Geometric 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.