# Kelly criterion for 3 outcomes

I have been exploring the Kelly criterion for optimizing the bet size for a two outcome bet situation. I'm having trouble applying this to a three outcome bet. I may refer to this excellent thread: Kelly criterion with more than two outcomes, answered by David Speyer. The thing that troubles me is that the result of his answer is a single number for all three bets combined. In my mind, there should be three different "optimal" bet sizes, depending which outcome you bet on.

Say the odds for a match is the following:

Team1 wins, 1.54 - probability 65%
Team2 wins, 4.00 - probability 25%
Draw, 10.00 - probability 10%

You can only bet on one outcome, and there is only one of the outcomes that happens.

Explanation much apreciated. Thanks!

• Evidently your "odds" are what I would call 1+odds. If I bet \$1 on team2 and win the I get back \$4 which is a profit of \$3 plus my bet is returned. Right? Also the odds are too generous (too high) as this would correspond to "negative commission". And also, the odds are very close to the true win probability. A more interesting case is where there is some difference between a wagering opportunity's odds and its true probability. And, finally, I do not understand the restriction of one bet. With only 3 alternatives it is likely that only 0 or 1 bet is optimal anyway. Jul 21, 2014 at 18:31 • Why can you only bet on one outcome? Ive never seen a market where you didn't have the choice to bet on all three. Jan 23, 2015 at 15:39 ## 2 Answers You are correct that there are three different optimal bet sizes. This is due to the requirement that the gambler commits to a specific outcome of the team game. Three are three outcomes to the team game: 1. Team1 wins 2. Team2 wins 3. Draw However, for any of those outcomes, there are only two outcomes to the gambler's bankroll: 1. The gambler bet on the correct outcome of the team game • The gambler's bankroll increases by the net odds received on the wager 2. The gambler bet on the wrong outcome of the team game • The gambler's bankroll decreases by the amount wagered The Kelly criterion is only concerned with the outcomes to the gambler's bankroll. Since the gambler is required to commit to a choice on the outcome of the team game, and this choice affects the outcomes to the gambler's bankroll, this choice must be determined before the evaluation of the Kelly criterion. There are thus three optimal bet sizes, each dependent on the gambler's respective commitment to the outcome of the team game: 1. Team1 wins • Net odds received on wager:$0.54$• Probability of winning:$0.65$• Probability of losing:$0.25+0.10 = 0.35$• Fraction of the current bankroll to wager: $$f^* = \frac{(0.54)(0.65)-(0.35)}{(0.54)} \approx 0.00185$$ 2. Team2 wins • Net odds received on wager:$3.00$• Probability of winning:$0.25$• Probability of losing:$0.65+0.10 = 0.75$• Fraction of the current bankroll to wager: $$f^* = \frac{(3.00)(0.25)-(0.75)}{(3.00)} = 0.00$$ 3. Draw • Net odds received on wager:$9.00$• Probability of winning:$0.10$• Probability of losing:$0.65+0.25 = 0.90$• Fraction of the current bankroll to wager: $$f^* = \frac{(9.00)(0.10)-(0.90)}{(9.00)} = 0.00$$ In the question "Kelly criterion with more than two outcomes" (where a colored jelly bean is grabbed at random from a bag of 10 colored jelly beans), there are three possible outcomes in the gamble: 1. Black Jelly Bean: no payout (i.e. simply lose wagered amount) 2. Blue Jelly Bean: net odds received on the wager =$10$3. Red Jelly Bean: net odds received on the wager =$30\$

Each of these outcomes affect the gambler's bankroll in a distinct way from the other outcomes. However, since no choice is involved in this gamble, there is only a single optimal bet size, that is determined by the three possible outcomes to the gambler's bankroll.

This is the difference between your team game example and the jelly bean bag example: choice. A choice must be committed before the Kelly criterion can be determined.

Say in general we have $$K$$ outcomes, and each with probability $$p_k$$. And say we also have a chance of $$p_0$$ that none of them occurs.

Say we will split the bet each time according to $$q_k$$, $$\sum_{k=1}^K q_k=1-q_{0}$$, and $$q_0$$ is the fraction that we don't bet on anything. If we have $$W$$ to begin with, after a single bet, we will assume the return if outcome $$k$$ happens can be written in a form $$Wo_k(q_0, q_k)$$. Note that $$o_k(\cdot,\cdot)$$, $$k=1,\cdots, K$$, is a function of $$q_0$$ and $$q_k$$ only and generally should be monotonically increasing with respect to both of them. And $$o_0$$ should just be $$q_0$$ (we don't gain or lose if we don't bet).

Now, say we have one dollar to start with and we bet for $$N$$ times, the approximate final net-worth will be approximately

$$W = q_{0}^{Np_{0}}\prod_{k=1}^K (o_k(q_{0,}q_k))^{N p_k} = \left( q_{0}^{p_{0}}\prod_{k=1}^K (o_k(q_{0,}q_k))^{p_k}\right)^N.$$

To maximize the net-worth, we should maximize $$J=q_{0}^{p_{0}}\prod_{k=1}^K (o_k(q_{0},q_k))^{p_k}$$ with respect to $${\bf q}=q_{0},q_1,\cdots,q_K$$. Let's try to maximize the log value instead, we have to maximize

$$\ln J =p_{0 } \ln q_{0}\sum_{k=1}^K p_k \ln o_k(q_0,q_k)$$ such that $$\sum q_k=1$$ and $$q_k \ge0$$, $$\forall k$$.

