Expected Value (EV) for Beginners: The Math of Every Bet

Expected value (EV) is the average result of a bet if you could repeat it forever. You multiply each possible outcome by its probability and add them up. In a casino nearly every EV is negative, and that one number, not luck or streaks or gut feeling, decides where your money goes over time.

The formula in one line

For any bet, list the outcomes, attach a probability to each, and take the weighted average.

Start with a fair coin. You win $10 on heads and lose $10 on tails.

EV = 0.5 × $10 − 0.5 × $10 = $0. Over many flips you expect to break even. That is what a fair bet looks like.

Now a friend offers $11 on heads while you still risk $10 on tails. EV = 0.5 × $11 − 0.5 × $10 = +$0.50 per flip. Flip 100 times and you expect to be up about $50. That half-dollar edge is exactly what casinos build into every game, pointed the other way.

A casino example: $10 on red

A European roulette wheel has 37 pockets: 18 red, 18 black, and a green zero. A red bet pays 1 to 1.

EV = (18/37 × $10) − (19/37 × $10) = −$10/37 ≈ −$0.27.

The zero is the whole story. Without it, red would be a coin flip. With it, you lose 27 cents per $10 bet on average, a 2.70% house edge.

Now put $10 on a single number, which pays 35 to 1:

EV = (1/37 × $350) − (36/37 × $10) = ($350 − $360) ÷ 37 ≈ −$0.27.

Same EV, very different ride. Red wins almost half the time for small amounts. A single number usually loses and occasionally pays big. EV measures the average; it says nothing about how bumpy the path is. That is the job of variance, covered in RTP vs. volatility.

On an American wheel with 0 and 00, the red bet becomes (18/38 × $10) − (20/38 × $10) ≈ −$0.53. One extra pocket doubles the expected loss.

EV of common bets, per $10

Any Seven is easy to check yourself: EV per $1 = (1/6 × $4) − (5/6 × $1) = −$1/6, or −16.67%. On $10 that is −$1.67. You can test two-dice probabilities in the free dice lab.

Bets with more than two outcomes

Many games have several prize levels. The formula is the same: weight every outcome by its probability, then subtract what you paid.

Take a made-up $1 scratch ticket. One ticket in 5 wins $2, one in 50 wins $10, one in 1,000 wins $100, and the rest win nothing.

Expected prize = (0.2 × $2) + (0.02 × $10) + (0.001 × $100) = $0.40 + $0.20 + $0.10 = $0.70. Subtract the $1 cost and EV = −$0.30 per ticket. The ticket returns 70 cents on the dollar, a 30% house edge.

Three mistakes trip people up here:

• Mixing up profit and payout. A prize of $2 on a $1 ticket is a $1 profit. Use profit with stake, or payout with cost, but never mix them.
• Confusing "to 1" with "for 1". 35 to 1 returns your stake plus $35. 35 for 1 returns $35 in total, which is only 34 to 1.
• Using hopeful probabilities. EV is only as good as the probabilities you feed it. In casino games those are known exactly; in your head they rarely are.

From one bet to a whole night

EV has a property that makes it powerful: it adds. The expected result of a session is just the EV per bet multiplied by the number of bets.

Make 200 bets of $10 on European red and you have wagered $2,000. Expected result: $2,000 × −1/37 = −$54.05.

It does not matter whether those bets go on red, on single numbers, or on a mix. Every European roulette bet carries the same −2.70%, so $2,000 of action costs $54.05 on average however you slice it.

It also does not matter what pattern you bet in. Raising after losses, lowering after wins, skipping spins: the formula only cares about the total wagered. That is why no betting system changes the expected result, as the betting systems breakdown shows in detail.

Expected value is not what will happen

Consider a dice game: roll one die, win $5 on a six, lose $1 on anything else.

EV = (1/6 × $5) − (5/6 × $1) = $0.83 − $0.83 = $0.

Yet no single roll ever produces $0. You get +$5 or −$1. EV is the center of gravity of the outcomes, not an outcome itself. Over many rolls your average result per roll gets closer and closer to EV, which is the law of large numbers at work.

Lottery tickets show the same gap. In a 6/49 game the jackpot odds are 1 in C(49,6) = 13,983,816. On a $2 ticket with a $5 million jackpot, the jackpot's contribution to EV is $5,000,000 ÷ 13,983,816 ≈ $0.36. Smaller prizes add a little back; taxes and shared jackpots take some away. The ticket is still deeply negative, even though someone eventually wins.

Using EV to spot a bad deal: blackjack insurance

When the dealer shows an ace, you may be offered insurance: a side bet of up to half your stake that pays 2:1 if the dealer has blackjack.

The offer feels protective, and that is the trick. It is a separate bet with a house edge more than ten times larger than basic-strategy blackjack itself. The same math applies to taking even money on your own blackjack.

When is EV ever positive?

Rarely, and never by accident.

• Card counting. As cards leave a blackjack shoe, a deck rich in tens and aces can tip EV slightly in the player's favor. The edge is small, swings are large, and casinos can shuffle early or ban counters. See card counting explained.
• Rare video poker pay tables. Full-pay Deuces Wild returns 100.76% with perfect strategy, an EV of about +$0.08 per $10. The pay table is hard to find, and small strategy mistakes can easily wipe out the edge.
• Your generous friend. The $11-for-$10 coin flip. Casinos do not offer it.

The bottom line

Expected value is the most useful number in gambling math. Multiply outcomes by probabilities, add them up, and you know what a bet costs on average, before you place it.

Almost every casino bet has negative EV, and session EV equals the house edge times everything you wager. Use EV to compare bets, to recognize bad side bets, and to price your entertainment honestly. If chasing that number back to zero ever feels urgent, support is available.