Expected Value Is Not Enough: Measure the Spread
A worked explanation of expected value is not enough measure the spread, including the key limitation and a checkable example.
Most people think that if two choices have the same average payoff, they are equally good. That mistake can cost you a bankroll. The truth is that expected value is not enough measure the spread. Two choices with identical expected value can wipe you out or leave you untouched, purely because of how far their results scatter. The question is not just what you average over a thousand plays but what happens on the next ten. That is why you need variance.
Why Two Choices With The Same Payoff Carry Different Risks
Choice A pays $100 with probability 0.5 and $0 with probability 0.5. Choice B pays $50 with probability 1. Both have an expected value of $50. If you only looked at expected value, they are identical. They are not.
Choice B is a sure thing. Every time you play, you get $50. There is no risk. Choice A is a coin flip. Half the time you get $100, half the time you get nothing. The average after many flips is $50, but on any single play you either double your money or lose it all. The spread of outcomes is what matters.
Variance measures that spread. For Choice A, variance = ($100 - $50)² × 0.5 + ($0 - $50)² × 0.5 = 2500 dollars². Standard deviation is the square root of variance, $50. For Choice B, variance = 0 and standard deviation = $0. That is the entire difference: one bet has a standard deviation of $50, the other has zero. The definition from Blitzstein and Hwang in Introduction to Probability, Chapter 5, is the standard: Var(X) = E[(X - μ)²] = Σ (x_i - μ)² × P(x_i).
Expected Value Is Not Enough Measure The Spread Example
Work a real case. Suppose you have $500 in your pocket. You are offered a bet that pays $1,000 with probability 0.5 and $0 with probability 0.5. Expected value is $500. The standard deviation is $500. On your first play you either walk away with $1,500 or you are broke. If you lose, you cannot play again. The expected value looks good, but the spread means a 50% chance of ruin on a single trial.
Compare a bet that pays $600 with probability 0.5 and $400 with probability 0.5. Expected value is still $500. Standard deviation is $100. The worst case on one play leaves you with $400, not zero. You can keep playing. Two choices with the same expected payoff carry very different risks because the standard deviation tells you the range of plausible outcomes. The first bet can end your session in one round. The second cannot.
That is why a variance calculator is the tool you reach for after you have computed expected value. It gives you the missing second dimension in a decision under uncertainty.
Calculation Steps
The formula for variance using the computational shortcut is Var(X) = E[X²] - (E[X])². For Choice A, E[X] = $50. E[X²] = ($100² × 0.5) + ($0² × 0.5) = 5000. Var = 5000 - 2500 = 2500 dollars². Standard deviation = $50. That matches the direct calculation.
For Choice B, E[X] = $50. E[X²] = $50² × 1 = 2500. Var = 2500 - 2500 = 0. Standard deviation = $0. The computational formula is faster for distributions with many outcomes, but the interpretation is the same: variance is in squared units, standard deviation is in the original units and is what you use to judge risk.
The Law of Large Numbers says that as the number of trials increases, the average outcome converges to the expected value. But that is a long-run result. For a small number of trials, variance dominates. A standard deviation of $50 on a $50 expected value means that roughly 68% of outcomes fall between $0 and $100, assuming a symmetric distribution. That is a huge range relative to the average.
Why the Method Works
Variance alone does not capture everything. Two distributions can have the same variance but different probabilities of extreme losses. One might have a fat tail on the loss side, the other a fat tail on the gain side. Variance is symmetric by construction: it squares deviations, so it treats a loss of $50 the same as a gain of $50. That is a limitation. Skewness and tail risk matter for decision-making, especially for bets with rare, large payoffs like a lottery.
For a lottery ticket with a $1 billion jackpot and a 1-in-300 million chance, the raw expected value is about $3.33 before taxes and splits. The variance is enormous because the jackpot outcome is so far from the mean. The standard deviation is in the tens of thousands of dollars. That tells you the outcome on a single ticket is nearly certain to be zero, and the average only converges after millions of tickets. The expected value is not enough measure the spread explained to a bettor who buys two tickets and expects the math to work on Saturday night.
Common Questions
Why can two choices with the same expected payoff carry very different risks?
Because expected value measures only the long-run average, not the spread of possible outcomes. Variance and standard deviation measure that spread. A bet with high variance can produce a losing streak that bankrupts a small bankroll even though its expected value is positive, while a bet with zero variance always pays the same amount.
What is the difference between variance and standard deviation?
Variance is the average squared deviation from the mean, expressed in squared units (e.g., dollars²). Standard deviation is the square root of variance, expressed in the same units as the original data (e.g., dollars). Standard deviation is more interpretable because it is on the same scale as the outcomes.
Can variance be zero?
Yes. If a choice pays the same amount every time, the variance is zero. All outcomes are identical to the mean, so there is no spread. A guaranteed $50 has variance zero. A coin flip that pays $100 or $0 has variance 2500 dollars².
Does variance alone tell you everything about risk?
No. Variance does not capture skewness or tail risk. Two distributions can have the same variance but very different probabilities of extreme losses. Variance treats deviations above and below the mean equally, so it does not distinguish between a risk of losing everything and a risk of gaining a windfall.
What is the failure case for using only expected value and variance?
A bet with expected value of $1 and standard deviation of $100 can produce a long losing streak that bankrupts a small bankroll despite having positive expected value. The expected value only holds over many trials, but the variance means that on any single trial you could lose $100 or more. If your bankroll is $200, you can be wiped out in two plays even though the expected value is positive.