Expected value in plain English

Expected value, often shortened to EV, is one of the most useful ideas in probability. It answers a simple question: if you could repeat a decision many times, what would you gain or lose on average each time? This article walks through the idea with everyday examples. It is educational, aimed at adults 18 and over, and is not a recommendation to gamble. Laws on gambling vary by country and U.S. state.

The basic formula

To find expected value, multiply each possible outcome by its probability and add the results. Gains count as positive numbers and losses as negative ones. If the total is zero, the game is fair. If it is negative, you lose on average; if it is positive, you gain on average.

Note the word average. EV says nothing about any single try. It describes the long-run pattern across a large number of repetitions.

Example 1: a fair dice game

Imagine you pay 1 unit to roll a die. If it lands on six, you receive a 5-unit profit, and otherwise you lose your 1-unit stake. The probability of six is 1/6, and the probability of anything else is 5/6.

EV = (1/6 x 5) + (5/6 x -1) = 5/6 - 5/6 = 0.

The game is perfectly fair. Over thousands of rolls you would expect to break even, though any short stretch could be up or down.

Example 2: the same game with a smaller prize

Now suppose a six pays only 4 units of profit. The calculation becomes:

EV = (1/6 x 4) + (5/6 x -1) = 4/6 - 5/6 = -1/6, or about -0.17 units per roll.

On average you lose about 17 cents for every dollar staked, or 16.7%. This small shortfall is the type of margin that makes real games profitable for the house.

Example 3: a lottery-style ticket

Say a ticket costs 2 units and offers a 1 in 300 chance of winning 500 units, and nothing otherwise. The average prize is 500 / 300, about 1.67 units, against a cost of 2. EV is about -0.33 units per ticket. It feels like a big chance at a big prize, but on average each ticket costs you a third of the price paid.

Example 4: a roulette bet

A 1-unit straight-up bet on a European roulette wheel pays 35 to 1 and wins with probability 1/37.

EV = (1/37 x 35) + (36/37 x -1) = (35 - 36)/37 = -1/37, about -2.70%.

On average, the bettor loses about 2.7 cents per unit. On an American wheel with an extra 00 pocket, the same bet averages (35 - 37)/38 = -2/38, about -5.26%, so the added pocket roughly doubles the average loss.

Why a negative EV still sometimes pays off

Even with a negative EV, a single try can win, and sometimes win big. That is what makes games of chance appealing. But as the number of repetitions grows, results drift toward the average, and the more you play, the more reliably the edge shows up. This is why no one can "beat" a negative-EV game through persistence alone. Our article on the gambler's fallacy explains why losing streaks do not create a later correction.

Using EV as a reality check

Expected value is most useful as a way to see through marketing. Ask what the prize is, how likely it is and what it costs. If the average return is below the price, the product is working against you, even if the headline prize is large. Pair this with a clear understanding of how prices are expressed, which our guide to American, decimal and fractional odds covers.

Limits of the idea

EV ignores variance, the size of the swings along the way. Two bets can have the same average but very different risk. It also assumes the probabilities are known exactly, which is not always the case outside simple games. Treat EV as one tool, not a prediction.

Frequently asked questions

Does a positive EV guarantee a win?

No. It describes the long-run average, not any single result.

Why do casinos care about EV?

Because a small negative EV for the player multiplied by a large number of bets reliably produces profit for the house.

Can I use EV to decide how much to bet?

EV tells you whether a bet is favorable on average, not how much you can afford to lose. That decision belongs to your budget.