The problem that I'm working on is very similar to MTT payouts, although this actually has nothing to do with MTTs, but I'm hoping somebody has solved optimal investment size for MTTs at some point or at least has some intuitions about how to do it.

Basically I have a distribution of payouts, which looks very similar to an MTT payout schedule with investment x, returns from -25x to 1000x with various probabilities (nonlinear and not fitted to any particular function) with a given exact and non-changing return/edge of Y.

I can use full kelly because I actually know exactly what my edge is, unlike poker or something where you have to use fractional kelly because you need to speculate.

To restate, the complexity arises from the fact that payouts are not binary outcomes, but rather aligned along a distribution of various payouts with various given probabilities which in sum give a net positive edge.

Based on this (or any other) given payout structure, I want to compute optimal X as a percentage of my bankroll using kelly. Any way to do this in excel with solver or crystal ball or something?

Maximize expected ln(roll) with excel solver.

I did that, but you can't take logs of negative numbers, so if some of the outcomes are negative I couldn't make it work?

Maybe re-calibrate so that the lowest negative outcome is 0 and the cost per game is higher (maximal loss)?

You can't have a negative bankroll

Yeah, I think I made some errors when I was doing the logs.

This example is very similar to crossbooking a losing player at MTTs. Most of the time you win 1 unit, sometimes you lose 2 or 5 or 10 units, sometimes you lose 100, 500 or 5000 units. Net on net, you have a small edge, and you know exactly what it is b/c you know exactly the prob distribution of hitting each outcome (unlike in an MTT).

So do you assume the bet size is 5000, your return when you lose 5000 is 0, your return when you lose 1000 is 80%, your return when you lose 2 is 99.whatever percent, and your return when you win 1 is 100.whatever percent?

I tried that with ( probability * ln ( bankroll * % return) ) summing them and then comparing to the natural log of the original bankroll, but I think I messed it up somehow because it just kept getting bigger the larger I made the bankroll # and there was no inflection point.

Let X be the portfolio, and let p_i be the probability of outcome e_i which are all in [0, N]. Suppose you bet some fraction r of your total bankroll of 1. Then the Kelly Criterion suggests finding the maximum value (for r in [0,1]) of:

E[ U_r(X) ] = sum_i p_i ln( 1-r + r*(e_i) )

To try to find the maximum wrt r, lets try to find when the derivative is 0:

d/dr E[ U_r(X) ] = sum_i p_i(e_i - 1) / (1 + r(e_i - 1) )

This is polynomial in r and can be solved with any reasonably robust calculator. Then for every root in [0,1] as well as 0 and 1, you check E[ U_r(X) ] and find the maximum.

You can get an approximation by dividing your edge by the expected squared outcome.

For example, you say the outcomes range from -25x to +1,000x. Say they are -25x with 90% probability, +25x 4% of the time, +100x 3%, +500x 2% and +1,000x 1%. Your edge is 1.5 and the expected squared outcome is 15,888. Divide and get an approximately optimal bet of 0.0094% of bankroll. The exact answer is 0.0102% of bankroll.

There's not much difference. After 10,000 plays with average luck you'll have 2.106 times your original bankroll with 0.0102% bets and 2.098 times with 0.0094%. In all practical applications there will be more important noise in the Kelly computation than that.

If your roll is 10000 and you have a 90% chance of winning 1u and a 10% chance of losing 5u, then you would be trying to maximize u in

.9*ln(10000+u) + .1*ln(10000-5u)