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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?

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  • $\begingroup$ My method of finding the $35$ gave an expected gain from the start with this strategy of about $14.215$ while your $(\frac56)^5 \cdot 35\approx 14.066$ $\endgroup$ Commented Oct 27, 2023 at 16:01
  • $\begingroup$ @Henry Where one can read on your method? $\endgroup$ Commented Oct 27, 2023 at 20:24
  • $\begingroup$ I did the same as Mike Earnest, and assumed you would stop at a very high level, saw that that give poor results until you took the money starting from $35$ or $36$ and then I read the corresponding expectation at the start. $\endgroup$ Commented Oct 27, 2023 at 21:18

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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.

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  • $\begingroup$ Is then 35 the optimal threshold? $\endgroup$ Commented Oct 27, 2023 at 20:39
  • $\begingroup$ @user The expectation is the same whether you re-roll at $35$ or take the money. Above that it is better to take the money unless you take additional pleasure from gambling to compensate for the reduced expectation; below it is better to re-roll unless you are risk averse. $\endgroup$ Commented Oct 27, 2023 at 21:20

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