Reading EV = Σ pᵢ · xᵢ
| X | The random quantity: a number you won't know until the event happens, such as the die roll. |
| i | A label for each possible outcome, counting 1, 2, 3 … For a die, i runs from 1 to 6. |
| xᵢ | The payoff of outcome i. For a die, x₁ = 1, x₂ = 2 … x₆ = 6. |
| pᵢ | The probability of outcome i, between 0 and 1. All the pᵢ together add to exactly 1. |
| pᵢ · xᵢ | That one outcome's contribution to the average. |
| Σ | Greek capital sigma, meaning "sum over every i": do the thing for each outcome and add the results. |
| EV, E[X], μ | Three names for the same answer. μ (mu) is the usual symbol once you use it inside other formulas. |
Why multiply by p? Picture rolling a die 600 times. You get about 100 of each face, so the total is about 100·1 + 100·2 + … + 100·6 = 2,100, and the average is 2,100 ÷ 600 = 3.5. Divide each count by 600 and the counts become probabilities: (1/6)·1 + (1/6)·2 + … + (1/6)·6. EV is an ordinary average where each outcome counts in proportion to how often it happens.
Reading Var(X) = Σ pᵢ · (xᵢ − μ)²
| μ | The EV from above: the centre of the outcomes. |
| xᵢ − μ | The deviation: how far outcome i lands from the centre. Negative means below the centre. |
| ( … )² | The squared deviation. Squaring makes every distance positive, and it punishes big misses far more than small ones (a miss of 3 counts 9, a miss of 1 counts 1). |
| pᵢ · ( … )² | Each squared miss, weighted by how often it happens. |
| Σ | Add them up. Variance is the average squared distance from the centre. |
| σ = √Var | Standard deviation (lower-case sigma). Taking the square root undoes the squaring, so the answer is back in the payoff's own units (pips, dollars) and you can compare it with the EV. |
Why square instead of just averaging the deviations? Because the plain average deviation is always exactly zero. The centre is defined as the balance point, so misses above and below cancel. The table below proves it: the p·(x − μ) column always sums to 0. You could use absolute distance |x − μ| instead, but squares have one property that makes everything later work: variances of independent quantities add. That is where the √n rule comes from.
Why the shortcut E[X²] − μ² works. E[X²] means "the average of the squared payoffs", Σ pᵢxᵢ². Expand (x − μ)² = x² − 2μx + μ², then average each piece: E[X²] − 2μ·E[X] + μ² = E[X²] − 2μ² + μ² = E[X²] − μ². For a die: 91/6 − 3.5² = 15.17 − 12.25 = 2.92.