Expected payoff of dice game 
You roll a fair $6$-sided die. For each roll, you're paid the face value. The game stops when you roll a $1,2,3$. If you roll a $4,5,6$, you can roll again and keep accumulating payments.

There are several ways to solve this problem. I tried the geometric variable approach, but I feel that my answer, while correct, is not complete, but I don't know what I'm missing.
Essentially, define $N$ as a geometric random variable for the number of rolls until the game terminates. $E[N] = 2$.
Define $X$ as the pay off per roll. $E[X] = 3.5$.
Each roll is independent. Define $P$ as the payoff. So $P = N * X$
So the expected pay off should be $E[P] = E[N] * E[X] = 7$. This is the step that I feel like I'm missing some justification on, and I don't feel it's complete.
Is it valid to define $P = N * X$? If so, then is this answer complete, or is there something missing?
 A: The result is correct, but it's hard to justify along the lines you proposed.
As an alternative method:
Consider the possible outcomes of the first roll, and note that, if the game does not end, then it restarts (and, of course, will have the same expectation, $E$, going forward).
We see that $$E=\frac 16 \times \left(1+2+3+(4+E)+(5+E)+(6+E)\right)\implies 6E=21+3E\implies E=7$$
A: While $P = NX$ is a valid definition, $N$ and $X$ are not independent, so you cannot say that $\mathbb E(P) = \mathbb E(N)\mathbb E(X)$, which is generally false for dependent variables. Additionally, I do not think that $\mathbb E(X) = 3.5$ -- but to get the right answer for your question, I think you'll want to take a different approach regardless.
A: To make your approach more formal, let $N$ be the number of rolls until the game stops, and $X_i$ the value of the $i$th roll ($i=1,2....$). The total payoff $P$ is $\sum\limits_{i=1}^\infty X_i$. Given $N$:
$$E(P\vert N)=E(\sum\limits_{i=1}^\infty$X_i\vert N)=\sum\limits_{i=1}^N E(X_i\vert N)=5\cdot (N-1)+2=5N-3$$
since the expected value of each roll that didn't end the game is $5$ (options are $4,5,6$) and the terminal roll's expectation is $2$.
From the tower law,
$$E(P)=E(E(P\vert N))=E(5N-3)=7$$
You can use this method to compute $E(X)=E(E(\tfrac{P}{N}\vert{N}))=5-3E(\tfrac{1}{N})$. Wolfram says the inside expectation is $0.69$ so $E(X)=2.92$ which is not $3.5$.
This makes sense in the following way: $E(NX)=7$ while $E(N)E(X)<3.5\cdot 2$ so $Cov(N,X)>0$ and indeed, higher values of $N$ tend to go with higher values of $X$, as there were many rolls of $4-6$ before the terminal $1-3$.
