Rolling a die until we have all the numbers- variance

Question

We roll a die until we obtain all numbers from $1$ to $6$. I found the expected value of rolls computing it like $X = X_1 + \dots + X_6$ where $X_i$ is number of rolls needed to obtain a result different from previous $i-1$ and using a geometric distribution. And my result is correct. But then I wanted to find a variance. Firstly I thought of doing it this way: $$\text{Var} (X_1 + \dots + X_6) = \text{Var}(X_1) + \dots + \text{Var}(X_6) + 2 \sum_{1\le i<j\le6} \text{Cov}(X_i, X_j),$$ but covariance is not easy to find here. Can somebody please show me how to find a variance of number of dice rolls?

Answer 1

The covariance is easy, your $X_i$ are independent.

Suppose you rolled the die until you had seen $i$ different numbers. The probability that you get one of the $6-i$ remaining numbers on the next roll is $1-\frac i6$. It doesn't matter which $i$ numbers you've seen or how many rolls it took.

Therefore the $X_i$ are independent shifted geometrics mean $\frac 1{p_i}$ and variance $\frac{1-p_i}{p_i^2}$ where $p_i = 1- \frac i6$.

As they are independent the covariances are $0$ and you can just sum the variances.