLE would be 1.21

    $150 * 1.21 = $181.5..... 3 bets of $60.50 each



    TRANSLATING TO FOOTBALL WAGERING

    For simplicity's sake.... all picks bet simultaneously have the SAME probability of winning


    Variable "X" would then be the probality of winning.

    Variable "Y" would be the number of betting events being wagered simultaneously

    Variable "Z" would be the differing VIGORISH extracted by the bookie,e.g....

    -110(0.91)
    -107(0.93)
    -105(0.95)

    The result of the equation would then presumably provide how much to wager on each bet OR provide the multiple.

    Thank you....

    If you'd like attribution by your username or real name, please mention it... because this is going to be posted on about 6 sports forums.

    Thank you,
    kahlmy_ishmael_xxiii
    Simultaneous bets in blackjack are different than in football because in blackjack the results of the bets are dependent from facing the same dealer hand. In football the results of simultaneous bets are independent. With independence, simultaneous bets can be sized essentially independently using the ordinary Kelly Criterion. I say essentially because the proof of this involves an approximation that can be significant.

    Let the result of a one-unit bet as a proportion applied to a bankroll be 1 + x. Let b be the size of the bet and B be the bankroll. The Kelly Criterion sets b to maximize

    U(b) = E \ln((1+bx)B) = E \ln(1+bx) + \ln(B).

    Using the Taylor's series approximation \ln(1+bx) \approx bx - b^2 x^2/2, and simplifying the expectation, we are to maximize

    U(b) = bE(x) - b^2 E(x^2)/2 + \ln(B).

    Setting the derivative with respect to b equal to zero yields the Kelly Criterion

    \hat{b} = E(x)/E(x^2) \approx E(x)/Var(x)

    when E(x) is small.

    Let 1+y be the result of n simultaneous, independent, one-unit bets with the same exprectation and variance as x. Then E(y) = nE(x) and because of the independence Var(y) = nVar(x). Applying the Kelly Criterion here

    \hat{b} \approx E(y)/Var(y) = nE(x)/nVar(x) = E(x)/Var(x).

    The optimal size \hat{b} of each of the n individual simultaneous bets is unchanged from the non-simultaneous case.

    This result is approximate because of the use of the Taylor's series approximation that is only accurate when by is small. So it should not be used to justify betting a large portion of the bankroll on a number of simultaneous bets.

    In the case of blackjack, simultaneous bets are not independent and the variance is Var(y) = nV + n(n-1)C , where V \approx 1.32 is the single hand variance and C \approx .48 is the covariance between hands. Then

    \hat{b} \approx E(y)/Var(y) = nE(x)/(nV + n(n-1)C) = E(x)/(V+(n-1)C).

    The optimal size of the individual simultaneous bets decreases with the number of simultaneous hands.
    Last edited by JakeD; December 22nd 2006 at 07:30 PM.
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  3. #3
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    Thank you for the reply, JakeD.
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