# Potential strategy in a random game

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?

• 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. – Don Thousand Nov 19 '18 at 0:00
• I am talking long term strategy though. – IAlreadyHaveAKey Nov 19 '18 at 0:23
• In that case, I see no difference. – Don Thousand Nov 19 '18 at 0:42
• 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. – IAlreadyHaveAKey Nov 19 '18 at 0:58
• Yes, but the strategy is independent of stage. – Don Thousand Nov 19 '18 at 2:07