Trading Expected Value Against Variance

An interviewer offers you a game: pay $1 to play, flip a coin, win $3 on heads and nothing on tails. The expected value is positive — you should obviously play, right? Now change the payoff: 1% chance of winning $10,000, 99% chance of losing $90. The expected value is still comfortably positive. Would you still play, as many times as they'd let you? Most people hesitate, and that hesitation is the entire point of the question — expected value alone doesn't tell you whether a bet is survivable.

The casino analogy

A casino's edge on a single roulette spin is razor-thin — a few percent, positive, in the house's favour. No individual spin is a "sure thing," and on any given spin the casino can lose a lot of money. The casino survives and gets rich not because variance doesn't matter, but because it controls the size of each bet relative to its bankroll and plays an enormous number of independent rounds, so the swings average out long before they threaten the business. A gambler who bets his entire net worth on one spin with the same positive edge is taking on far more risk per unit of expected profit than the casino ever does. Same edge, wildly different outcome, because size and repetition weren't managed.

The two numbers you need

Expected value is the probability-weighted average outcome:

$EV = \sum_i p_i \cdot x_i$

Here $p_i$ is the probability of outcome $i$ and $x_i$ is its payoff; you multiply each possible result by how likely it is and add them up. In words: it's the average result if you played the game an enormous number of times.

Variance measures how spread out the actual outcomes are around that average:

$\text{Var}(X) = E[X^2] - (E[X])^2$

In words: square each outcome's distance-flavoured contribution, average that, and subtract the square of the mean you already found. The square root of variance, the standard deviation, tells you the typical size of a swing away from the average, in the same units as the payoff itself — which is the number your gut actually reacts to.

Worked example 1: two bets with the same EV

Bet A: win $2 with probability 0.5, lose $0 (i.e., win $0) with probability 0.5. $EV = 0.5(2) + 0.5(0) = 1$, i.e., $1.

Bet B: win $1,000,000 with probability 0.000001, lose $0.0001 otherwise (rounding for simplicity, lose nothing meaningful). Constructed so $EV \approx 1$, i.e., about $1, as well.

Both have identical expected value, $1. But Bet A's standard deviation is $1 (it swings between $0 and $2), while Bet B's standard deviation is close to $1,000, because almost all of its "average" comes from an event that almost never happens. If you can only play once, Bet A gives you something close to $1 with near-certainty; Bet B gives you almost-certainly nothing and, one time in a million, a fortune. Same EV, completely different bet.

Worked example 2: repetition changes the answer

Take the $1-to-play, 1% chance of +$10,000, 99% chance of −$90 game from the opening. $EV = 0.01(10{,}000) + 0.99(-90) = 100 - 89.1 = 10.90$, i.e., $10.90 per play — strongly positive. Standard deviation per play works out to roughly $995, since almost the entire spread comes from that rare $10,000 outcome. Play it once with a $1,000 bankroll and you have a 99% chance of ending up down $90 (9% of your bankroll) — survivable, but the point of the exercise is that a single play tells you almost nothing about the $10.90 average. Play it 500 times with the same $1,000 bankroll staked a few dollars at a time, and the sum of 500 draws concentrates around $500 \times 10.90 = 5{,}450$, i.e., $5,450, with a standard deviation that grows only with $\sqrt{500} \approx 22$, not with 500 — the average outcome per play becomes far more reliable as you repeat it, because independent swings partly cancel while the average keeps compounding.

Drag the trial count and probability above and watch how a rare, large payoff keeps the mean fixed while the shape stays a tall spike at zero with a thin tail — that thin tail is where all of Bet B's expected value quietly lives.

What this means in practice

Trading desks and market makers face this daily: many small, positive-EV trades with modest variance are the business model, not one huge positive-EV bet. A quant's real job in a "would you take this bet?" question is rarely just computing the sign of the EV — it's asking how many times you get to play, what fraction of your capital is at risk per play, and whether a run of bad luck before the averages kick in could wipe you out first. That last question is risk of ruin, and it's why position sizing (how much of your bankroll goes on each play) is treated as at least as important as finding positive-EV opportunities in the first place.