Originally posted by 8lrr8 via PM
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) )
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 ≈ 1.88496
For full-Kelly we have:
X ≈ 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%) ≈ 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 ≈ 6.3813%
W<sub>BE</sub> = 600*log(1-6.3813%) / (log(1-6.3813%) - log(1+(1.88496-1)*6.3813%) ≈ 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) ≈ 7.352%
So for half-Kelly the probability of failing to make a profit is =binomdist(327,600,59%,1) ≈ 1.4213%