Statistical Variance: Samples, Uncertainty, and Bankroll Swings
Variance measures dispersion around an expectation. Its units matter: variation in the number of wins is different from variation in net bankroll return. The following examples isolate those quantities under a declared hypothetical model.
Win counts and bankroll units
Assume N independent wagers, constant win probability p = 0.55, a fixed one-unit stake, and -110 odds on every wager. There are no pushes, voids, fees, or changing prices. A win profits b = 100/110 = 10/11 units; a loss costs one unit. Let W be the win count. Then W is binomial, with mean Np and standard deviation √(Np(1−p)) wins.
Net profit X = (1+b)W − N = (21/11)W − N. Therefore E[X] = 0.05N units and SD(X) = (21/11)√(N × 0.55 × 0.45) units. Variance is SD squared, measured in squared units. Multiplying win-count SD by the payout difference is essential.
| Wagers | Expected profit (units) | Win-count SD (wins) | Profit SD (units) |
|---|---|---|---|
| 100 | 5.00 | 4.97 | 9.50 |
| 500 | 25.00 | 11.12 | 21.24 |
| 1000 | 50.00 | 15.73 | 30.03 |
| 2500 | 125.00 | 24.87 | 47.49 |
Source: NIST binomial distribution; profit transformation derived here.
Horizon and uncertainty
At N = 100, the expected profit plus or minus two standard deviations is approximately −14.00 to +24.00 units. This is a model dispersion range, not a confidence interval for an unknown edge, and the discrete outcome distribution does not give it exact 95% coverage. Positive expected return does not ensure profit at any finite horizon.
For independent observations with the same p, the standard error of the win proportion is √(p(1−p)/N). Multiplying N by four halves this quantity. The law of large numbers concerns convergence of averages under its assumptions; it does not promise recovery of losses or convergence of the bankroll to a fixed amount. Absolute profit SD grows with √N even as average-return uncertainty shrinks.
Real selections can share teams, injuries, market conditions, and outcomes. For dependent returns, variance of their sum includes twice every pairwise covariance. Different prices or stakes also change the calculation. Use a record that includes all selections and a method appropriate to that dependence; no fixed bet count confirms an edge.
CLV and sizing are separate diagnostics
A closing-price benchmark supplies information beyond a win/loss result, but this guide has no dataset establishing that it detects an edge with fewer observations. Specify the operator, market, close time, normalization, and sample selection. CLV does not establish that the closing estimate is correct.
Kelly sizing is based on expected net return and payout, not the win-count SD. For a binary wager its fraction is (b × p − (1−p))/b; it depends on the assumed probability. Fractional sizing changes exposure but cannot repair a biased forecast. Stress-test stake policies against changing probabilities, correlated losses, and limits before drawing conclusions from a simulation.
Continue reading: Bankroll Management · Building Your Own Forecasting Model.
Frequently Asked Questions
Is 4.97 the bankroll standard deviation over 100 wagers?
No. Under the stated 55%, independent, one-unit, -110 model, 4.97 is the standard deviation of the win count. Net-profit standard deviation is about 9.50 units.
Does the law of large numbers guarantee profit?
No finite-horizon profit is guaranteed. Convergence of averages requires assumptions about the observations, and the expectation itself may be negative or estimated incorrectly.
How many observations prove an edge?
There is no universal count. Prices, stake variation, dependence, selection methods, effect size, and the uncertainty standard all affect the evidence required.