probability - Fair dice thrown until each number is obtained at least once

A fair die is thrown until each number is rolled at least once, but we throw at most seven times. I need to find a probability function, X describes how many fours are thrown.

My problem is that I can either throw 6 times (when I get one of the 6! permutations of (1,2,3,4,5,6)) or 7 times (when one number appears at least twice).

I think the sample space must be sth. like this (where $\omega_i\in\{1,..,6\} \forall i)$:

$\Omega=\{\omega : \omega=(\omega_1,...,\omega_6), \omega_i \not= \omega_j\ \forall i\not=j, \}\cup A$ ... A is the set of all $(\omega_1,...,\omega_7)$ where the first six components $\omega_1,...,\omega_6$ are different numbers because otherwise we would be in the first set again. Maybe this is even too complicated, i don't know, but how does P(X=k) look like?

Answer

Throw the die seven times whatever happens during the six first throws and compare $X$ to the number $Y$ of fours during these seven throws.

One knows that $Y$ is binomial Bin$(7,\frac16)$, that is, $p_k=P(Y=k)$ is $p_k={7\choose k}\left(\frac16\right)^k\left(\frac56\right)^{7-k}$ for every $0\leqslant k\leqslant 7$.

The only case when $Y\ne X$ is if $X=1$ and $Y=2$. This happens if the six first throws produce every result exactly once and if the seventh throw produces a four, thus $P(X\ne Y)=P(X=1,Y=2)$ is $q=\frac{6!}{6^6}\cdot\frac16$.

Finally, $P(X=1)=p_1+q$, $P(X=2)=p_2-q$, and, for every $0\leqslant k\leqslant 7$ not $1$ or $2$, $P(X=k)=p_k$.