Probability of multiple independent events occurring after repeated attempts

Question

Suppose there is a bag of 10 identical octahedrons, one of which is a fair eight-sided die numbered 1-8.

Randomly draw the one of the octahedrons out of the bag. If you draw the die, you must roll it. Put the octahedron back into the bag before drawing again.

What is the probability of you getting rolls of each number at least once in $n$ tries?

I realize this question wasn't very elegantly put, and that's because this was a question I asked myself when I was playing a game. The principle is the same but the question was changed so that game-specific vocabulary don't confuse people.

I first tried $(0.1^8) (8!/8^8) \,{_n}C_8$ but then quickly realized my mistake as $n<8$ didn't give $P=0$ and it was also possible for $P$ to be $>1$.

After a few more failed attempts and some research I realized that this question might be beyond my current knowledge and that this rabbit hole might be deeper than I initially thought. It is a rather weird question, but I was hoping if I can get some answers here.

Edit: It would seem that my question ended up being confusing after all, so let me clarify. The octahedrons are shaped the exact same but only one of them is actually a die. Or really, the point was that every time there's only a 0.1 chance that you get to roll the die.

Answer 1

If I understand the question correctly, you have a coupon collector's problem with $m=8$ coupon types, but you're only drawing a coupon with probability $\frac1r$ on each try, with $r=10$. Given the probability of having a complete coupon collection after drawing $k$ coupons,

$$\def\stir#1#2{\left\{#1\atop#2\right\}}\frac{m!}{m^k}\stir km$$

(see here), and using the binomial distribution for the number of coupons you actually draw, the desired probability is

\begin{align} &r^{-n}\sum_{k=0}^n\binom nk(r-1)^{n-k}\frac{m!}{m^k}\stir km\\ ={}&r^{-n}\sum_{k=0}^n\binom nk(r-1)^{n-k}\frac1{m^k}\sum_{j=0}^m(-1)^{m-j}\binom mjj^k \\ ={}&r^{-n}\sum_{j=0}^m\binom mj(-1)^{m-j}\sum_{k=0}^n\binom nk(r-1)^{n-k}\frac{j^k}{m^k}\\ ={}&r^{-n}\sum_{j=0}^m\binom mj(-1)^{m-j}\left(r-1+\frac jm\right)^n\\ ={}&\sum_{j=0}^m\binom mj(-1)^j\left(1-\frac j{rm}\right)^n\;. \end{align}

We could have obtained the same result directly by inclusion-exclusion; the last factor is the probability that any particular $j$ of the numbers have not been rolled after $n$ tries.