I roll 6-sided dice until the sum exceeds 50. What is the expected value of the final roll? I roll 6-sided dice until the sum exceeds 50. What is the expected value of the final roll? 
I am not sure how to set this one up. This one is not homework, by the way, but a question I am making up that is inspired by one. I'm hoping this will help me understand what's going on better.

The purpose of this answer here is to convince readers that the distribution of the roll 'to get me over $50$' is not necessarily that of the standard 6=sided die roll.
Recall that a stopping time is a positive integer valued random variable $\tau$ for which $\{\tau \leq n \} \in \mathcal{F}_n$, where $\mathcal{F}_n = \sigma(X_0, X_1, \cdots, X_n)$ is the canonical filtration with respect to the (time-homogeneous) markov chain $X_n$.
The Strong Markov Property asserts (in this case) that conditioned on the event $\tau < \infty$, the random variables $X_1$ and $X_{\tau + 1} - X_{\tau}$ are equidistributed. Letting $\tau = -1 + \min\{k \mid X_k > 50\}$ ought to prove the equidistribution with a regular die roll, right?
Well, no. What happens is that $\tau + 1$ is a stopping time, but $\tau$ is not, because it looks into the future one timestep. This is just enough to throw off the SMP.
