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?
probability game-theory gambling
asked Nov 18 '18 at 23:34
IAlreadyHaveAKey
539310
|
show 2 more comments
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?
probability game-theory gambling
asked Nov 18 '18 at 23:34
IAlreadyHaveAKey
539310
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
1
Yes, but the strategy is independent of stage.
– Don Thousand
Nov 19 '18 at 2:07
|
show 2 more comments
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?
probability game-theory gambling
asked Nov 18 '18 at 23:34
IAlreadyHaveAKey
539310
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?
probability game-theory gambling
probability game-theory gambling
asked Nov 18 '18 at 23:34
IAlreadyHaveAKey
539310
asked Nov 18 '18 at 23:34
IAlreadyHaveAKey
539310
asked Nov 18 '18 at 23:34
IAlreadyHaveAKey
539310
asked Nov 18 '18 at 23:34
IAlreadyHaveAKey
539310
asked Nov 18 '18 at 23:34
IAlreadyHaveAKey
539310
539310
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
1
Yes, but the strategy is independent of stage.
– Don Thousand
Nov 19 '18 at 2:07
|
show 2 more comments
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
1
Yes, but the strategy is independent of stage.
– Don Thousand
Nov 19 '18 at 2:07
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
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
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
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
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
1
1
Yes, but the strategy is independent of stage.
– Don Thousand
Nov 19 '18 at 2:07
Yes, but the strategy is independent of stage.
– Don Thousand
Nov 19 '18 at 2:07
|
show 2 more comments
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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
1
Yes, but the strategy is independent of stage.
– Don Thousand
Nov 19 '18 at 2:07