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Suppose there are two betting games $G_1$ and $G_2$, the outcomes of which are entirely independent of player input.

The expected return for a single iteration of $G_1$ is $r_1$ and for $G_2$ is $r_2$, these returns being a percentage of your bet.

Design a game $G$ where you start by choosing either $G_1$ or $G_2$, and you have a chance of getting to play again, and you again choose between the two games.

Suppose if you choose $G_1$ the chance of playing again is $p_1$ and for $G_2$ is $p_2$.

If you take the strategy of always choosing $G_1$, then the overall expected return of $G$ is $$\frac{r_1}{1-p_1}.$$

If we assume that $$\frac{r_1}{1-p_1} = \frac{r_2}{1-p_2},$$

Is this enough to conclude that $G$ has no optimal strategy and that the overall return is the same regardless of your choice at each stage?

If not, what conditions can we place on $r_1,r_2,p_1,p_2$ to ensure that $G$ has no optimal strategy?

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  • $\begingroup$ Long term, both options are equivalent, but that certainly is not true short term. More specifically, the game with less variance of results will most likely do better short term. $\endgroup$ – Don Thousand Nov 19 '18 at 0:00
  • $\begingroup$ I am talking long term strategy though. $\endgroup$ – IAlreadyHaveAKey Nov 19 '18 at 0:23
  • $\begingroup$ In that case, I see no difference. $\endgroup$ – Don Thousand Nov 19 '18 at 0:42
  • $\begingroup$ How do we know that there is no optimal strategy though, like choosing some pattern, say $G_1$, $G_2$, $G_1$, $G_2$, ... which will result in a higher expected return in the long run? You get to choose which game to play at every stage. $\endgroup$ – IAlreadyHaveAKey Nov 19 '18 at 0:58
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    $\begingroup$ Yes, but the strategy is independent of stage. $\endgroup$ – Don Thousand Nov 19 '18 at 2:07

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