The geometric distribution
Repeat independent trials with success probability p. The number of trials up to and including the first success is geometric.
E[trials] = 1/p
So a fair coin takes 2 flips to the first head on average; a die takes 6 rolls to the first six. Derivable in one line by conditioning on the first trial:
E = p(1) + (1-p)(1 + E), which rearranges to E = 1/p.
Variance is (1-p)/p^2, which is large for small p - waiting times are highly variable and the mean alone is a poor description.
The exponential distribution
The continuous analogue. Time until an event arriving at rate lambda, with mean 1/lambda.
Memorylessness
Both distributions have the defining property:
P(X > s + t | X > s) = P(X > t)
Having waited already tells you nothing about how much longer you will wait.
The interview form: "A fair coin has come up tails ten times. What is the expected number of further flips to a head?" The answer is 2, exactly as it was before the ten tails. Candidates who hesitate here are usually pattern-matching to the gambler's fallacy.
These are the only memoryless distributions - geometric in discrete time, exponential in continuous. That uniqueness is itself an occasional interview question.
The inspection paradox
The follow-up that catches almost everyone.
Buses arrive as a Poisson process, one every 10 minutes on average. You arrive at a random time. How long do you wait?
The answer is 10 minutes, not 5. And the expected gap between the bus before you and the bus after you is 20 minutes, not 10.
The reason: arriving at a random moment, you are more likely to land inside a long gap than a short one, precisely because long gaps occupy more of the timeline. You are sampling gaps in proportion to their length, so the gap you land in is biased upward.
Getting this right, and being able to say why, is a strong signal.
Practise in random variables.