Mixing
Duel question, why shoot when $P_1 (s)+P_2 (s)=1$
$begingroup$
There are two duelists with one shot each, their probability to hit from a distance $x$ is given by two functions $P_1 (x)$, $P_2 (x)$ that are both continuously and strictly decreasing from $P(0)=1, P(L)=0$. They walk towards each other from $L$ to $0$, and are each free to shoot whenever.
Our payouts: hit is 1, both miss or both die is 0 and die is -1.
We can logically come to the conclusion that they will shoot simultaneously, but we're having trouble showing that they will shoot from a distance $s$ when $P_1 (s) + P_2 (s) = 1$. Any help here would be greatly appreciated.
Assuming P1 is the better player we believe that P1 will shoot when when his payout is maximized (forcing P2 to shoot), this occurs at $max(P_1 (s) - P_2 (s))$.
probability game-theory
asked Jan 23 at 12:56
armaraarmara
1367
$endgroup$
|
show 9 more comments
$begingroup$
There are two duelists with one shot each, their probability to hit from a distance $x$ is given by two functions $P_1 (x)$, $P_2 (x)$ that are both continuously and strictly decreasing from $P(0)=1, P(L)=0$. They walk towards each other from $L$ to $0$, and are each free to shoot whenever.
Our payouts: hit is 1, both miss or both die is 0 and die is -1.
We can logically come to the conclusion that they will shoot simultaneously, but we're having trouble showing that they will shoot from a distance $s$ when $P_1 (s) + P_2 (s) = 1$. Any help here would be greatly appreciated.
Assuming P1 is the better player we believe that P1 will shoot when when his payout is maximized (forcing P2 to shoot), this occurs at $max(P_1 (s) - P_2 (s))$.
probability game-theory
asked Jan 23 at 12:56
armaraarmara
1367
$endgroup$
$begingroup$
Interesting. Why would the inferior shooter wait to allow the superior shooter to get to their preferred distance? If that would be positive expectation for the superior shooter, then the inferior one would be better off shooting at distance $0$, and thereby guaranteeing a $0$ outcome. Or am I misunderstanding?
$endgroup$
– lulu
Jan 23 at 13:21
$begingroup$
To be clear: my argument, if correct, would show that if $P_1(x)>P_2(x)$ for all $x>0$ then shooter $2$ should just shoot immediately to guarantee a $0$ outcome. Waiting at all leads to a negative expectation for the inferior shooter.
$endgroup$
– lulu
Jan 23 at 13:22
$begingroup$
Hey, sorry for not being clear. I've updated the question. Both duelists only have 1 shot each. Also, if player 2 gets hit and misses his own shot, he gets -1 while player 1 gets +1. If player 2 shoots too early with low probability, player 1 can just wait until he has 100% chance to hit, and then shoot.
$endgroup$
– armara
Jan 23 at 13:28
1
$begingroup$
Ah, maybe. Here's an interesting thing: If I return fire, your expectation is $P_1(x)-P_2(x)$. If I don't, your expectation is $2P_1(x)-1$. These are equal when $P_1(x)-P_2(x)=2P_1(x)-1implies P_1(x)+P_2(x)=1$. So I think the trick is to show that equilibrium occurs when those two expectations coincide.
$endgroup$
– lulu
Jan 23 at 14:06
1
$begingroup$
Well, glad you like it, but I am not persuaded that this really does provide an equilibrium. It seems likely, but I think further argument is required. Anyway I'll post it and edit as I think of more.
$endgroup$
– lulu
Jan 23 at 14:11
|
show 9 more comments
$begingroup$
There are two duelists with one shot each, their probability to hit from a distance $x$ is given by two functions $P_1 (x)$, $P_2 (x)$ that are both continuously and strictly decreasing from $P(0)=1, P(L)=0$. They walk towards each other from $L$ to $0$, and are each free to shoot whenever.
Our payouts: hit is 1, both miss or both die is 0 and die is -1.
We can logically come to the conclusion that they will shoot simultaneously, but we're having trouble showing that they will shoot from a distance $s$ when $P_1 (s) + P_2 (s) = 1$. Any help here would be greatly appreciated.
Assuming P1 is the better player we believe that P1 will shoot when when his payout is maximized (forcing P2 to shoot), this occurs at $max(P_1 (s) - P_2 (s))$.
probability game-theory
asked Jan 23 at 12:56
armaraarmara
1367
$endgroup$
There are two duelists with one shot each, their probability to hit from a distance $x$ is given by two functions $P_1 (x)$, $P_2 (x)$ that are both continuously and strictly decreasing from $P(0)=1, P(L)=0$. They walk towards each other from $L$ to $0$, and are each free to shoot whenever.
Our payouts: hit is 1, both miss or both die is 0 and die is -1.
