You are offered a game where you roll 2 fair 6-sided die and add the sum to your total earnings. You can roll as many times as you'd like however, in the case where both die land on the same face, the games stops and you lose everything you gained until that point. How much would you pay to play this game?
By forming an inequality of expected value of re-rolling given our current earnings is $x$, we get $\frac16(0) + \frac56(x+7) > x$ or $x < 35$. This means optimal strategy is for us to re-roll any value where $x < 35$ and keep our earnings when $x > 35$. That being said, I am wondering if there is a computationally simple way to find the expected earnings of this game given our strategy. If $35$ is our indifference point, I think $(\frac56)^5 \cdot 35$ is a pretty good approximation since on average you may expect to roll 5 times given an average sum of 7 but this is the best I was able to come up with. Anything that might be better?
I do not know of a speedy way to compute the expected earnings.
Let $E_n$ be the expected earnings this strategy produces, starting from $n$ points. You want to determine $E_0$. Whenever $n\ge 35$, $E_n=n$, because the strategy stipulates that you stop on these numbers. For $n<35$, $$ E_n=\sum_{k=3}^{11} P(\text{two dice are unequal and sum to $k$})\cdot E_{n+k} $$ This allows you to compute $E_n$ for all $0\le n\le 34$ by starting from $n=34$, and working backwards to $n=0$, using the previously computed numbers for $E_{n+k}$. For example, $$ \begin{align} E_{34}&=\frac{2}{36}E_{37}+\frac{2}{36}E_{38}+\frac{4}{36}E_{39}+\frac{4}{36}E_{40}+ \frac6{36}E_{41}\\ &\quad+\frac{4}{36}E_{42}+\frac{4}{36}E_{43}+\frac{2}{36}E_{44}+\frac{2}{36}E_{45}. \\\\&=\frac{2}{36}\cdot{37}+\frac{2}{36}\cdot{38}+\frac{4}{36}\cdot{39}+\frac{4}{36}\cdot{40}+ \frac6{36}\cdot{41}\\ &\quad+ \frac{4}{36}\cdot{42}+\frac{4}{36}\cdot{43}+\frac{2}{36}\cdot{44}+\frac{2}{36}\cdot{45}. \\\\&=34+\tfrac16. \end{align} $$Using a computer to do these tedious calculations, I found that $E_0=\frac{5481504540706117}{385610460475392}$, which agrees with the approximation in Henry's comment.