I'm trying to obtain a more generic version of the Kelly criterion for when we have simultaneous independent events to bet on,

I'm going to focus on the case where we just have 2 different events.

In the case where we have just one event the objective is to maximize the function:

$$f(x) := p\log(1+bx-x) + (1-p)\log(1-x)$$

$b$ are the odds received and $x$ the fraction of our total wealth that we bet.

In the case where we have two events to bet on, I suppose that the function to maximize would be of the form:

$$p_1p_2\log(1+x_1b_1+x_2b_2 - x_1-x_2) + p_1(1-p_2)\log(1+x_1b_1- x_1-x_2)+(1-p_1)p_2\log(1+x_2b_2 - x_1-x_2)+(1-p_1)(1-p_2)\log(1- x_1-x_2)$$

As it is differentiable, the problem is equivalent to find the solution of the system of equations:

$$p_1p_2\frac{b_1-1}{1+x_1b_1+x_2b_2 - x_1-x_2} + p_1(1-p_2)\frac{b_1-1}{1+x_1b_1- x_1-x_2}=(1-p_1)p_2\frac{1}{1+x_2b_2 - x_1-x_2}+(1-p_1)(1-p_2)\frac{1}{1- x_1-x_2}$$

$$p_1p_2\frac{b_2-1}{1+x_1b_1+x_2b_2 - x_1-x_2} + (1-p_1)p_2\frac{b_2-1}{1+x_2b_2 - x_1-x_2}=p_1(1-p_2)\frac{1}{1+x_1b_1- x_1-x_2}+(1-p_1)(1-p_2)\frac{1}{1- x_1-x_2}$$

However, this seems completely intractable to solve explicitly. I'm asking for your help to do it. Maybe there is some trick to solve it, or I'm not attacking it in the right way.

Thank you in advance.

  • $\begingroup$ User @RichardSantiago (unable to Comment yet) asks if by "odds" you mean "decimal odds" or "fractional odds"? He guesses that your meaning is decimal odds, where 4 : 1 is quoted as 5.00, rather than fractional odds where 4 : 1 is quoted as 4/1, because of the expression $1+bx-x$ within your definition of $f(x)$. $\endgroup$
    – hardmath
    Sep 2, 2017 at 14:30

3 Answers 3


Any elaborate answer appears to be helplessly convoluted. Nonetheless, Withrow (2007) concluded:

When the number of bets is small, the optimal sizes of bet seem to be almost exactly proportional to the Kelly stakes on individual bets. This appears to be true for all such sets of data that the author has so far examined although only one example is presented here. A proof of this hypothesis should perhaps be attempted as part of future work. At present, it is left as an interesting but tentative observation. As the number of bets increases and the sum of stakes approaches 100% of the bankroll, the optimal bet sizes are somewhat different, although roughly monotonically related to the Kelly stakes. In fact, a better approximation to the optimal size of bet for a given event in this case is to use the edge $p-\tilde p$ of that bet alone. These intriguing relationships were obtained empirically and it is unclear yet whether these or other relationships may be deduced rigorously (p. 622).

Where $p$ is the 'real' probability of a single outcome inferred from probability theory or some statistical model and $\tilde p$ is the probability of the outcome implied by its monetary return as defined by the author, based on bookmarker odds: $$\tilde p=\frac{1}{decimal\space odds}$$

Notice: $$decimal\space odds=net\space profit\space per\space unit\space stake+1 $$ $$fractional\space odds=decimal\space odds-1$$


Whitrow, C. (2007), Algorithms for optimal allocation of bets on many simultaneous events. Journal of the Royal Statistical Society: Series C (Applied Statistics), 56: 607–623. doi:10.1111/j.1467-9876.2007.00594.x

  • 1
    $\begingroup$ Given the money at stake, it's quite surprising that this problem hasn't been solved with more depth and accuracy. Or maybe it has, and people aren't willing to disclose it. $\endgroup$
    – justinpc
    Mar 27, 2019 at 10:11

For two independent events with symmetric payoff (biased coin tossing: win or lose 1 unit) you can find the answer on page 19 of Thorp's paper: http://www.eecs.harvard.edu/cs286r/courses/fall12/papers/Thorpe_KellyCriterion2007.pdf

'The Kelly criterion in blackjack sports betting, and the stock market'

Example 6.1. Suppose we bet simultaneously on two independent favorable coins with betting fractions f1 and f2 and with success probabilities p1 and p2, respectively. Then the expected growth rate is given by ... (see page 19):


Approximately proportional to p-b, spread across all bets, but always lower than the sequential, one-event optimal kelly bet. Pick the sequential kelly bet if there aren't enough bets. Where b is the implied odds and p is what you think the odds are.

Via https://vegapit.com/article/numerically_solve_kelly_criterion_multiple_simultaneous_bets

To add my intuition: We bet less than single event kelly to avoid ruin. If there are say only two bets, each with optimal sequential kelly of 0.1 (let's say they are 0.1 chance of winning a big lottery prize), you don't wanna bet half your money on each, because you could very likely lose everything. We bet proportional to the probabilistic edge because that's how much better our odds are versus the game.

Example: We have 3 independent bets

100:1 odds 0.1 chance of winning -- kelly & net odds around 0.091
500:1 odds 0.1 chance of winning - kelly around 0.098
0.1:1 odds 0.99 chance of winning (say this is a corporate bond) -- kelly around 0.890

In this case we bet about 9% of our money on each bet (slightly more for the higher implied odds bet), and the rest in the "risk free" (almost) bond.

I'm pretty sure the actual odds and payouts of the bets if they're say, option contracts, are gonna be rough estimates anyway, so a rough estimate of the optimal bet would probably suffice.

Otherwise you could do a numerical regression with gradient descent like described here since the function is convex https://vegapit.com/article/numerically_solve_kelly_criterion_multiple_simultaneous_bets


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