I am interested in the solution of the following exercise:
One spins a wheel with fields marked 1 to 50. One has $N \in\mathbb{N}$ turns. After every turn on can either stop and receive the amount shown in the field or spin once more. The probability that the wheel stops at any number $k$ is $\frac{1}{50}$.
What is the optimal stopping time for maximal payout? What is the exact rule for $N = 3$?
I am able to solve it by setting up a suitable filtration on $\{ 1,\dots,50 \}^N$ and then calculate the Snell envelope for the process $\xi_i$ which corresponds to the result of the $i$-th turn. This leads to an optimal stopping time. But even for $N = 3$ it is quite tedious to calculate the exact stopping rule.
Therefore I am interested in any other way to solve this. Perhaps there is a way to simplify the Snell envelope argument.
$\newcommand{\P}{\mathbb{P}}\newcommand{\E}{\mathbb{E}}$I would say the approach for optimal play should be something like this, assuming we want to maximise expected earnings. Let $X$ be the random variable of the value on any given spin (here $X\sim \mathrm{Unif}\{1,\ldots,50\}$). Let $\mu = \E[X] = 51/2$. If $N=1$, then your expected earnings is $v_1=\mu$. If $N=2$, this is what you should do: if your first spin is greater than $v_1$ ($=\mu$), stop and take it. Otherwise, spin again and on average earn $v_1=\mu$. Thus your expected earnings if $N=2$ is $$v_2 = \E[X\mid X > v_1]\P(X > v_1) + v_1\P(X\leq v_1).$$
In general with $n\geq 2$ spins, similar logic applies: if you first spin higher than $v_{n-1}$, then you should stop and take the result. Otherwise spin again and earn on average $v_{n-1}$. Thus we have a recursive formula for the expected earnings with $n\geq 2$ spins: $$\color{blue}{v_{n} = \E[X\mid X > v_{n-1}]\P(X > v_{n-1}) + v_{n-1}\P(X \le v_{n-1})}.$$ Using this and $v_1=\mu$, you can recursively compute the expected earnings of the game for $n$ turns with $n=1,\ldots,N$, because the required conditional expectations and probabilities are quite easy to calculate here. (This is because all probability calculations are for a uniform distribution and the conditional distributions are uniform.)
You can also work out the expected number of spins you will do. Once you work out the values of $v_1,\ldots,v_{N-1}$, if you call $E_{N}$ the expected number of spins you will do if you start with $N$ spins allowed, we have the recursive formulation $$\color{blue}{E_{N} = 1 + E_{N-1}\cdot \P(X\leq v_{N-1})} \quad \text{for }N\geq 2, \text{with }E_1 = 1.$$
(This is because you stop if you first spin greater than $v_{N-1}$, otherwise you spin again with $N-1$ rolls remaining.) You should be able to use this to work out $E_{3}$ for example.