### The Kelly criterion for three-outcome gambles

By Elliot Noma, Yu Bai

Optimal bet sizing for a three outcome gamble can an be solved using an extensions of the two-outcome analysis for the Kelly criterion.

We assume that gambles have three outcomes: a_{1,} a_{2} and a_{3} with known probabilities p_{1}, p_{2} and p_{3}. We want to determine the amount of our portfolio to invest in each gamble to maximize the long term cumulative outcomes:

.

To avoid degenerate solutions, we assume that the expected value of the gamble is positive and at least one of the outcomes is negative.

To solve the optimal leverage when wagering on three-outcome gambles, we use an extension of the standard analysis for two-outcome gambles. To find the S that maximizes z, we log transform both sides of the equality and set the first derivative to zero:

or

This equality may be written in the form AS^{2} + BS + C = 0 where

,

,

Expected Value

The two solutions for S are

and

.

However, only S_{–} maximizes the long run expected return. To see this, we rewrite the quadratic equation and then consider the two cases when A < 0 and A > 0. We also make use of the fact that C equals the expected value of the gamble and we restrict our strategies to gambles with a positive expected value.

First, applying the fundamental theorem of algebra, we rewrite the quadratic equation as the product of two factors times the leading coefficient:

which becomes

.

We consider the two cases A < 0 and then A > 0.

Case 1: A < 0

Since , C > 0 and A < 0, we know that one root is negative and one root is positive. However, we know that only the positive root is relevant to our solution and by examining the formulas we can see that S_{–} > S_{+}. Therefore, when A < 0, is the solution.

Case 2: A > 0

In this case since A > 0. We know that there is at least one positive root, so both roots must be positive. We also know that only the smaller root is the solution and by examining the formulas for S_{–} and S_{+ }we see that . To see that the smaller root is the solution, consider the second derivative of log(z):

The second derivative is always negative in the range 0 to . This means that there is only one maximum value in this range and since the two roots are positive, only the lowest one can be in this range. So when A > 0, is the solution.

As in the two-outcome gamble, we know that the optimal leverage increases as the expected value increases, the minimum return outcome moves closer to zero, or the highest outcome increases in value. To get a more general feel for this optimization we fix the three randomly-chosen outcomes and vary the probabilities for each outcome. The following graphs show how the leverages and optimal returns are affected by changing the probabilities of the outcomes.

In this set of graphs there are three outcomes, two that are negative and one that is positive. At the top of the triangle the first outcome, which is negative, has probability of one, so the expected return from the gamble is negative and the optimal solution is to not bet. Similarly, the lower right vertex occurs when the third outcome, which is also negative, has probability one. Here too, the optimal strategy is to not bet. Only when the second outcome, denoted by B, has a high probability, does the optimal strategy employ betting on the gamble. The more probable B is, the higher leverage and the higher the long-run expected return.

The next pair of graphs show how the leverage and long-run return changes for a different set of outcomes. In this case, there are two positive outcomes. Outcomes B and C are positive and increasing the probabilities of either motivates a more levered strategy and higher returns. The optimal strategy is not to bet only when the negative outcome, A, has a high probability.

To see how the three-outcome and two-outcome solutions relate to each other, see our post.