Expectation of a posterior distribution I'm trying to solve the following problem (under which is my attempt at it)

I'm confused on how to solve for the expectation here in a non-conditional manner (summing over all values of n) without doing it manually. Any help would be great! :)
 A: There is a probability of $\frac16$ that the game ends right away with a $6$ and a probability of $\frac56$ that you flip two fair coins and are then back in the initial situation. The expected number of heads from two fair coins is $1$. Thus, the expected number of heads $E$ in the initial situation satisfies
$$
E = \frac16\cdot0+\frac56(1+E)\;,
$$
with solution $E=5$.
A: To use the same symbol to refer both to a random variable and to the argument to its probability mass function is atrocious notation. Also, the notation should be established before the list of multiple-choice answers appears, i.e. it should say something like "Let $X$ be the number of heads that appear, and let $N$ be the number of rolls before a $6$ appears."
Now we can say things like $\operatorname E(X\mid N=6) = \frac 1 2 \cdot 12 = 6.$
And $\operatorname E(X\mid N) = N.$
And finally, $ \displaystyle \operatorname E(X) = \operatorname E( \operatorname E(X\mid N)) = \operatorname E(N) = \sum_{n=0}^\infty n\Pr(N=n).$
Notice the difference between capital $N,$ the random variable, and lower-case $n,$ the thing that runs through the list of possible values of (capital) $N.$ Without that distinction, one cannot even understand something like $\Pr(N=n).$
