I knew it smelled like a hypergeometric setup but blanked on the exact expected value formula mid-explanation.
Model the selection process as a geometric distribution where each round is an independent Bernoulli trial with success probability p = 10/1000 = 0.01. The expected number of rounds until first success is 1/p = 100, so on average, you would be selected in the 100th round. Clearly state assumptions (random selection with replacement, independence) and explain the memoryless property.
Pro tip: Mention that the expected round is 100, but the distribution is highly skewed: the median is about 69 rounds, and there's a 36.6% chance of waiting more than 100 rounds. This shows deeper understanding and avoids the common misconception that the average is the most likely outcome.
Let X be the round number of your first selection. Each round, the probability of being selected is p = 10/1000 = 0.01, assuming independent draws with replacement.
X follows a geometric distribution with success probability p = 0.01, since we are waiting for the first success in a sequence of independent Bernoulli trials.
The expected value of a geometric distribution is E[X] = 1/p = 1/0.01 = 100. So on average, you expect to be selected in the 100th round.
Mention that the distribution is right-skewed: the median is about 69 rounds, and the probability of waiting more than 100 rounds is (0.99)^100 ≈ 0.366. This highlights that the average may not represent a typical wait time.
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