Geometric Distribution vs Poisson Distribution

The geometric distribution describes the trial on which the first success arrives, in a run of independent trials that each succeed with the same probability $p$. Its values are 1, 2, 3 and onward with no upper limit, the probabilities are $P(X = k) = (1-p)^{k-1}p$, and the summaries are $\mu_X = \frac{1}{p}$ (mu-X, the mean) and $\sigma_X = \frac{\sqrt{1-p}}{p}$ (sigma-X).

Roll a fair die until the first 6, so $p = 1/6$. The first 6 lands on roll 3 with probability $(5/6)^2(1/6) = 25/216 \approx 0.1157$, the mean is $\mu_X = 6$ rolls, and the standard deviation is $\sigma_X = \frac{\sqrt{5/6}}{1/6} \approx 5.48$ rolls, nearly as large as the mean itself.

"The mean is 6, so the first 6 usually turns up somewhere around roll 6" is wrong three times over. The most likely single trial is the first one, $P(X = 1) = 0.1667$, and every later trial is less likely than the one before it. The median is 4, since $P(X \le 3) = 0.4213$ and $P(X \le 4) = 0.5177$. And the mean sits above the median because the right tail runs forever: by roll 6, the mean itself, only $1 - (5/6)^6 = 0.6651$ of runs have finished.

The distribution also forgets. Roll ten times with no 6 and the count of further rolls you need still has mean 6, exactly as it did at the start, because the die keeps no record. "I am due for a 6" has no arithmetic behind it.

College Board removed the geometric distribution from AP Statistics with the Fall 2026 revision, on its published list of removed topics, so it is not assessed on the exam that first runs in May 2027. It stays standard college material, and reading a waiting-time setting correctly is still how you rule out a binomial one, which is assessed at topic 2.10 of Unit 2.

The Poisson distribution counts occurrences in a fixed window of time or space, on the assumption that events arrive independently at a constant average rate. One parameter sets the whole thing: $\lambda$ (lambda), the mean number of occurrences per window. The probabilities are $P(X = k) = \frac{\lambda^k e^{-\lambda}}{k!}$ for $k = 0, 1, 2$ and upward with no ceiling, and both the mean and the variance equal $\lambda$.

Take a help desk averaging 3 calls an hour, so $\lambda = 3$. Then $e^{-3} \approx 0.0498$, and $P(X = 2) = \frac{3^2 e^{-3}}{2!} = \frac{9(0.0498)}{2} \approx 0.2240$. The count 3 is exactly as likely, since $P(X = 3) = \frac{27(0.0498)}{6} \approx 0.2240$ too. A quiet hour of at most one call has probability $0.0498 + 3(0.0498) \approx 0.1991$.

"The mean and the variance are both 3, so the standard deviation is 3" is the slip that equality invites. The variance is 3; the standard deviation is $\sqrt{3} \approx 1.732$. For a positive rate the two agree at $\lambda = 1$ and nowhere else, so the coincidence you are remembering is between the mean and the variance, never between the mean and the spread.

The window is part of the parameter rather than part of the background. Three calls an hour means $\lambda = 1.5$ over half an hour and $\lambda = 24$ over an eight-hour day, so carrying 3 into a half-hour question doubles where the answer is centered. The constant-rate assumption is the first thing to break in practice: calls that bunch up after an outage vary more than a Poisson allows, and the tell is a sample variance sitting well above the sample mean.

No topic in the Fall 2026 AP Statistics course covers the Poisson distribution. The count model the course does use is the binomial, topic 2.10 of Unit 2, which fixes the number of trials at $n$ instead of letting the count run without a bound.