Expected number of strikes to kill a $3$-headed dragon

Question

You want to slay a dragon with $3$ heads. There is $0.7$ chance of destroying a head and $0.3$ chance of missing. If you miss, a new head will grow. $X$ is a random variable for the number of rounds until you slay all $3$ heads. Find $E[X]$.

I get the following pmf that $P(X = n) = {?}\ 0.7^{k} 0.3^{k-3}$ where $n$ is the number of slays ($3$, $5$, $7$...) and $k$ is the number of strikes that destroy a head. I am struggling with coming up with a coefficient for the expression or number of ways to permute successes and failures. I understand that the missing strikes cannot be at the end, and there also cannot be more than $2$ successful strikes before the 1st miss. How to think of a expression to capture the coefficient?

Answer 1

To expand on the discussion in the comments:

Let $E_-$ denote the expected number of rounds it takes to lower the number of heads by $1$. This is, of course, independent of the number of heads you are currently facing. It is also clear that the expected number of rounds needed to remove $n$ heads is just $nE_-$.

Considering the results of the next round we see that $$E_-=.7\times 1+.3\times (1+2E_-)=1+.6E_-\implies E_-=2.5$$

It follows that you expect it to take $3E_-=7.5$ rounds to slay the dragon.

Answer 2

Let $\epsilon_1,\ldots,$ be iid random variables such that $P(\epsilon_1=1) = 0.3$ and $P(\epsilon_1=-1) = 0.7$.

Defining the random walk $S_0=3$, $S_n = 3+\sum_{k=1}^n \epsilon_k$, and the hitting time $\tau$ where $S_{\tau}=0$, you're looking for $E[\tau]$.

It is well-known that $E[\tau] = \frac{3}{0.7-0.3}=7.5$.