On a 60% Coin the Right Bet Is 20%, and 40% Turns a Winning Game Into a Losing One

Here is the question, near enough to the words an interviewer uses: a coin comes up heads sixty percent of the time, and you are told that. You start with twenty five dollars and you get three hundred flips. On each flip you bet whatever you like on heads at even money, so a correct call wins your stake and a wrong one takes it. What fraction of your money do you put on each flip?

It gets asked on trading and market making seats, usually in a middle round once the arithmetic screens are behind you and somebody senior wants to hear you reason about risk out loud. Risk and portfolio research desks use it too, as a warm-up before questions about position limits. The interviewer knows the number already. What they are listening for is whether you notice that the expected value is pointing somewhere useless, and whether you say what your answer assumes before they have to ask.

This has been played for real money. Sixty one people, mostly students of economics and finance or young professionals at investment firms, were each handed twenty five dollars and half an hour with a coin biased sixty to forty. Twenty eight percent of them finished under two dollars. Eighteen of the sixty one put everything on a single flip at some point. Forty one of the sixty one bet on tails at some point, on a coin they had been told lands heads six times in ten. Five of the sixty one had ever heard of the rule below.

Start with the expectation, and watch it fail

Bet a fraction $f$ and one flip multiplies your money by $1+f$ with probability $p$ and by $1-f$ otherwise. The expected multiplier is $p(1+f) + (1-p)(1-f) = 1 + f(2p-1)$, and because flips are independent the expectation compounds:

With $p > \tfrac12$ that is strictly increasing in $f$. So if the objective really is expected final wealth, the calculus has a clear opinion: put everything on every flip. At $f = 1$ the expected multiplier over three hundred flips is a number with twenty four digits, and it is honestly computed.

Now look at where that number lives. Betting everything means one tail ends the game, so surviving three hundred flips requires three hundred heads:

The identity is exact, and it says something blunt. The single all-heads path carries the entire expectation. Delete that one branch and the expected wealth of the all-in strategy is exactly zero. The average was never wrong. It just describes a player who does not exist.

What survives three hundred multiplications

Your money is not built by adding and subtracting, it is built by multiplying, and multiplication does not care about order. If $K$ of the $n$ flips come up heads, then whatever order they arrived in,

Two multipliers and a count. Nothing else about the sequence matters at all. Take logarithms to turn the product into a sum, which is the one move that lets the law of large numbers reach the problem, and divide by $n$:

The log multipliers are independent, identically distributed and bounded whenever $f < 1$, so the strong law applies to them directly. The head frequency $K/n$ goes to $p$ almost surely, and with it:

One number per fraction. It is worth being precise about what it does and does not say. This is an almost sure statement about a limit, and no simulation can establish it, because a finite run cannot distinguish a limit from a long detour. It is the strong law that does the work, and the boundedness of $\ln(1\pm f)$ for $f<1$ is the hypothesis that lets it in.

One derivative, and the answer is the edge

Write $q = 1-p$ and differentiate (4):

Set it to zero. Then $p(1-f) = q(1+f)$, so $p - q = f(p+q) = f$, which is as clean as a first-order condition gets:

The fraction to bet is exactly the edge. Nothing more elaborate is going on. And the stationary point is the maximum rather than some other critical point, because

everywhere on $(-1,1)$, so $g$ is strictly concave and the maximum is unique. At $p = \tfrac35$ formula (6) returns $f^\star = \tfrac15$ exactly, and the growth rate there is

Compounded over three hundred flips that is a factor of $e^{300g} = 420.17$, so twenty five dollars becomes about ten thousand five hundred. Four hundred and twenty times your money, out of a coin that leans your way by ten percentage points.

Double it, and the growth rate goes negative

Take the right answer and be twice as confident. Same coin, same edge, same three hundred flips, forty percent of your money on the table instead of twenty:

A favourable game turned into a losing one by bet size alone. Nobody changed the coin.

Where exactly does it turn? Set $g(f) = 0$. With $p = \tfrac35$, multiplying by five gives $3\ln(1+f) + 2\ln(1-f) = 0$, so

The root at $f=0$ is the player who does not bet. The one that matters is the quartic root in $(0,1)$, and it sits at $f_0 = 0.3893906833$. It is algebraic of degree four with no closed form worth writing down, and, more to the point, it is below double. The game turns against you a whisker before you have doubled the correct bet, not at the doubling.

The shape around the peak is the part worth carrying out of the room. Bet half of what you should, ten percent instead of twenty, and $g(\tfrac1{10})/g(\tfrac15) = 0.747$: three quarters of the growth rate survives. Bet double and you keep less than nothing. Timidity costs a quarter of the winnings, boldness costs the entire proposition, and the curve knew that before you did.

