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Kelly Criterion Interview Questions: Expected Value, Bet Sizing, and Uncertain Odds

Practice Kelly criterion interview questions with a 55% even-money game, log-growth derivation, fractional sizing, and uncertain odds.

The Kelly criterion answers a narrower question than “Is this bet profitable?” It chooses a fraction of current wealth to maximize expected logarithmic growth under a specified payoff model and known win probability. In an interview, define the payout, calculate expected value, derive the fraction, and then challenge the probability estimate before suggesting a stake. A positive edge is not permission to bet the whole bankroll.

Evidence boundary: Kelly's original 1956 paper establishes the long-run growth objective for repeated bets. The numerical games below are original teaching examples, not reported questions from a named employer. An uncertain probability or a finite personal risk limit can justify a different decision from the textbook optimum.

Here we slow the reasoning down so you can defend each step aloud.

First Fix the Odds and the Objective

Suppose a $1 stake wins $b in net profit with probability $p$ and loses the full $1 otherwise. The losing probability is $q=1-p$ . Decimal odds are sometimes quoted as gross return, including the original stake: if decimal odds are 2.5, then $b=1.5$ , not 2.5. Ask which convention the interviewer uses before doing arithmetic.

Expected net profit per dollar staked is $pb-q$ . This is the edge. The bet has positive expected value when $pb>q$ , equivalently $p>1/(b+1)$ . But expected value alone says nothing about how much of your bankroll to expose. If you maximize expected wealth after one bet in this simple model, the mathematical answer is often to stake everything on a positive-edge bet. A single loss then makes wealth zero.

Kelly uses a different objective. Let $f$ be the fraction of current wealth staked, with $0\le f<1$ for a loss that would otherwise wipe out the bankroll. Wealth becomes $W(1+bf)$ after a win and $W(1-f)$ after a loss. The expected log change is

g(f) = p log(1 + b f) + (1 - p) log(1 - f).

This is a model of repeated proportional betting and compounding. It does not promise the highest outcome in every finite sequence, minimize drawdown, or set a suitable personal risk budget. Kelly's paper explicitly distinguishes exponential growth from maximizing expected capital.

Derive the Binary Kelly Fraction

Differentiate inside the feasible interval. The first derivative is $pb/(1+bf)-q/(1-f)$ . Setting it to zero gives $pb(1-f)=q(1+bf)$ , hence $pb-q=bf(p+q)=bf$ . Since $p+q=1$ ,

f* = (pb - q) / b = (p(b+1) - 1) / b.

The second derivative, $-pb^2/(1+bf)^2-q/(1-f)^2$ , is negative when $0 < p < 1$ and $b>0$ . Therefore an interior stationary point is a maximum. If $pb\le q$ , the unconstrained expression is nonpositive; a long-only bettor who may keep cash chooses $f=0$ . If a rule caps the stake at $c$ , choose the feasible maximum, normally min(c, max(0, f*)) under these assumptions.

Do not memorize the formula without its domain. If a loss is only partial, payouts vary continuously, several bets are correlated, or borrowing is allowed, the wealth expression and feasible set change. A derivative copied from the binary, full-loss case does not solve those variants.

Work an Even-Money Example All the Way Through

Imagine a hypothetical game with $p=0.55$ and even-money net odds $b=1$ . The edge per $1 stake is $0.55-0.45=0.10$ . The full-Kelly fraction is $0.10$ , so a bankroll of 1,000 game units implies a 100-unit stake under the stated model. A win leaves 1,100; a loss leaves 900.

Fraction of bankroll	Stake from 1,000	Expected wealth after one round	Expected log change per round
0%	0	1,000	0
5%	50	1,005	approximately 0.003753
10%	100	1,010	approximately 0.005008
20%	200	1,020	approximately -0.000138

The last row is the interview trap. Expected wealth after one round rises as the stake rises because this toy bet has positive expected net profit. Expected log growth falls after the optimum and is already negative at 20%. “The larger stake has higher EV” and “the larger stake is worse for repeated log growth” are both true. State the objective before deciding which statement answers the question.

At $f=1$ , a loss produces zero wealth and log(0) is undefined in the limiting calculation; with any positive loss probability, expected log growth is negative infinity. That is the mathematical reason all-in betting is incompatible with this repeated-growth objective. It is not a claim that a particular interview game will continue forever.

What if the Win Probability Is Only an Estimate?

