Value of $x$ for a probability game to be fair To make it clear, I am not asking the following question, though it is related.
My question is instead the following.

A person pays $k$  dollars to play a game. In the game, they roll a single 6-sided die. If it lands anywhere from $1$ to $5$ inclusive, they win exactly that many dollars. If they land on $6$, then the game terminates.
What is the value of $x$ so that the game is fair?

I am stuck for the following reason. It is clear that the expected number of rolls to land on a specific value is $6$. However, my concern is the journey it took to finally land on that $6$ to terminate the sequence. The money won for a $(1,1,1,6)$ journey is very different from the money won on a $(5,5,5,6)$ journey.
I know that we want the expectation to be zero, but I am not sure what to define as the random variable $X$, much less so how to compute $E(X)$.

Let $N$ denote the number of rolls needed to arrive at result $6$ and let $X_i$ denote the gain at the $i$-th roll. Then the total gain can be expressed as:$$X=X_1+\cdots+X_{N-1}$$Here: $$\mathbb EX_i=\frac16[1+2+3+4+5]=\frac52$$
Then for positive integer $n$: $$\mathbb E[X\mid N=n]=\mathbb E[X_1+\cdots+X_n]=\frac{5n}2$$
Applying the law of total expectation we find:$$\mathbb EX=\sum_{n=1}^{\infty}P(N=n)\mathbb E[X\mid N=n]=\sum_{n=1}^{\infty}P(N=n)\frac52n=\frac52\mathbb EN=\frac52\cdot6=15$$

Another route (suggested by @lulu).
If $\mu$ denotes $\mathbb EX$ then:$$\mu=\frac16\cdot0+\sum_{j=1}^5\frac16(j+\mu)=\frac52+\frac56\mu$$telling us directly that $\mu=15$.
