Explainer
Slot volatility explained: what the average misses
The short answer
Volatility concerns variation around an average. A mean on its own cannot tell you how tightly outcomes cluster or how much rare, larger outcomes contribute to the total.
In this article
Imagine two measurement processes, each with an average output of one unit. Process A always produces one unit. Process B produces zero 90% of the time and ten units 10% of the time. Their averages match, while their individual outputs look almost nothing alike. That missing distinction is the reason to examine spread.
The same mean can hide different distributions
Illustrative example
| Measure | Process A | Process B |
|---|---|---|
| Possible outputs | 1 only | 0 or 10 |
| Expected output | 1 | 1 |
| Variance | 0 | 9 |
| Standard deviation | 0 | 3 |
For B, the mean is 0 × 0.9 + 10 × 0.1 = 1. The squared distance from the mean is 1 for a zero and 81 for a ten. Weighting those distances gives 1 × 0.9 + 81 × 0.1 = 9. Its square root is a standard deviation of 3. Process A never leaves its mean, so both measurements of spread are zero.
A label is less precise than a measure
“Low,” “medium,” and “high” are classifications, not complete probability distributions. To compare two labels, you would need to know how each was assigned, whether the thresholds match, and what each label includes. A rating without its method cannot be converted into a numerical standard deviation.
OpenStax’s discussion of expected value and standard deviation provides the mathematical background. Our two distributions are intentionally simple enough to recalculate by hand. Real systems can have many more outcomes and states.
What spread still does not tell you
Even a standard deviation does not list every possible outcome or establish the probability of a specific result. Different distributions can share a mean and a variance. To answer a question about the tails—the least common outcomes—you need more of the distribution, not just another adjective.