Use the Lagrange multiplier method, let

$$L = p_0 \ln q_0+\sum_{k=1}^K p_k \ln o_k(q_0,q_k) + \lambda \left(\sum_{k=0}^{K}q_k- 1\right)+\sum_{k=0}^K \mu_kq_k,$$

The optimal solution should satisfy the KKT conditions as below

\begin{align*} \frac{\partial L}{\partial q_k}&=0, \forall k, \\ \mu_k &\ge 0, \forall k\\ \mu_k q_k &=0, \forall k\\ q_k &\ge 0, \forall k,\\ \sum_{k=0}^K q_k &=1. \end{align*} Consider the first equation, $$\frac{\partial L}{\partial q_k}=0$$, finally we will get

\begin{align*} \frac{p_k}{o_k}\frac{\partial o_k}{\partial q_k} + \mu_k &=\frac{p_0}{q_0}+\sum_{k'=1}^K\frac{p_{k'}}{o_{k'}}\frac{\partial o_{k'}}{\partial q_0}+ \mu_0, \forall k\ge 1 \\ \mu_k &\ge 0, \forall k\\ \mu_k q_k &=0, \forall k\\ q_k &\ge 0, \forall k,\\ \sum_{k=0}^K q_k &=1. \end{align*}

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For sanity check, take the original Kelly's criterion as an example, say with probability $$p$$, the net odd is $$b$$. So we have

$$p_1 = p$$, $$p_0 = 1-p$$, and $$o_1(q_1,q_{0})=b q_1 + 1$$.

For $$q_0,q_1 > 0$$, we have $$\mu_0=\mu_1=0$$, thus we have optimum allocation if

$$\frac{p_1}{o_1}\frac{\partial o_1}{\partial q_1}= \frac{p_0}{q_0}+ \frac{p_1}{o_1}\frac{\partial o_1}{\partial q_0}\Rightarrow \frac{bp}{b q_1 +1}= \frac{1-p}{q_0}+0\Rightarrow \frac{bp}{b q_1 +1}= \frac{1-p}{1-q_1}\Rightarrow q_1 =\frac{b p-(1-p)}{b}$$ as in the original formula.

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Now, back to the question, say we will split our bets into $$q_1$$ for Team 1, $$q_2$$ for Team 2, $$q_3$$ for draw, and $$q_0$$ for leaving behind.

we have

$$p_0=0, p_1=0.65$$, $$p_2=0.25$$, and $$p_3=0.1$$. And

$$o_1=r_1 q_1 + q_0, o_2 = r_2 q_2 + q_0$$, and $$o_3 = r_3 q_3 +q_4$$, where $$r_1=1.54$$, $$r_{2}=4$$, and $$r_3=10$$.

From above discussion, let's try to find a solution that satisfies all the above KKT conditions. Let's first assume $$q_1,q_2,q_3 > 0$$, this will mean $$\mu_1=\mu_2=\mu_3 =0$$. And hence we have

$$\frac{p_1 r_1}{r_1 q_1+q_{0} } = \frac{p_2 r_2}{r_2 q_2 + q_0} = \frac{p_3 r_3}{r_3 q_3 + q_0}=\sum_{k=1}^3 \frac{p_k}{r_kq_k +q_0}+\mu_0$$

Note that here $$r_1 p_1 =r_2 p_2 =r_3 p_3 \triangleq R$$, which is rather common in practice. Consequently, $$r_1q_{1}=r_2 q_2 = r_3 q_3$$ or $$\frac{q_1}{p_1}=\frac{q_2}{p_2}=\frac{q_3}{p_3}$$ and \begin{align} \frac{R}{r_3 q_3 + q_0}= \frac{1}{r_3q_3 +q_0}+\mu_0 \tag{1} \label{eq:qall} \end{align} Note that when $$R<1$$, $$\mu_0 <0$$ and thus we don't have a valid solution. And when $$R>1$$, $$\mu_0>0$$ and $$q_0=0$$ from the KKT conditions (more precisely, the complementary slackness condition $$\mu_0q_0=0$$) and so we should all in. And when $$R=1$$, that is what we have in this problem, $$\mu_0=0$$, and there is no constraint on $$q_0$$ as long as $$0 \le q_0 \le 1$$. This is because there is no expected gain or loss from the bet. It does not matter how much one puts in. But if we ever put in anything (say all-in), the money should be split proportional to $$p$$'s values. That is, \begin{align*} q_1=p_1=0.65\\ q_2=p_2=0.15\\ q_3=p_3=0.1 \end{align*}

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One may wonder why we will have all-in in this case. This is precisely because there is no chance of losing in the example ($$p_0=0$$). Say, if there is a chance the dealer will run away $$(p_0>0)$$, then \eqref{eq:qall} will become \begin{align} \frac{R}{r_3 q_3 + q_0}= \frac{p_0}{q_0}+\frac{1-p_0}{r_3q_3 +q_0}+\mu_0 \end{align} Note that $$q_3=(1-q_0)\frac{p_3}{p_1+p_2+p_3}=\frac{p_3(1-q_0)}{1-p_0}$$, since we expect $$q_0$$ should not be $$0$$, we expect $$\mu_0=0$$. Thus, we can solve out $$q_0$$ as $$q_0 = \frac{R p_0 }{(R-1)(1-p_0)+R p_0}$$ and $$q_k =p_k\frac{1- q_0}{1-p_0}, \forall k \ge 1.$$