The Kelly Criterion in Quant Interviews

The Kelly question usually arrives right after you have priced a biased coin. The interviewer nods at your expected value, then asks the question that actually separates candidates: you can play this coin repeatedly, so how much of your bankroll do you bet each flip? Plenty of people can recite a formula. The interviewer is listening for whether you know where it comes from, what it assumes, and where it breaks, because that is the difference between memorizing a result and understanding sizing.

The setup, and why logs

You hold a bankroll and face a repeatable bet. Define the pieces precisely, because interviewers grade the setup as much as the answer:

• $f$ is the fraction of your current bankroll you stake each round
• $p$ is the probability of winning, and $q = 1 - p$ the probability of losing
• $b$ is the payoff in b-to-1 terms: a winning $1 stake returns the stake plus $b$ dollars of profit

After one round your wealth is multiplied by $1 + fb$ on a win or $1 - f$ on a loss. Over many rounds, wealth is a product of these factors, and the logarithm turns that product into a sum. By the law of large numbers, the average of those log factors converges to its expectation, so what controls your long-run compounded growth is the expected log growth per bet:

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

Why not just maximize expected wealth? Because expectation is linear: the one-round expected multiple is $1 + f(pb - q)$, which increases in $f$ whenever the edge $pb - q$ is positive. The expected-value maximizer therefore stakes everything, every round, and goes broke at the first loss, which arrives with probability one if you play long enough. The log is not a stylistic preference. It is the objective that matches multiplicative compounding and prices in ruin.

The formula in one line of calculus

Differentiate $g$ and set it to zero:

$$g'(f) = \frac{pb}{1 + fb} - \frac{q}{1 - f} = 0$$

Cross-multiply: $pb(1 - f) = q(1 + fb)$, so $pb - pbf = q + qbf$, which collects to $pb - q = bf(p + q) = bf$. Hence

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

Every symbol again: $f^*$ is the growth-optimal stake fraction, $p$ the win probability, $q = 1 - p$, and $b$ the net odds per unit staked. The numerator $pb - q$ is your edge, the expected profit per dollar staked. The denominator is the odds. Kelly is edge over odds, and that phrase is worth saying out loud in the interview. The sanity checks interviewers like to hear: zero edge means $pb = q$ and $f^* = 0$; as $b$ grows huge, $f^* \to p$; and a negative $f^*$ means do not bet.

A fully worked example: the 60/40 coin at even money

Take $p = 0.6$, $q = 0.4$, $b = 1$. Then $f^* = 0.6 - 0.4/1 = 0.2$: bet 20% of bankroll each flip. Now compute the growth rate at half Kelly, full Kelly, and double Kelly, showing every step:

$$g(0.1) = 0.6 \ln 1.1 + 0.4 \ln 0.9 = 0.6(0.09531) + 0.4(-0.10536) = 0.05719 - 0.04214 \approx 0.0150$$ $$g(0.2) = 0.6 \ln 1.2 + 0.4 \ln 0.8 = 0.6(0.18232) + 0.4(-0.22314) = 0.10939 - 0.08926 \approx 0.0201$$ $$g(0.4) = 0.6 \ln 1.4 + 0.4 \ln 0.6 = 0.6(0.33647) + 0.4(-0.51083) = 0.20188 - 0.20433 \approx -0.0024$$

Read the table the way a desk would, using the median wealth multiple after 100 flips, which is $e^{100g}$:

Stake	Growth $g(f)$ per flip	Median wealth after 100 flips
Half Kelly, $f = 0.1$	$+0.0150$	about 4.5x
Full Kelly, $f = 0.2$	$+0.0201$	about 7.5x
Double Kelly, $f = 0.4$	$-0.0024$	about 0.78x

Three things deserve emphasis. Half Kelly keeps roughly 75% of the growth ($0.0150 / 0.0201 \approx 0.75$) while halving the stake and dramatically shrinking the swings. Double Kelly does not merely grow slower: its growth rate is negative, so the typical path loses about a quarter of a percent per flip and is down roughly 22% after 100 flips, on a coin where you win 60% of the time. And the curve keeps falling: here $g$ crosses zero near $f \approx 0.39$, just shy of double Kelly, and at $f = 1$ a single loss is ruin, so $g(1) = -\infty$. Meanwhile the arithmetic expected value $1 + 0.2f$ kept rising the whole way. That divergence between expectation and growth is the entire lesson. Overbetting a winning coin destroys you in the precise sense that your wealth's typical trajectory compounds downward while its average is dragged up by vanishing lottery paths.

Why interviewers love this question

One prompt tests four skills at once. It tests expected value, because you must extract $pb - q$ cleanly before anything else. It tests comfort with logs and concavity, because the candidate who cannot explain why log wealth is the right objective is pattern-matching. It tests risk, because the overbetting cliff is a statement about ruin, not about preferences. And it tests practical judgment, because the strong answer ends with what you would actually do, which is never full Kelly on an estimated edge. It also maps directly onto market-making: when you quote a market you are choosing a size as well as a price, and a quote sized past your edge is the double-Kelly column wearing work clothes.

The variants you should expect

Different payoff odds

The formula $f^* = p - q/b$ moves fast when $b$ moves. A fair coin paying 2-to-1 gives $f^* = 0.5 - 0.5/2 = 0.25$ despite zero probability edge: the edge lives in the odds. The breakeven win probability at b-to-1 is $1/(b+1)$, one third for $b = 2$, and quoting that line shows you can invert the formula, not just apply it.

