Calculating breakeven probability (received via PM)

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  • Ganchrow
    SBR Hall of Famer
    • 08-28-05
    • 5011

    #1
    Calculating breakeven probability (received via PM)
    Originally posted by 8lrr8 via PM
    Ganchrow:

    what's the formula to compute probability one will be down after X number of bets, given a fixed winrate, line (e.g. -110), and kelly bet size?

    e.g.: suppose i place 600 bets, all at a 59% winrate, all at -113. what's the probability i'll have less than my starting bankroll after 600 bets if i bet full kelly? what's the probability if i bet 70% of full kelly?
    Let's look at the general case first, and then from there we can easily solve for the specific case you've given.

    Before can figure out the probability of breaking even we first need to determine what our breakeven win rate actually is.

    Were we betting a fixed amount per bet this would be quite simple (ignoring the possibioity of bankruptcy). The breakeven win rate would simply be the reciprocal of the decimal odds. So for US odds of -110 = decimal odds of 1.909091, the breakeven win rate would be 1/1.909091 ≈ 52.381%. For those lacking a surfeit of free time my odds converter will perform these calculations for you.

    When staking a percentage of bankroll (as with Kelly), however, the situation become slightly more complex.
    Let N = # of bets
    Let W = number of bets won, ≤ N
    Let d = decimal odds
    Let X = % of bankroll wagered on each bet (this could be any multiple of Kelly)
    Let B<sub>0</sub> = starting bankroll
    Let B<sub>N</sub> = bankroll after N trials

    B<sub>N</sub> = B<sub>0</sub> * (1+(d-1)*X)<sup>W</sup> * (1-X)<sup>N-W</sup>

    So to find the breakeven numbers of wins we set B<sub>N</sub> = B<sub>0</sub> and solve for the breakeven win level W<sub>BE</sub>.

    1 = (1+(d-1)*X)<sup>W<sub>BE</sub></sup> * (1-X)<sup>N-W<sub>BE</sub></sup>

    taking the logarithm of both sides:
    W<sub>BE</sub> * log(1+(d-1)*X) + (N-W<sub>BE</sub>) * log(1-X) = 0

    and solving for W<sub>BE</sub>:

    W<sub>BE</sub> = N*log(1-X) / (log(1-X) - log(1+(d-1)*X) )

    So given your initial figures we have:
    B<sub>0</sub> = 1
    N = 600
    p = 59%
    X = B<sub>0</sub> = 1
    d = 213/113 &asymp; 1.88496

    For full-Kelly we have:
    X &asymp; 12.6700%

    Using the forumula from above:

    W<sub>BE</sub>= 600*log(1-12.6700%) / (log(1-12.6700%) - log(1+(1.88496-1)*12.6700%) &asymp; 336.24 wins

    This means that betting full-Kelly over 600 bets, a player would end up with less than his starting bankroll were he to win 336 or fewer of his bets.

    For half-Kelly we have:
    X &asymp; 6.3813%

    W<sub>BE</sub> = 600*log(1-6.3813%) / (log(1-6.3813%) - log(1+(1.88496-1)*6.3813%) &asymp; 327.31 wins.

    This tells us that betting half-Kelly over 600 bets, a player would end up with less than his starting bankroll were he to win win 327 or fewer of his 600 bets.

    The next issue is calculating the likelihood of winning W or fewer bets out of N trials. This is a simple application of the binomial distribution and is easily done using Excel's binomdist() function.

    =binomdist(W,N,p,1)

    And for full-Kelly the probability of failing to make a profit is =binomdist(336,600,59%,1) &asymp; 7.352%

    So for half-Kelly the probability of failing to make a profit is =binomdist(327,600,59%,1) &asymp; 1.4213%
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