The Kelly Criterion

A bet in your favour, repeated, with a fixed share of your bankroll on each one. The average bankroll rises with every increase in the stake. Watch what happens to the players who make up that average, and find the stake that is best for any one of them.

Worked example — The Kelly Criterion

Kelly's fraction	0.200
Growth per bet at Kelly's fraction	+0.02014
Share of Kelly's growth kept	100.0%
Stake where growth falls to zero	0.3894
Average gain per bet, share of bankroll	+4.00%
Average bankroll after 100 bets	50.50
Median bankroll after 100 bets	7.49
Chance of ending below the start	17.9%

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — The Kelly Criterion

The Kelly criterion chooses the share f of the bankroll to stake on a repeated bet so as to maximise the long-run growth rate g(f) = p ln(1 + bf) + q ln(1 − f), which is the expected logarithm of the bankroll's growth factor per bet. For a bet won with probability p and paying b to 1 the maximum is at f* = (bp − q)/b; at even money f* = p − q, the edge. By the law of large numbers one player's bankroll after n bets is close to e^ng(f) on a log scale, while the average bankroll is (1 + (bp − q)f)ⁿ, which grows with every increase in f.

What each symbol means

p: the chance of winning each bet; q = 1 − p
b: what one unit staked wins, on top of the stake returned
f: the share of the current bankroll staked on each bet
g(f): one player's growth rate per bet, in natural-log units

Assumes: The same bet with the same p and b, independent each time, and a bankroll that can be split as finely as the stake requires. The stake is reset to f of the current bankroll before every bet.

The average bankroll soars while the typical player goes broke 🖖

Press All in. The coin lands your way 60% of the time and pays even money, so every bet returns 20% of the stake on average, and staking everything every time makes the average bankroll as large as it can be: 1.2¹⁰⁰ = 8.28×10⁷ times the start after 100 bets. That average is carried by one outcome, winning all hundred, whose chance is 0.6¹⁰⁰ = 6.53×10⁻²³. Every other player is broke, as are all 200 in this draw. John Kelly put it in 1956: such a gambler "would be broke with probability one if he continued indefinitely". Staking less does not make the problem go away. Press Twice Kelly, 40% of the bankroll a bet: after 100 bets the average is 2,199.76, the median 0.78, and 53.8% of players are below where they started. The <a href="/stat/martingale-betting">doubling system</a> had the mirror image of this, an average that loses made of evenings that almost all win.

Every bet in your favour, and poorer with probability one 🖖

On the upper chart the dashed ghost is the log of the average's growth factor, ln(1 + 0.2f), and it rises with every bit of stake you add. The solid curve is g(f) = 0.6 ln(1 + f) + 0.4 ln(1 − f), the rate at which one player's bankroll grows, bet after bet, with probability one. It peaks at Kelly's fraction, 0.200, and is back at zero at 0.3894, 1.947 times Kelly, where every bet still gains 7.79% of the bankroll on average. Past that stake the player's fortune goes to zero with probability one, however favourable each bet; Edward Thorp states it as part (ii) of his first theorem on the subject. At even money the zero comes before twice Kelly whatever the chance of winning. Press Twice Kelly, 1,000 bets: the average bankroll is 2.65×10³³ and the median 0.0866. Near f = 0 the gap between the two curves is about half the variance of one bet's return, the half-variance drag that drawdown and volatility drag measures on real portfolios; at Kelly's fraction it is already about as large as the growth itself.

Half Kelly keeps three quarters of the growth 🖖

Press Half Kelly. Staking 0.1 of the bankroll instead of 0.2 halves every bet and keeps 74.7% of the growth rate, because g(f) is flat at its peak and the first step down from the top costs little. The table shows what the step buys: after 100 bets the chance of being below the start falls from 17.9% to 6.4%. Kelly's own result still holds, that his bettor "will, with probability one, eventually get ahead and stay ahead" of any other fixed stake, but eventually is slow: on the same 100 outcomes the Kelly bettor finishes ahead of the half-Kelly bettor 69.7% of the time, and after 1,000 bets 94.3%. Thorp's figure for half Kelly is three quarters of the growth, and at even money it never keeps more. He learned of Kelly's paper from Claude Shannon at MIT in 1960, used it at the blackjack tables, and brought it to gamblers in Beat the Dealer in 1962, noting that in practice the stake is generally adjusted down. Kelly's fraction is also the stake that maximises expected log-wealth, Daniel Bernoulli's measure from the St Petersburg game.