Several simultaneous independent bets

Offered ten independent 60/40 coins at once, you do not bet 20% on each. The right intuition: splitting a fixed total stake across independent bets keeps the expected value and cuts the variance, and lower variance means less drag on log growth, so the optimal total deployed rises above single-bet Kelly while the stake per coin falls below it. Interviewers want the direction and the diversification logic, not an exact solution.

Correlated bets

Correlation removes the diversification that justified the larger total. Ten perfectly correlated bets are one bet at ten times the size, so the per-bet fraction must divide by ten. Ten tickers driven by one factor are one bet wearing ten hats, and sizing them as independent is overbetting in disguise.

Edge uncertainty, the real argument for fractional Kelly

You never know $p$; you estimate it. Errors are asymmetric in consequence: underbetting costs a sliver of growth, overbetting walks you toward the cliff. Suppose you stake 0.20 believing the coin is 60/40, but the truth is 55/45. The true optimum is $0.55 - 0.45 = 0.10$, so you are unknowingly at double Kelly: $g = 0.55 \ln 1.2 + 0.45 \ln 0.8 = 0.55(0.18232) + 0.45(-0.22314) = 0.10028 - 0.10041 \approx -0.0001$. Five points of overconfidence converted a strong edge into zero growth. Since estimation error only ever threatens you from one side, professionals shade down, and half Kelly is the standard compromise.

Where candidates lose the thread

Maximizing expected value instead of growth. The most common failure: computing $1 + f(pb - q)$, noticing it increases in $f$, and concluding you should bet big. On the 60/40 coin the all-in bettor survives ten flips with probability $0.6^{10} \approx 0.006$. If your sizing logic implies a 99.4% chance of ruin inside ten rounds of a great bet, the logic is wrong.

Running Kelly on a negative edge. Offered a 45% coin at even money, the formula says $f^* = 0.45 - 0.55 = -0.10$. That minus sign is the answer: stake zero, and if the structure lets you take the other side, recompute with the other side's $p$ and $b$. Announcing a 10% bet after computing a negative fraction is an instant red flag.

Fuzzy bankroll definitions. The $f$ in Kelly is a fraction of capital you can genuinely lose and still keep playing. It is not gross notional under leverage, not money you cannot redeploy, and on a desk it is the risk capital allocated to the strategy, not the firm's balance sheet. Kelly is also a repeated-game answer; for a one-shot decision the question becomes utility and risk tolerance, and saying so is a mark of maturity, not evasion.

Practice set

Question 1. A bet pays 3-to-1 and wins 30% of the time. What is the Kelly stake?

Answer. $f^* = p - q/b = 0.30 - 0.70/3 = 0.30 - 0.2333 = 0.0667$, one fifteenth of bankroll. Check it as edge over odds: $(0.30 \times 3 - 0.70)/3 = (0.90 - 0.70)/3 = 0.20/3 \approx 0.0667$.

Question 2. You have a 55/45 coin at even money and your desk mandates half Kelly. What do you stake, and what share of full-Kelly growth do you keep?

Answer. $f^* = 0.55 - 0.45 = 0.10$, so half Kelly stakes 5%. Growth at the two stakes: $g(0.05) = 0.55(0.04879) + 0.45(-0.05129) = 0.02683 - 0.02308 = 0.00375$, and $g(0.10) = 0.55(0.09531) + 0.45(-0.10536) = 0.05242 - 0.04741 = 0.00501$. The ratio is $0.00375 / 0.00501 \approx 0.75$: half the size, three quarters of the growth.

Question 3. A counterparty offers you 5-to-1 against an event you believe happens 15% of the time. Do you back it, and at what size?

Answer. $f^* = 0.15 - 0.85/5 = 0.15 - 0.17 = -0.02$, so you do not back it: the breakeven probability at 5-to-1 is $1/6 \approx 16.7\%$, above your 15%. But if you can take the other side, you risk 5 to win 1, so $p = 0.85$ and $b = 0.2$, giving $f^* = 0.85 - 0.15/0.2 = 0.85 - 0.75 = 0.10$: risk 10% of bankroll laying the event. A price that is wrong one way is an opportunity the other way, which is the market-making instinct in one line.

FAQ

Do interviewers expect the full derivation?

They expect the one-line calculus on request and fluency without it. Edge over odds plus the sanity checks, $f^* = 0$ at zero edge, $f^* \to p$ as $b$ grows, never stake on a negative edge, covers most interviews. What gets candidates rejected is quoting the formula and then being unable to say why log wealth is the objective.

Is half Kelly always better than full Kelly?

It is not better, it is a different point on a tradeoff. Half Kelly keeps about 75% of the growth with far smaller drawdowns, and in the continuous approximation a full-Kelly bettor has a 50% chance of seeing half the bankroll gone at some point. With a perfectly known edge, full Kelly maximizes growth by definition. Since edges are estimated and overbetting is the catastrophic side, fractional Kelly is the professional default. The interview answer is to articulate that tradeoff, not to pick a slogan.

Does Kelly apply to markets, or just coin flips?

The continuous analogue invests $f^* = \mu / \sigma^2$, expected excess return over variance. The geometry transfers exactly: there is a growth-optimal exposure, growth falls off asymmetrically past it, and correlated positions shrink it. Coin-flip Kelly is the toy model interviewers use because every number is computable at the table.

Size your prep like a position

Kelly questions reward candidates who have computed, not skimmed. Build the expected-value reflexes the formula sits on, work the sizing-and-quoting problems in the market-making track, then put yourself under the clock with the timed screens and the full question bank. When the 60/40 numbers in this guide feel like recall instead of derivation, you are ready for the question in the room.