### The set of gambles with the same Kelly bet size

The Kelly criterion calculates the bet size that maximizes the long run return over repeated gambles. To better understand the relationship between bet size, gamble outcomes and the probability of winning, we explore the space of gambles having the same optimal leverage.

To do this we consider a gamble (p, a, b) which has probability of winning, p, a positive outcome, a, and a negative outcome, b. This gamble has an optimal leverage denoted by S. We now find the set of two-outcome gambles (P, A, B), whose optimal size matches that of the original gamble. As in our previous posts, the long-run return of a two-outcome gamble:

$Z = (1+Sa)^p(1+Sb)^q$

As in our previous posts, we optimize Z by optimizing log(Z):

$log(Z) = p*log(1+Sa)+q*log(1+Sb)$

To do this we take the derivative of log(Z) with respect to S and solve when it equals 0:

$\frac{pa}{1+Sa}+\frac{qb}{1+Sb}=0$

From this we get the optimal size of two-outcome gamble (p, a, b):

$S(p, a, b) = -\frac{ap+(1-p)b}{ba}$

Similarly, the optimal gamble size of  gamble (P, A, B) is:

$S(P,A,B)= -\frac{AP+(1-P)B}{BA}$

If these gambles have the same optimal bet size, they satisfy:

$-\frac{ap+b(1-p)}{ba} = -\frac{AP+(1-P)B}{BA}$

which can be rewritten as:

$\frac{P}{B}+\frac{1-P}{A}=\frac{p}{b}+\frac{1-p}{a}$

To improve our intuition about this equality we consider two examples. In the first example we fix the positive outcome and show how the negative outcome and the probability of the negative outcome change to keep the bet size constant.

1. fix the positive outcome of the gamble and determine the set of negative outcomes and probabilities of losing that have the same optimal bet size.  Here we assume $a = A$ and solve for the relationship between P and B which retain the a constant optimal bet size.

$\frac{P}{B}+\frac{1-P}{a}=\frac{p}{b}+\frac{1-p}{a}=-S$

represent P by B:

$P = -\frac{B(1+Sa)}{a-B}$

represent B by 1-P:

$B = \frac{a((1-P) -1)}{Sa+(1-P)}$

To illustrate this relationship, we select a (p = 0.5, a = 0.8, b = -0.1) and determine the size of the optimal bet, which is 4.375. We next find the set of all gambles which have the same positive outcome of 0.8 and the same optimal bet size of 4.375. We plot this set in purple in the preceding graph showing the relationship between 1-P and B.  In a similar fashion we calculate the optimal bet size for other gambles in which p= 0.5 and b = -0.1 for a range of values for the winning outcome, a. We plot each set of gambles in a different color for combinations of B and 1-P.  Since all lines trace arcs from the upper right to the lower left, this graph complies with our intuition that the lower the probability of losses, the more we can afford to lose when we lose. Comparing among different values of a, as our winning outcome gets smaller, the curvature increases. We next examine the behavior of the curves as the probabilities approach 0 and 1.

1-p approaches 1. As the probability of losing approaches to 1 the losing outcome approaches to 0. To see this we use the previously derived values of B as a function of 1-P

$B = \frac{a((1-P )-1)}{Sa+(1-P)}$

So, $\lim_{(1-P\to1)}B=0$

If the losing outcome B approaches to 0, this gamble will be tremendously attractive and the optimal bet size approaches infinity. However, at the same time, as the probability of losing approaches one, the gamble becomes one that we will definitely lose. These two factors offset each other.

1-p approaches zero. At the other extreme,

$\lim_{(1-P\to0)}B = -\frac{1}{S}$

Even as the probability of a loss approaches zero, the size of the losing outcome is bounded. This is because, as long as there is any chance of a losing outcome, we can not afford to take a loss greater than our total equity and still continue gambling.

This means that the losing outcome can never exceed -1/S. We can see the orange line as example. The optimal bet size of the orange line is 2.5 and when the probability of losing approaches to 0, the losing outcome approaches to -0.4, which is exactly -1/2.5.

In the next analysis, we consider how the two outcomes balance each other to keep a constant optimal bet size.

2. Fix the probability and determine the positive and negative outcomes that have the same optimal bet size.

From the formulas above, if we keep probability p and optimal bet size S:

$A=-\frac{B(1-p)}{p+BS}$

This is equivalent to:

$B = \frac{A((1-p )-1)}{SA+(1-p)}$

As in the first example, we calculate the optimal leverage for a gamble, in this case S(p=0.25, a = 4 , b = -1) which equals  0.0625. We then determine the set of gambles with p=0.25 and an optimal bet size of 0.0625 varying the size of the positive and negative outcomes. We plot these gambles in red in the graph above. We draw lines of different colors for sets of gambles with a fixed optimal bet size for various other values of p, the probability of winning. Notice all the curves intersect at point (A = 4 , B = -1), because this is how we generate those gambles: each line stands for gambles with same optimal bet size as gamble (a given P, A = 4, B = -1).

All lines have negative slopes so we can fix the optimal bet size and increase the magnitude of both the winning and losing outcomes.  We next consider the extreme values for B and A.

The limit when B becomes more negative. As B becomes more negative, A must increase to keep the bet size constant. Since the following relationship holds between A, B and p,

$A=-\frac{B(1-p)}{p+BS}$

we can see that for specific values of p, B and S, A can increase to infinity. Since S and p are fixed, the denominator goes to zero when

$B\rightarrow-\frac{p}{S}$

This means that B is bounded below by the value

$-\frac{p}{S}$

The limit as A and B approaches zero. As the values of A and B shrink toward zero, the value of -A/B converges to the odds of losing.  Given a optimal bet size S and a probability p: $\frac{p}{B}+\frac{1-P}{A}=S$

which can be written: $-\frac{A}{B}=\frac{1-p}{p+SB}$

So when losing outcome B approaches 0, -A/B converges to:

$\lim_{(B\to0)}-\frac{A}{B}=\frac{1-p}{p}$

which is the odds of losing in this gamble. The graph above shows this convergence. On the red line for instance, the probability of winning is 0.25 and -A/B converges to the odds of losing, (1-0.25)/0.25 = 3, as B approaches zero.  As in the previous plot, all curves intersects at point (-A/B = 4 , B = -1) since all gambles are generated by (a given P, A = 4, B = -1).

In future posts we will talk about the different statistical  characteristics for gambles that share an optimal bet size.