Is X a negative binomial? ( Tosses of dice)

Question

Two dice are tossed independently until each pair has appeared at least once. Find the expected number of tosses needed and the standard deviation of this number of tosses.

My attempt

Let X be the random variable that counts the number of tosses needed until each pair appears at least once. Let $X_1$ be a random variable that counts the number of tosses until the first "new" pair appear, $X_2$ counts the number of adittional tosses, after the first "new" pair until the second "new" pair appears and so on until $X_{36}$. Then,

$$X=X_1+X_2+...+X_{36}$$

We can see that each $ X_i$ is a geometric random variable with $p=\frac{1}{36}$ (that count the number of trials).And since the tosses are independent, I thought that I could say that they are independent. But I am not sure about the last affirmation. If they are indeed independent, I could say that $X$ is negatove binomial, right?

Thanks in advance!

Answer 1

$X_i$ is a geometric random variable with $p_i = \left(36 - (i - 1)\right)/36$.

$X_i$ is the number of tosses until the next new pair; any new pair counts. Until $X_i$ you have observed $i - 1$ distinct pairs, and you are waiting for any of the $36 - (i - 1)$ remaining pairs to roll up.

For example, take $i = 1$, then $p_1 = 1$, because in the very first toss any pair is a new pair.

So, no! this is not negative binomial. Yes, the $X_i$ are independent random variables, but the $p_i$ are not the same.