In the geometric distribution, why is the mean number of trials until the first success 1/p?

## Setup Run independent trials that each succeed with the same probability p. The **geometric** distribution describes how long you wait for the first success. There are two conventions, and confusing them is the single most common error: - **Trials convention** (used here): X = number of trials up to *and including* the first success. Support {1, 2, 3, ...}, `P(X = k) = (1-p)^(k-1) p`, mean `1/p`. - **Failures convention**: Y = number of failures *before* the first success. Support {0, 1, 2, ...}, `P(Y = k) = (1-p)^k p`, mean `(1-p)/p`. The two differ by exactly 1, since X = Y + 1. Both have variance `(1-p)/p²`, because shifting a variable by a constant leaves its spread unchanged. State your convention before you state your answer. ## Why the mean is 1/p — the restart argument Let E be the expected number of trials until the first success. Run one trial. That trial always costs you 1. With probability p it succeeds and you stop; with probability (1-p) it fails and — crucially, because the trials are independent and identically distributed — you face exactly the same problem you started with. So `E = 1 + (1-p) · E`. Solving: `E - (1-p)E = 1`, so `pE = 1` and `E = 1/p`. This argument is worth memorising because it generalises to many waiting-time problems and needs no sum manipulation. ## Why it is 1/p — the direct sum The brute-force route: `E[X] = Σ_{k≥1} k (1-p)^(k-1) p`. Writing q = 1-p, this is `p · Σ_{k≥1} k q^(k-1) = p · 1/(1-q)² = p/p² = 1/p`, using the standard derivative-of-a-geometric-series identity. ## Variance and skew `Var(X) = (1-p)/p²`, so the standard deviation is `sqrt(1-p)/p`, which for small p is very close to the mean 1/p itself. That near-equality is a warning: the geometric is heavily right-skewed, with a long tail of unlucky waits. Take a click-through probability p = 0.02: - Mean: 1/0.02 = **50** impressions. - Standard deviation: sqrt(0.98)/0.02 ≈ **49.5**. - Median: the smallest k with P(X ≤ k) ≥ 0.5, which works out to **35**. - P(X > 50) = 0.98^50 ≈ **0.364**. So the "expected" wait of 50 is exceeded more than a third of the time, and the typical wait is nearer 35. Treating the mean as a typical outcome, or as a planning guarantee, is a real mistake in campaign or funnel work: quoting "about 50 impressions per click" hides that a quarter of the time you will need more than 68 impressions. (P(X > 68) = 0.98^68 ≈ 0.25.) ## Memorylessness The geometric is the **only** discrete distribution that is memoryless: `P(X > m + n | X > m) = P(X > n)`. After 30 straight failures, the expected number of *further* trials is still 1/p. Nothing is "due". This falls straight out of independence — the coin does not know its history — but it contradicts strong intuition, so interviewers like to test it. ## Extending to the r-th success The **negative binomial** counts trials needed to reach the r-th success. Under the trials convention its mass function is `P(X = k) = C(k-1, r-1) p^r (1-p)^(k-r)` for k ≥ r, with - mean `r/p`, - variance `r(1-p)/p²`. These follow immediately from the fact that the total wait is a sum of r independent geometric waits: means add, and variances add because the segments are independent. At r = 1 it reduces to the geometric. So with p = 0.02, waiting for 3 clicks takes 150 impressions on average. ## The assumptions All of this needs independent trials with a **constant** p. If the click probability decays as a user sees the same ad repeatedly, or if impressions are served to a heterogeneous audience with different propensities, the geometric will misdescribe the tail — typically the pooled wait is more variable than the model says, because the low-p users dominate the long waits. ## Common errors Quoting the mean as p rather than 1/p; using the two conventions interchangeably within one answer; assuming the median equals the mean when the distribution is strongly skewed; and expecting a run of failures to make success more likely.