### Relating the three-outcome and two-outcome Kelly solutions

In our previous article, we derived the general solution for the optimal bet size of a three-outcome gamble. In this article, we will focus on the three outcome gamble whose middle outcome is 0. Adjusting for the probability of the zero-outcome, the optimal bet size for this three-outcome gamble is the same as that of a corresponding two-outcome gamble.

For a three-outcome gamble with middle outcome zero, the outcomes are $a_1$ < 0, $a_2$ = 0, $a_3$ > 0 with a positive expected value.

From our previous post, the optimal sizing of a three-outcome gamble is the solution of the equality:

$AS^2+BS+C=0$

Given the fact that $a_2$=0 we have:

$A=a_1a_2a_3=0$ ,

$B=(p_1+p_2)a_1a_2+(p_1+p_3)a_1a_3+(p_2+p_3)a_2a_3=(p_1+p_3)a_1a_3$,

$C=p_1a_1+p_2a_2+p_3a_3=$ Expected Value

So the quadratic equation becomes a linear equation.

$BS+C=0$

Solving this equation we find:

$S_3=-\frac{C}{B}=-\frac{p_1a_1+p_3a_3}{(p_1+p_3)a_1a_3}$

Dividing the numerator and denominator by $p_1$ makes

$S = -\frac{a_1+\frac{p_3}{p_1}a_3}{(1+\frac{p_3}{p_1})a_1a_3}$

This result shows two important properties of a three-outcome gamble whose middle outcome is 0:

1. The probability of outcome 0 does not impact the size of the optimal bet.
2. The optimal bet sizing only depends on the value of positive and negative outcomes and the ratio of the probability of winning and the probability of losing.

Next we compare results with those for the two-outcome gamble in the first post discussing the Kelly criterion. Suppose outcome -1 happens with probability q and outcome b happens with probability p. The optimal bet size is:

$S = \frac{bp-q}{b}$

Using the same assumptions as in our blog on two-outcome gambles, we can create the following set of variables.

$a_1=-1,a_3=b,\frac{p}{q}=\frac{p_3}{p_1}$.

Plugging these variables into the formula for the three-outcome, optimal bet we obtain:

$S = -\frac{a_1+\frac{p}{q}a_3}{(1+\frac{p}{q})a_1a_3}$

Multiplying the numerator and denominator by q we get:

$S = \frac{bp-q}{b}$

The formulas for the optimal bet size for the two-outcome gamble is the same as for the three-outcome gamble with one zero-outcome .

This completes the proof relating the three-outcome gamble to its corresponding two-outcome gamble.

An alternative way to view the zero, three-outcome gamble is to think of it as a compound gamble consisting of two gambles, the first gamble determining whether the second gamble is played.

The first gamble has two outcomes. One of the outcomes terminates the game, so that there is no return (the gambler gets his principal back). The alternative outcome specifies that that the second gamble is to be played. The second gamble is a two-outcome gamble that returns either a positive $a_3$ or negative $a_1$ amount.

We have shown above that the optimal bet size only depends on the outcomes of the second gamble. The probability associated with the first gamble is not relevant to the leverage employed.

Finally, we consider how the expected long-run return

$z=(1+Sa_1)^{p_1}(1+Sa_3)^{p_3}$

decreases as the probability of the zero-outcome, $p_2$ increases from zero to one.

We can rewrite the long-run return as

$z=((1+Sa_1)^{\frac{p_1}{p_1+p_3}}(1+Sa_3)^ {\frac{p_3}{p_1+p_3}})^{(p_1+p_3) }$

If the relative sizes of $p_1$ and $p_3$ are fixed, the optimal bet size is unaffected by the size of $p_2$, but the outer-most exponent decreases as $p_2$ increases.

Therefore, the long-run return is an exponential function of $p_1+p_3=1-p_2$ with a maximum value when $p_2$=0 and approaches one as $p_2$ approaches one. The highest expected long-run return occurs when the gamble has only two outcomes.