In a real decision, $p$ may come from a small sample, a model, or a subjective judgment. The expression f* is sensitive to that estimate. With even-money odds, changing $p$ from 0.55 to 0.53 changes full Kelly from 10% to 6%; if the true $p$ is 0.49, the bet has a negative edge and the long-only fraction should be zero.

An interviewer may give a point estimate and then ask how you would defend it. Ask where it came from, whether predictions were calibrated on comparable events, whether outcomes are independent, and whether odds or costs can change before settlement. If the stated 55% is an estimate with a plausible lower scenario of 49%, write both cases. The lower scenario is an assumption for stress testing, not a statistically justified confidence bound unless a sampling model supports it.

Believed win probability at even money	Point-estimate Kelly fraction	Decision consequence
55%	10%	Positive edge if estimate and payout are correct
53%	6%	Less room for estimation error
50%	0%	Fair bet before costs
49%	0% when cash is allowed	Negative edge; declining is permitted

Editorial inference: fractional Kelly, a cap, or abstaining can be sensible ways to manage model error and drawdown, but there is no universal “always use half Kelly” rule. Half Kelly at the 55% point estimate is 5%, not a new estimate of $p$ . It reduces exposure while sacrificing some modeled log growth. A formally robust strategy needs an explicitly defined uncertainty set and objective; simply replacing $p$ with an arbitrary pessimistic number is a scenario check, not a proof of optimality.

Five Interview Questions and Answer Checks

1. Why can a positive-EV bet have a zero Kelly stake? If $pb\le q$ , it cannot in this binary model: positive EV and a positive interior Kelly numerator have the same sign. But a zero implemented stake can arise from a stricter risk policy, uncertain odds, limited liquidity, or costs that erase the edge. Identify which assumption changed rather than contradicting the formula.

2. What happens when odds are quoted as gross payout? Subtract the returned stake to obtain net odds. At gross odds 2.5 and $p=0.45$ , $b=1.5$ , edge is $0.45(1.5)-0.55=0.125$ , and $f*=0.125/1.5\approx0.0833$ . Using 2.5 as net odds would materially overstate the edge and stake.

3. Why is 20% a bad stake in the 55% even-money game? Compute 0.55 log(1.2) + 0.45 log(0.8), approximately -0.000138. The result is below both no bet and the 10% optimum, despite higher one-round expected dollars. A good answer names the objective and shows one numerical check.

4. Does a losing streak disprove the model? No. With $p=0.55$ , two losses have probability $0.45^2=0.2025$ if rounds are independent. A streak is possible even if the model is correct. But repeated poor outcomes should trigger calibration and data-quality checks, not automatic doubling to recover losses.

5. Can you add separately calculated Kelly fractions for several bets? Generally no. Bets that settle on the same dice roll or market factor have correlated outcomes. Write joint wealth for each possible state and maximize expected log wealth over a vector of stakes subject to nonnegative wealth and position constraints. Sum-of-standalone fractions can violate that constraint or double-count the same edge.

Handle Constraints Before You Optimize

Real interview prompts may impose a maximum stake, a minimum chip, limited rounds, or a need to finish above a target rather than maximize long-run growth. State the rule explicitly. A 10% Kelly fraction rounded to a legal chip size may become 9% or 11%; evaluate the permitted choices using the actual objective rather than silently treating the continuous solution as executable.

If the bankroll is essential living money, the log-growth toy model is not a personal financial plan. If outcomes have a small chance of losses larger than the stake, the wealth multiplier must include that tail. If bets share a hidden factor, an independence assumption can be dangerously optimistic. A strong answer does not abandon the mathematics; it tells the interviewer which model component must be revised.

In a short interview, prioritize a consistent sequence: payout convention, fair probability, net edge, feasible wealth outcomes, objective, optimal fraction, and sensitivity. That sequence remains useful even when the final numerical stake changes. If time is limited, show the equation and one stress scenario before giving a confident-looking decimal.

Close With a Defensible Decision

The best Kelly interview answer does not end at the fraction. It explains why that fraction solves the stated log-growth problem, how it differs from maximizing expected dollars, and what happens when odds, probability, constraints, or objective change. In the 55% even-money example, 10% is a correct model solution; 5% is a deliberately reduced exposure; 0% is defensible if the edge is not established. None of those numbers is a universal betting recommendation.

Write the wealth outcomes before the derivative, then explain one reason your implemented stake might differ from the mathematical optimum.