What losing actually looks like over three hundred flips

Here is where the usual telling of this overreaches, so it is worth being careful. Over three hundred flips, betting forty percent is not a loss you can watch. The drift against you is tiny and the noise is not, so about forty eight percent of players still finish ahead, and it takes something like forty seven thousand flips before losing is the obvious outcome. Anybody who tells you that doubling the bet produces a visible bleed over three hundred flips has not run it.

The statement that is both exact and damning is about the median. Wealth in (3) is strictly increasing in $K$, so the median wealth is the wealth at the median head count. And $K \sim \mathrm{Bin}(300, \tfrac35)$ has $np = 180$, an integer, which puts the binomial median exactly there. So the middle run of the pack is the run with exactly one hundred and eighty heads:

The two expressions coincide because $180/300 = \tfrac35$ exactly, which makes the median an exact value here rather than an approximation. That agreement is specific to three hundred flips and breaks at two hundred and ninety nine or three hundred and one. Evaluate it at both fractions:

Twenty five dollars handed back as twelve. Same coin, same edge, same luck, one decision changed. So it is not a loss you observe over three hundred flips, it is a loss you can prove, and that gap between provable and observable is precisely what makes it dangerous. The player who overbets gets a run of results that looks survivable and a position that is not.

The cleanest version of the same divergence needs no bias at all. Flip a fair coin: heads your money doubles, tails it halves. The expected multiplier is $\tfrac12 \cdot 2 + \tfrac12 \cdot \tfrac12 = \tfrac54$ per flip, which compounds to $(5/4)^{300}$ over the session. Yet $2 \cdot \tfrac12 = 1$, so any run with equal counts ends exactly where it started, and the middle of the pack ends at the money it began with. Both numbers are correct at once, on the same coin, at the same time.

Mean minus half the variance, and where it stops being true

Let the bets get small and frequent instead of large and rare, which is closer to what a real position looks like, and (4) collapses into the formula every textbook hands you. Writing $X$ for the return of one bet, so $X = \pm f$, and expanding $\ln(1+x) = x - \tfrac{x^2}{2} + O(x^3)$:

The first term is your edge. The second is the price of bouncing around, subtracted whether you like it or not, and it is the reason a fund can post a healthy average return every year and still hand its investors back less than they put in. The same half-variance term shows up with its sign flipped when you go the other way, which is why the average of an exponential sits above the exponential of the average rather than on top of it.

Now the part almost nobody tells you. The familiar form of (11) is $\mu - \tfrac12\sigma^2$, with the variance in the second slot rather than the second moment. Those are not the same thing here:

In continuous time the gap vanishes and the two agree, which is where $f^\star = \mu/\sigma^2$ comes from. For this discrete coin it does not vanish, and the mean-variance curve is $0.2f - 0.48f^2$, whose maximum sits at $0.2/0.96 = 0.2083$ rather than at $0.2$. A four percent error in the answer, which you would never notice.

Apply the same approximation at the fraction we just condemned and it returns $+0.0032$. Positive. It says doubling your bet makes you money, when the exact value is $-0.0024469$. Not a small error, the opposite sign. Anyone holding both the approximation and the warning about forty percent is holding two statements that contradict each other, and it is the approximation that is wrong.

A truncated expansion is faithful where the discarded terms are small, and the terms it discards grow like $f^3$. It stops being faithful right about where the curve turns over. That is the difference between using a formula and knowing what it costs you.

The one number the formula trusts completely

Everything above assumed $p$ is known exactly. Six in ten, handed over as a fact. Nobody has ever handed anyone that.

Suppose you believe six in ten, bet your twenty percent, and the coin is really five and a half in ten. Still favourable. Still an edge:

Three hundred flips and you end up where you began. Believing 0.60 when the coin is 0.55 has spent the entire advantage on bet size, and nothing about the run would have told you. Now halve the fraction and repeat the experiment: betting ten percent against the same true five and a half in ten gives $+0.00501$ a flip, which is $\times 4.49$ over three hundred flips. And if the coin really was six in ten, ten percent still returns $0.747$ of the growth rate, about ninety one times your money instead of four hundred and twenty.

So halving costs you a large multiple when you are right about the coin, and rescues the whole exercise when you have overestimated it by five points. That asymmetry, and not caution as a personality trait, is why people who size positions for a living sit below the line. They are buying insurance against being wrong about the one input the formula never questions.

