Standard deviation and sample size
How far results wander from their average, and why it takes so many hands to know anything.
6 minute read
Repeat a spot n times with loss probability p. The expected number of losses is n × p, but the actual number scatters around it with a standard deviation of √(n × p × (1 − p)). For 100 aces-vs-kings spots that is √(100 × 0.18 × 0.82) ≈ 3.8: eighteen losses is the centre, but fourteen or twenty-two would be unremarkable.
| Spot (loss rate) | Lose ≥ 1 of 1 | Lose ≥ 2 of 5 | Lose ≥ 4 of 10 | Lose ≥ 8 of 20 | Lose all 5 |
|---|---|---|---|---|---|
| AA vs KK preflop (18%) | 18.0% | 22.2% | 8.8% | 1.8% | 0.02% |
| Set vs flush draw on the flop (26%) | 26.0% | 38.8% | 24.8% | 12.2% | 0.12% |
| Pair vs two overcards preflop (45%) | 45.0% | 74.4% | 73.4% | 74.8% | 1.85% |
Binomial probabilities of losing at least 40% of n independent repetitions. Loss rates are rounded from the exact matchups on the encyclopedia pages.
Two things shrink with sample size, and they shrink slowly. The spread of the loss count grows like √n while the count itself grows like n, so the proportion tightens — but only as 1/√n. To halve your uncertainty about a frequency, you need four times as many hands.
What this means for reading your results
A few hundred hands say almost nothing about whether a decision is good. The math of the decision — equity against a range, price, EV — is knowable now. The results take thousands of repetitions to catch up with it.
Quick check
You expect to lose 25 of 100 spots. What is the standard deviation of the number of losses?
Show answer
√(100 × 0.25 × 0.75) ≈ 4.3. Losing between about 21 and 29 is the normal range.