Problem solved in full

You think the coin is 60–40. It is 55–45. 4 steps
You estimate p = 0.6 and bet Kelly's fraction for it at even money. The true chance is 0.55. What happens to your growth rate, and what would half Kelly have done? Press Kelly's fraction first: the page gives the stake you chose and the growth you expect.

Open this problem in the tool
1. The stake you chose, and the growth you expect, from the page:
  
  $f^{*} = p - q = 0.6 - 0.4 = 0.2, g (f^{*}) = + 0.02014$
2. The true Kelly fraction is p − q = 0.10, so 0.2 is twice the right stake. The growth at the true p:
  
  $0.55 ln 1.2 + 0.45 ln 0.8 = - 0.00014$
3. Half of your Kelly, 0.1, is exactly the true Kelly:
  
  $0.55 ln 1.1 + 0.45 ln 0.9 = + 0.00501$
4. And if you had been right about 0.6, half Kelly would still have kept
  
  $\frac{0 . 6 ln 1 . 1 + 0 . 4 ln 0 . 9}{0 . 02014} = \frac{0 . 01504}{0 . 02014} = 74 . 7%$
Answer

At the true p, full Kelly grows at −0.00014 a bet: a bet with a 10% edge, staked at what you believed was the optimum, makes you poorer with probability one. Half Kelly grows at +0.00501, the best any stake can do at p = 0.55, and costs you a quarter of the growth if your estimate was right. Overestimating the chance of winning by five points is enough to put full Kelly past the zero-growth stake, which at p = 0.55 is 0.1987. That asymmetry is the usual reason for betting a fraction of Kelly.

$p = 0.55 : g (0.2) = - 0.00014 < 0 < g (0.1) = 0.00501; f_{c} = 0.1987 < 0.2$

Learning path

Fair games that ruin you

References (2)

The paper the page is named after: a gambler betting a fraction ℓ of his capital on an even-money channel symbol has growth G = q log(1 + ℓ) + p log(1 − ℓ), maximised at 1 + ℓ = 2q, that is ℓ = q − p; betting everything maximises the expected capital, "(2q)^N V₀", but "he would be broke with probability one if he continued indefinitely"; and the maximising gambler "will, with probability one, eventually get ahead and stay ahead" of one using any other fraction. Kelly's p is the chance of an error and q of a correct symbol, the reverse of the page's p and q: J. L. Kelly, Jr., A new interpretation of information rate. Bell System Technical Journal 35(4), 917–926 (July 1956). doi:10.1002/j.1538-7305.1956.tb03809.x
The practitioner's account: the coin-tossing case with g(f) = p log(1 + f) + q log(1 − f), f* = p − q, the zero-growth stake f_c with f* < f_c < 1, Theorem 1 (part (ii): if g(f) < 0 the bankroll tends to 0 almost surely; part (iv): Kelly beats any essentially different strategy), betting everything as the 'bold' criterion that makes ruin almost sure, half Kelly's growth reduced to 3/4 of the maximum in the continuous approximation, and Thorp's own history: introduced to Kelly's paper by Claude Shannon at MIT in 1960, used it in play, and introduced it to gamblers in the first edition of Beat the Dealer (1962), with stakes 'modified somewhat in practice (generally down)': E. O. Thorp, The Kelly criterion in blackjack, sports betting, and the stock market. Chapter 9 in S. A. Zenios and W. T. Ziemba (eds.), Handbook of Asset and Liability Management, Volume 1, pp. 385–428. Elsevier, 2006. doi:10.1016/S1872-0978(06)01009-X

By the 42Math team · Reviewed against published sources. Corrections welcome.