Maximising the log is a choice, not a law

Everything above maximised the expected logarithm of wealth. That is an objective somebody picked, and it is worth knowing why. Kelly, who wrote the growth rate down in 1956 while studying the capacity of a noisy communication channel, was explicit that his gambler maximises the log for a reason that has “nothing to do with the value function which he attached to his money, but merely with the fact that it is the logarithm which is additive in repeated bets and to which the law of large numbers applies.” The log is in the argument because products become sums, not because anyone measured how much people enjoy money.

What is proved about it is narrower than the folklore. Breiman established in 1961 that this strategy maximises the asymptotic growth rate and minimises the expected time to reach a receding target, and wrote in the same paper that “in the finite case, it is suboptimal”. Three hundred flips is a finite case: full size beats half size only about eighty one percent of the time over that horizon. The threshold behind Fig. 1, that wealth goes to infinity almost surely below the critical fraction and to zero above it, is due to Dubins and Savage, and Breiman credits it to them. It is not Kelly’s result and it is worth attributing correctly if the question comes back at you.

Then there is the objection, which is published and deserves better than a footnote. Samuelson granted the theorem and refused the conclusion: “He who acts in N plays to make his mean log of wealth as big as it can be made will, with odds that go to one as N soars, beat me who acts to meet my own tastes for risk. Who doubts that? What we do doubt is that it should make us change our views on gains and losses.” His charge is that maximising mean log of wealth describes “one odd (thin!) point on the line of all the tastes for risk”, and that winning with probability approaching one says nothing about what the losses are worth to you. He also claimed, in the closing line of that paper, to have argued the whole case in prose of but one syllable save for the last word.

The list of things the model quietly grants is short and every item is load-bearing: $p$ known exactly, money infinitely divisible, a bet compulsory every round, no cap on winnings, no minimum stake, flips independent. Move any of them and the answer moves. The real-money experiment above put a two hundred and fifty dollar cap on the account, and under a cap with a target, a constant ten percent reaches it more often than twenty percent does, so twenty percent is not the optimum of that game. Which is the honest reading of the whole result: it answers the question it was asked and not a neighbouring one.

What survives all of the caveats is one sentence, and it is the sentence to say out loud in an interview. Betting more than $f^\star$ is dominated. It lowers the growth rate and raises the risk at the same time, so no set of preferences makes it right for anybody. Every remaining argument is about how far below the line you should sit.

Sources and further reading

• J. L. Kelly Jr., “A New Interpretation of Information Rate”, Bell System Technical Journal 35 (1956), 917–926 (doi). The setting is a communication channel, not a casino, and the phrase “Kelly criterion” appears nowhere in it.
• L. Breiman, “Optimal Gambling Systems for Favorable Games” (1961), 65–78. Where the asymptotic optimality is proved, where the finite-case caveat is stated, and where the zero-growth threshold is credited to Dubins and Savage.
• P. A. Samuelson, “Why we should not make mean log of wealth big though years to act are long”, Journal of Banking and Finance 3 (1979), 305–307 (doi). The objection quoted above, in full.
• Ziemba, “A Response to Professor Paul A. Samuelson’s Objections to Kelly Capital Growth Investing” (2016). The reply, for the other half of the argument.
• V. Haghani and R. Dewey, “Rational Decision-Making Under Uncertainty: Observed Betting Patterns on a Biased Coin” (2016), arXiv:1701.01427. The sixty one participants, the twenty five dollars and the two hundred and fifty dollar cap. It reports no proportion of players who sized correctly, so nobody should quote one.
• Dinis, Unterberger and Lacoste, “Phase transitions in optimal betting strategies”, Europhysics Letters 131 (2020) 60005. The same criterion seen as a phase transition, if you want the physics reading.
• Background: Law of large numbers and Binomial distribution, whose median is what makes equation (10) exact.

On how the numbers here were checked. The calculus and the exact arithmetic went through a computer algebra system, one isolated run per claim. Separately, a simulator plays the game from the rules and never calls $g$ at all, replaying one fixed block of flips at every fraction so that a change in the picture is a change in bet size and never in luck; its geometric mean tracks $e^{300g}$ to within 1.9 percent at every fraction, and it was mutation tested. A second derivation redid the optimum in exact rational arithmetic before seeing any of the above, and a third pass was told to break every claim. One honest limit is worth stating: simulation cannot confirm the expectations at large $f$. At $f=1$ the empirical mean is zero against a closed form of $1.42\times10^{25}$ dollars, and at $f = 0.4$ six different seeds spread the estimate over a factor of one hundred and fifty. Those figures rest on exact arithmetic alone, which is the whole reason equation (2) is written as an identity rather than measured.