My friends and I invented a dice game called 18s to pass the time in between warhammer games.
The premise is simple.
Roll 18 dice. Once rolled, remove all 6s and roll the remaining dice.
Keep doing this until all dice are gone OR you don't roll a 6.
My question is: what are the odds of leaving yourself with 0 dice?
I've been trying to solve it, but it is proving too much.
Let $p={1\over6}$ be the probability of rolling a $6$ and let $q=1-p={5\over6}$ be the probability of rolling anything else. If you start with $n$ dice, the probability of winning the game (i.e., rolling all $6$'s on a single turn before rolling all non-$6$'s) is given by recursive formula:
$$P(n)=\sum_{k=0}^{n-1}{n\choose k}p^{n-k}q^kP(k)$$
with $P(0)=1$. If I've done the arithmetic correctly, the first few of these are
$$\begin{align} P(0)&=1\\ P(1)&=p\\ P(2)&=p^2(1+2q)\\ P(3)&=p^3(1+3q+3q^2+6q^3)\\ P(4)&=p^4(1+4q+6q^2+16q^3+12q^4+12q^5+24q^6) \end{align}$$
A patient calculator should be able to extend this up to $P(18)$, which will have the form $p^{18}Q_{18}(q)$, where $Q_{18}$ is a polynomial of degree $18\cdot17/2=153$ of the form $1+18q+153q^2+\cdots+18!q^{153}$. I don't recommend doing it by hand!
It's worth noting that if you let $p={1\over2}$ instead of $1\over6$ -- i.e., if you play the game with coins instead of dice, say winning if you toss all heads before all tails (and removing all coins that come up heads on each turn) -- then everything simplifies extremely nicely:
$$p={1\over2}\implies P(n)={1\over2}\quad\text{for all }n\ge1$$
This can be proved by induction from the recursive formula, or, more simply, by noting that no matter which round the game ends on, it ends either in a win where the coins all come up heads, or in a loss where the coins all come up tails, each of which is equally likely.