# Probability of multiple independent events occurring after repeated attempts

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.

• The solution involves a fairly messy Inclusion/Exclusion. For information about a closely related problem, please see the Coupon Collector's Problem (Wikipedia). Commented Mar 27, 2016 at 14:59
• It may be that I'm not understanding something here, but shouldn't the number of dice in the bag not matter if all of them are identical? No matter which die I draw, I will still roll a number from 1 to 8 and which die I rolled will not be taken into account. Commented Mar 27, 2016 at 15:40
• How can only one of $10$ identical octahedrons be a fair eight-sided die numbered $1$-$8$? If they're identical, either all or none of them are a fair eight-sided die numbered $1$-$8$, no? Commented Mar 27, 2016 at 22:01
• @AndréNicolas: It's not messy; it's a rather straightforward application of inclusion-exclusion. Commented Mar 27, 2016 at 22:45
• @joriki: Messier than the expressions OP was experimenting with. Commented Mar 27, 2016 at 22:54

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