The without-replacement case clicked pretty fast once I thought about it as a uniform distribution over 100 days.
Clarify the sampling process and define the random variable for the day a specific person is first selected. For without replacement, use the fact that each person is equally likely to be in any position in a random permutation; for with replacement, model each day's selection as a Bernoulli trial with probability 10/1000.
Pro tip: State your assumptions explicitly (e.g., days are independent, draws are uniform) and connect the result to practical implications like expected wait time and fairness in sampling.
Confirm that each day 10 distinct names are drawn, and we want the expected day (1-indexed) when a specific person is first selected. For with replacement, each draw is independent and a person can be selected multiple times.
In a random permutation of 1000 people, each person is equally likely to be in any position. The specific person's position is uniformly distributed from 1 to 1000, so the expected position is (1000+1)/2 = 500.5. Since selections occur in blocks of 10 per day, the day is ceil(position/10). Compute the expected day by averaging over positions.
Calculate E[day] = (1/1000) * sum_{k=1}^{1000} ceil(k/10). This sum equals 10*(1+2+...+100) = 10*5050 = 50500, so E[day] = 50500/1000 = 50.5 days.
Each day, the probability the person is selected at least once is 1 - (1 - 10/1000)^10? Wait, careful: each day 10 draws with replacement from 1000. The probability a specific person is not drawn in a single draw is 999/1000. Over 10 independent draws, probability not drawn is (999/1000)^10. So probability drawn at least once per day is p = 1 - (999/1000)^10. The number of days until first selection follows a geometric distribution with success probability p, so expected day = 1/p.
Calculate p = 1 - (0.999)^10 ≈ 1 - 0.99004488 = 0.00995512. Then expected day = 1/p ≈ 100.45 days.
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