Betting mathematics
How to calculate the expected value of a bet
Expected value (EV) is the average win or loss per unit staked if the same bet were repeated many times: EV = p × odds − 1. Odds of 2.50 with an estimate of p = 0.45 give EV = +0.125, but if the true probability is 0.38, the EV is −0.05. EV is only as good as the estimate of p.
Contents
Key terms
- Expected value (EV)
- The average result of a bet per unit staked, if the same bet were repeated many times.
- Probability estimate (p)
- Your or a model's estimate of how likely the outcome is. The odds are the price; p is the estimate.
- Break-even probability
- The probability at which EV is exactly zero: 1 / odds.
- Positive EV (+EV)
- A bet whose average result is above zero, provided the estimate of p is right.
The formula
A bet has two outcomes. With probability p you win odds − 1 per unit staked (net), and with probability 1 − p you lose the stake. Simplified, that leaves a short form.
EV = p × (odds − 1) − (1 − p)
EV = p × odds − 10.45 × (2.50 − 1) − (1 − 0.45) = 0.675 − 0.55 = 0.125
Odds of 2.50 and an estimate of p = 0.45 give EV = +0.125 per unit, or +12.5%. On a stake of 100 that is +12.50 per bet on average, over a long run of identical bets.
Break-even: when EV is zero
EV is zero when p × odds = 1, that is when p = 1 / odds. This is the same implied probability as in the previous guide. For odds of 2.50 the break-even point is 0.40: the bet has a positive EV only if the true probability is above 40%.
| p | EV per unit | On a stake of 100 |
|---|---|---|
| 0.35 | −0.125 | −12.50 |
| 0.38 | −0.050 | −5.00 |
| 0.40 | 0.000 | 0.00 |
| 0.42 | +0.050 | +5.00 |
| 0.45 | +0.125 | +12.50 |
| 0.50 | +0.250 | +25.00 |
At odds of 2.50, every percentage point of the estimate moves EV by 0.025. An error of just 2 points turns +0.050 into zero.
When the estimate is wrong
The same odds of 2.50, but the true probability is 0.38, not 0.45:
0.38 × 2.50 − 1 = 0.95 − 1 = −0.05
A bet that showed +12.5% on paper actually loses 5% of the stake on average. The odds did not change; only the estimate did.
EV is not a property of the odds but of the pair of odds and estimate. The key question is how accurate the estimates are, and that is measured with the Brier score and calibration.
Why positive EV does not mean a profit
EV is an average. A single bet either wins or loses the whole stake, and over a short run the result can drift far from the average.
100 bets at odds of 2.50, each with a true probability of 0.45 (EV = +0.125 per bet):
- the expected result is +12.5 units;
- the standard deviation is about 1.24 units for one bet and about 12.4 units for 100 bets;
- the probability of being behind after 100 bets (39 winners or fewer) is 13.4%;
- after 1,000 such bets that probability falls to about 0.06%.
The figures are calculated exactly from the binomial distribution, with no simulation, and hold only if p really is 0.45.
EV after the margin
Bookmaker odds already include a margin. If p is taken as the margin-free probability from the same odds, EV is negative for every outcome and equals 1 / sum − 1. For the market in the implied probability guide (sum 104.81%) that is 1 / 1.0481 − 1 = −4.59% on each outcome.
A positive EV therefore exists only when the estimate of p is better than the market. AI model performance compares the model's probabilities with margin-free odds on the same matches. It does not give the EV of a single bet, but it shows how reliable the estimate is, and every EV calculation depends on that.
Limitations
- EV refers to a large number of identical bets. A single bet always either wins or loses the whole stake.
- p is an estimate. A small error in p can flip the sign of EV, and the true probability is never observed directly.
- The examples ignore tax on winnings, stake limits and odds movement before kick-off.
- A positive EV on paper does not mean the odds will still be there when you place the bet.