[Ganchrow]

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 B₀ = starting bankroll
Let B_N = bankroll after N trials

B_N = B₀ * (1+(d-1)*X)^W * (1-X)^N-W

So to find the breakeven numbers of wins we set B_N = B₀ and solve for the breakeven win level W_BE.

1 = (1+(d-1)*X)^W_BE * (1-X)^N-W_BE

taking the logarithm of both sides:
W_BE * log(1+(d-1)*X) + (N-W_BE) * log(1-X) = 0

and solving for W_BE:

W_BE = N*log(1-X) / (log(1-X) - log(1+(d-1)*X) )

So given your initial figures we have:
B₀ = 1
N = 600
p = 59%
X = B₀ = 1
d = 213/113 ≈ 1.88496

For full-Kelly we have:
X ≈ 12.6700%

Using the forumula from above:

W_BE= 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_BE = 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%

[HedgeHog]

Ganch:

What are the probabilities of losing your entire bankroll with the Full Kelly compared to Half Kelly, given your example. Even with a 59% win average, the 12.67% of bankroll per bet could lead to a disaster in the early going.

[Ganchrow]

Originally Posted by HedgeHog

Ganch:

What are the probabilities of losing your entire bankroll with the Full Kelly compared to Half Kelly, given your example. Even with a 59% win average, the 12.67% of bankroll per bet could lead to a disaster in the early going.

Your probability of losing your entire bankroll would be zero in each case. (Although as a practical matter your bankroll could potentially drop to the point where you couldn't even afford a minimum size bet size.)

Using similar logic as in the post above, the probability of losing let's say 90% of your bankroll after 600 bets would be 1.1494% with full Kelly and 0.0088% with half-Kelly.

[HedgeHog]

That's right. I lost sight of the fact that as your bankroll diminishes during a losing streak so does your bet. I gain 10 IQ points every time I read one of your posts.