Pricing a bet: expected value and the noise around it
The fair price of a bet is what it pays, weighted by how often. What that number does not tell you is how bumpy the road is, and how many plays it takes for the average to show up.
- Compute the expected value of a game net of its fee, and the edge
- Compute the standard deviation of one play, and why the fee does not appear in it
- Solve for the fair probability or the fair prize
- Scale the mean and the standard deviation over n plays and say what repetition does
The intuition
A game pays $100 with probability 20%, nothing otherwise, and costs $25 to play. On average it pays $20, so each play loses $5: the game is unfavourable by 20% of the fee. That is the expected value, and it is the only thing that decides whether the game is worth entering at that fee. Played forever, it is what you earn per play.
It is not the whole story. The outcome of one play is $100 or nothing, with a standard deviation of $40, eight times the size of the edge. Over 100 plays the expected loss is $500 and the standard deviation is $400: the dollars at risk grew, but the noise shrank relative to the edge by √100 = 10. Repetition makes a positive edge reliable and a negative one certain. It never turns a bad bet into a good one.
EV per play = p × W − c. SD of one play = W √(p(1 − p)); the fee adds nothing. Fair fee = p × W; fair probability = c ÷ W; fair prize = c ÷ p. Over n plays: EV_n = n × EV, SD_n = √n × SD. P(win at least once in n) = 1 − (1 − p)ⁿ.
Why it works
- The conventions here: one prize W with probability p, nothing otherwise. The fee c is paid up front and not returned. Plays are independent.
- Expected value is a weighted average. p × W is what the game pays on average; subtract the fee. Divide by the fee for the edge, positive to the player or to the house.
- The fee is not risk. It is paid every time, so it shifts every outcome by the same amount and leaves the spread alone. Only the uncertain part, whether the prize arrives, has variance: p(1 − p)W².
- Fair means EV = 0. Solve p × W = c for whichever piece is missing: the probability c ÷ W, or the prize c ÷ p. A bookmaker sets the fee so the break-even probability sits above the true one.
- Means add, variances add. Over n independent plays the mean is n times bigger and the standard deviation only √n times bigger. The ratio of noise to edge falls by √n, which is why a casino wants millions of small bets.
- Same EV, different risk. Two games with the same expected value can have very different standard deviations. A risk-averse player prefers the lower one; the house, playing millions of times, does not care.
| Expected prize: 20% × $100 | $20.00 |
| EV per play: $20 − $25 | −$5.00, an edge of −20% to the player |
| SD of one play: $100 × √(20% × 80%) | $40.00 |
| Fair probability: $25 ÷ $100 | 25% |
| Fair prize: $25 ÷ 20% | $125.00 |
| 100 plays: EV 100 × (−$5); SD √100 × $40 | −$500.00; $400.00 |
One play: noise 8 times the edge. A hundred plays: 0.8 times. The loss became reliable, not smaller.
The formulas
The average payout less the fee.
A Bernoulli scaled by the prize; the fee drops out.
Set EV to zero and solve.
Means add; variances add, so standard deviations grow with the root.
Worked example
Probability times prize, minus the fee. The follow-up asks whether you should play, and the answer depends on the sign.
See it move
Same game. Change the win probability, the prize, the fee and how many times you play.
- Raise the win probability. The expected prize and the expected value rise; watch the standard deviation, which peaks at a 50% chance.
- Raise the fee. The expected value falls dollar for dollar and the standard deviation does not move: a constant is not risk.
- Play more times. The total's standard deviation grows with the square root of the plays while its expected value grows in proportion, so the noise shrinks relative to the edge, whichever sign the edge has.
Run it backwards
The fee and the prize are known. What win probability would make the game exactly fair?
Fair means EV = 0, so p × W = c and p = c ÷ W. Compare it with the true probability to see who holds the edge.
The follow-up is how a bookmaker uses the number: in reverse, setting the price so the break-even probability sits above the true one.
Traps
Say it in the interview
“This game pays $100 one time in five and costs $25. Would you play?”
Check yourself
5 fresh questions, with new numbers. Answer each one correctly to finish the chapter. Get one wrong and you will see the full working, then you can try it again with new numbers.
Answers within 1% are marked right. Type the number; $, %, x and M are fine. First tries count toward Learned: the topic is Learned once every chapter is done and 75% of first tries were right.
- EV = p × W − c; the edge is EV ÷ fee.
- SD = W √(p(1 − p)); the fee is not risk.
- Fair probability = c ÷ W; fair prize = c ÷ p.
- n plays: EV × n, SD × √n. Repetition makes the edge reliable, whichever sign it has.