Suppose that you are trying to collect a complete set of $n$ baseball cards. Suppose you buy them one at a time and each time you get a randomly chosen card. Let $N_n$ be the number of cards you have to buy to get the complete set. Show $N_n/(n \ln n) $ converges in probability to 1.
(homework) (convergence in probability)
Some hints: 1) Express the waiting time as a sum of $n$ independent random variables, each with a geometric distribution. 2) Find the expectation and variance of that sum 3) combine the above information to get to an answer
Zombie question, but in case anyone else reads this looking for the answer, I'm going to expand a little bit on the answer given by Kjetil. He gives you the right starting point with writing $N_n$ as a sum of independent geometric r.v.'s, $$N_n = T_1 + T_2 + ... + T_{n-1} + T_n,$$ where $T_i$ is geometric with parameter $i/n$, for $1 \leq i \leq n$. As such, $$\Bbb{E}[N_n] = \sum \Bbb{E}[T_i] \text{ and } \operatorname{Var}(N_n) = \sum \operatorname{Var}(T_i).$$ You can look up formulas for geometric expectation and variance here. (It will also be useful to know the asymptotics of the nth harmonic number, and that $\sum 1/n^2 = \pi^2/6$.)
Those are the parts of the answer Kjetil addresses, but the more involved part that his answer glosses over is how to use the expectation and variance of $N_n$ to show that the related r.v. $$X_n := \frac{N_n}{n \ln n}$$ converges to 1 in probability, i.e. $$\Bbb{P}(|X_n - 1| > \epsilon) \to 0$$ for any fixed $\epsilon > 0$. To show $X_n \to 1$ in probability, we first need to show $\lim_{n \to \infty} |\mu_n - 1| = 0,$ where $$\mu_n = \Bbb{E}[X_n] = \frac{\Bbb{E}[N_n]}{n \ln n}.$$ Then we have to use the triangle inequality $|X_n - 1| \leq |X_n - \mu_n| + |\mu_n - 1|,$ together with Chebyshev's inequality: $$\Bbb{P}(|X_n - \mu_n| > d) \leq \frac{\sigma_n^2}{d^2},$$ where $$\sigma^2_n = \operatorname{Var}(X_n) = \frac{\operatorname{Var}(N_n)}{n^2 \ln^2 n},$$ to prove the desired convergence in probability.
