Kelly Sizing in a Coin-Flip Game

You have a coin that lands heads with probability 0.6, tails with probability 0.4. You start with $100 and can bet any fraction of your current bankroll on each flip, repeated many times: win doubles your bet, lose forfeits it. What fraction of your bankroll should you bet each time to maximize your long-run growth rate?

Have a guess before reading on — most people's instinct ("bet a lot, you have an edge!") and the correct answer are further apart than they expect.

Why maximizing expected value per bet is the wrong target

The tempting approach is to maximize expected bankroll after one flip: betting your entire bankroll each time has the highest expected value per flip (since $0.6 \times 2 - 0.4 \times 1 >$ betting nothing), but bet everything repeatedly and you go broke with probability 1 the first time you lose — which happens 40% of the time on flip one alone. Expected value per flip and long-run survival are different objectives, and over many repeated flips, the quantity that actually matters is the expected growth rate of your bankroll — the expected value of the logarithm of your wealth, not the expected value of the wealth itself. Log-growth is what compounds; a single huge expected-value bet that also has a real chance of ruin compounds to zero.

The Kelly formula

For a bet that wins with probability $p$ and pays even money ($1 won per $1 bet, wagering fraction $f$ of the bankroll), the expected log-growth per flip is:

$g(f) = p \ln(1+f) + (1-p)\ln(1-f)$

In words: with probability $p$ your bankroll multiplies by $(1+f)$, with probability $1-p$ it multiplies by $(1-f)$, and you're averaging the logarithms of those multipliers because logs of compounding factors add up over repeated flips. Maximizing this over $f$ (take the derivative, set to zero) gives the famous Kelly fraction:

$f^* = p - (1-p) = 2p - 1$

For even-money bets, the optimal fraction is simply your edge — the amount your win probability exceeds 50%, doubled.

Solving the coin game

With $p = 0.6$: $f^* = 2(0.6) - 1 = 0.2$. Bet 20% of your current bankroll on every flip.

Check the growth rate at this fraction versus a naive alternative. At $f=0.2$:

$g(0.2) = 0.6\ln(1.2) + 0.4\ln(0.8) = 0.6(0.1823) + 0.4(-0.2231) = 0.1094 - 0.0893 = 0.0201$

So your bankroll grows by a factor of $e^{0.0201} \approx 1.0203$ per flip on average — about 2% per flip, compounding. Compare to betting 50% of the bankroll each time: $g(0.5) = 0.6\ln(1.5)+0.4\ln(0.5) = 0.6(0.405)+0.4(-0.693) = 0.243 - 0.277 = -0.034$ — a negative growth rate, meaning that despite having a real edge on every single flip, betting too large a fraction guarantees your bankroll shrinks to zero over time. That's the entire lesson of the puzzle: having an edge is necessary but nowhere near sufficient for a bet size to be safe.

A second example: unequal payout odds

Same 0.6-probability coin, but now a win pays 3-to-1 (bet $1, win $3) instead of even money. What's the Kelly fraction?

The general Kelly formula for win probability $p$, loss probability $q=1-p$, and payout odds $b$-to-1 (win $b$ per $1 risked) is:

$f^* = \frac{bp - q}{b} = p - \frac{q}{b}$

Here $b=3$, $p=0.6$, $q=0.4$: $f^* = 0.6 - 0.4/3 = 0.6 - 0.1\overline{3} = 0.4\overline{6}$. Bet about 46.7% of the bankroll. Notice the better payout odds justify a much larger fraction than the even-money case's 20% — the formula scales the bet with both how often you win and how much you win when you do.

Where this generalizes

Kelly sizing is the standard answer to "how big should I bet/size a position given an edge and repeated opportunities," and it shows up directly in position-sizing discussions for any strategy with a known or estimated edge and payout structure — not just coin flips, but option strategies, market-making inventory limits, and portfolio leverage. The reusable insight is to always separate two questions that feel like one: "do I have an edge" (a sign question) and "how much should I risk on it" (a magnitude question, answered by maximizing log-growth, not raw expected value).