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As the title suggests, I want to calculate the Kelly criterion for a bet with 3 possible outcomes: win, fractional (partial) loss, or full loss.

Here is the proof of the Kelly criterion for a bet with 2 possible (binary) outcomes (win or lose) from the proof section of the Kelly criterion Wikipedia article:

Heuristic proofs of the Kelly criterion are straightforward. The Kelly criterion maximizes the expected value of the logarithm of wealth (the expectation value of a function is given by the sum, over all possible outcomes, of the probability of each particular outcome multiplied by the value of the function in the event of that outcome). We start with 1 unit of wealth and bet a fraction $f$ of that wealth on an outcome that occurs with probability $p$ and offers odds of $b$. The probability of winning is $p$, and in that case the resulting wealth is equal to $1+fb$. The probability of losing is $q=1-p$, and in that case the resulting wealth is equal to $1-fa$. Therefore, the expected geometric growth rate $r$ is $$r=(1+fb)^p\cdot(1-fa)^q$$ We want to find the maximum r of this curve (as a function of f), which involves finding the derivative of the equation. This is more easily accomplished by taking the logarithm of each side first. The resulting equation is: $$E=\log(r)=p\log(1+fb)+q\log(1-fa)$$ with $E$ denoting logarithmic wealth growth. To find the value of $f$ for which the growth rate is maximized, denoted as $f^\ast$, we differentiate the above expression and set this equal to zero. This gives: $$\frac{dE}{df}\rvert_{f-f^\ast}=\frac{pb}{1+{f^\ast}b}+\frac{-qa}{1-{f^\ast}a}=0$$ Rearranging this equation to solve for the value of $f^\ast$ gives the Kelly criterion: $$f^\ast=\frac pa-\frac qb$$ Notice that this expression reduces to the simple gambling formula when $a=1=100$%, when a loss results in full loss of the wager.


Here is as far as I've gotten trying to calculate the Kelly criterion for a bet with 3 possible outcomes: win, fractional (partial) loss, or full loss:

We start again with 1 unit of wealth and bet a fraction $f$ of that wealth on an outcome that occurs with probability $p$ and offers odds of $b$. The probability of winning is $p$, and in that case the resulting wealth is equal to $1+fb$. The probability of a fractional loss is $q$, and in that case the resulting wealth is equal to $1-fa$. I'll call the probability of a full loss $l$, and in that case the resulting wealth is equal to $1-f$. Therefore, the expected geometric growth rate $r$ is:

$$r=(1+fb)^p\cdot(1-fa)^q\cdot(1-f)^l$$

where $p+q+l=1$.

Following the same steps above, I want to find the maximum r of this curve (as a function of f), by taking the logarithm of each side first. I think (but I'm not sure) the resulting equation would be:

$$E=\log(r)=p\log(1+fb)+q\log(1-fa)+l\log(1-f)$$

with $E$ denoting logarithmic wealth growth. To find the value of $f$ for which the growth rate is maximized, denoted as $f^\ast$, I want to differentiate the above expression and set this equal to zero. Once again, I think (but I'm not sure) this would give:

$$\frac{dE}{df}\rvert_{f-f^\ast}=\frac{pb}{1+{f^\ast}b}+\frac{-qa}{1-{f^\ast}a}+\frac{-l}{1-{f^\ast}}=0$$

From here I am not sure how to rearrange this equation to solve for the value of $f^\ast$ and give the Kelly criterion.

I have 2 questions:

  1. How can this equation be rearranged to solve for the value of $f^\ast$ and give the Kelly criterion for a bet with 3 outcomes: win, fractional loss, or full loss?

  2. The accepted answer to a similar question (Kelly criterion with more than two outcomes) indicates that there is not a simple formula for the root of that equation. Is there a simple formula for the root of the above equation that I am trying to solve?

Thank you in advance for any help you can offer!

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    $\begingroup$ Multiplying through by $(1+f^* b)(1-f^* a)(1-f^*)$ will give a quadratic equation for $f^*$, meaning the solution can be expressed through the quadratic formula. $\endgroup$ Jun 6, 2023 at 21:10
  • $\begingroup$ @eyeballfrog thanks for the response. Unfortunately it's been quite some time since I've worked with the quadratic formula. Can you please share what the quadratic equation for $f^\ast$ would look like? $\endgroup$
    – Iceman6425
    Jun 7, 2023 at 17:26

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