We can logically come to the conclusion that they will shoot simultaneously, but we're having trouble showing that they will shoot from a distance $s$ when $P_1 (s) + P_2 (s) = 1$. Any help here would be greatly appreciated.
Assuming P1 is the better player we believe that P1 will shoot when when his payout is maximized (forcing P2 to shoot), this occurs at $max(P_1 (s) - P_2 (s))$.
probability game-theory
probability game-theory
asked Jan 23 at 12:56
armaraarmara
1367
asked Jan 23 at 12:56
armaraarmara
1367
edited Jan 23 at 13:26
armara
asked Jan 23 at 12:56
armaraarmara
1367
asked Jan 23 at 12:56
armaraarmara
1367
asked Jan 23 at 12:56
armaraarmara
1367
1367
$begingroup$
Interesting. Why would the inferior shooter wait to allow the superior shooter to get to their preferred distance? If that would be positive expectation for the superior shooter, then the inferior one would be better off shooting at distance $0$, and thereby guaranteeing a $0$ outcome. Or am I misunderstanding?
$endgroup$
– lulu
Jan 23 at 13:21
$begingroup$
To be clear: my argument, if correct, would show that if $P_1(x)>P_2(x)$ for all $x>0$ then shooter $2$ should just shoot immediately to guarantee a $0$ outcome. Waiting at all leads to a negative expectation for the inferior shooter.
$endgroup$
– lulu
Jan 23 at 13:22
$begingroup$
Hey, sorry for not being clear. I've updated the question. Both duelists only have 1 shot each. Also, if player 2 gets hit and misses his own shot, he gets -1 while player 1 gets +1. If player 2 shoots too early with low probability, player 1 can just wait until he has 100% chance to hit, and then shoot.
$endgroup$
– armara
Jan 23 at 13:28
1
$begingroup$
Ah, maybe. Here's an interesting thing: If I return fire, your expectation is $P_1(x)-P_2(x)$. If I don't, your expectation is $2P_1(x)-1$. These are equal when $P_1(x)-P_2(x)=2P_1(x)-1implies P_1(x)+P_2(x)=1$. So I think the trick is to show that equilibrium occurs when those two expectations coincide.
$endgroup$
– lulu
Jan 23 at 14:06
1
$begingroup$
Well, glad you like it, but I am not persuaded that this really does provide an equilibrium. It seems likely, but I think further argument is required. Anyway I'll post it and edit as I think of more.
$endgroup$
– lulu
Jan 23 at 14:11
|
show 9 more comments
$begingroup$
Interesting. Why would the inferior shooter wait to allow the superior shooter to get to their preferred distance? If that would be positive expectation for the superior shooter, then the inferior one would be better off shooting at distance $0$, and thereby guaranteeing a $0$ outcome. Or am I misunderstanding?
$endgroup$
– lulu
Jan 23 at 13:21
$begingroup$
To be clear: my argument, if correct, would show that if $P_1(x)>P_2(x)$ for all $x>0$ then shooter $2$ should just shoot immediately to guarantee a $0$ outcome. Waiting at all leads to a negative expectation for the inferior shooter.
$endgroup$
– lulu
Jan 23 at 13:22
$begingroup$
Hey, sorry for not being clear. I've updated the question. Both duelists only have 1 shot each. Also, if player 2 gets hit and misses his own shot, he gets -1 while player 1 gets +1. If player 2 shoots too early with low probability, player 1 can just wait until he has 100% chance to hit, and then shoot.
$endgroup$
– armara
Jan 23 at 13:28
1
$begingroup$
Ah, maybe. Here's an interesting thing: If I return fire, your expectation is $P_1(x)-P_2(x)$. If I don't, your expectation is $2P_1(x)-1$. These are equal when $P_1(x)-P_2(x)=2P_1(x)-1implies P_1(x)+P_2(x)=1$. So I think the trick is to show that equilibrium occurs when those two expectations coincide.
$endgroup$
– lulu
Jan 23 at 14:06
1
$begingroup$
Well, glad you like it, but I am not persuaded that this really does provide an equilibrium. It seems likely, but I think further argument is required. Anyway I'll post it and edit as I think of more.
$endgroup$
– lulu
Jan 23 at 14:11
$begingroup$
Interesting. Why would the inferior shooter wait to allow the superior shooter to get to their preferred distance? If that would be positive expectation for the superior shooter, then the inferior one would be better off shooting at distance $0$, and thereby guaranteeing a $0$ outcome. Or am I misunderstanding?
$endgroup$
– lulu
Jan 23 at 13:21
$begingroup$
Interesting. Why would the inferior shooter wait to allow the superior shooter to get to their preferred distance? If that would be positive expectation for the superior shooter, then the inferior one would be better off shooting at distance $0$, and thereby guaranteeing a $0$ outcome. Or am I misunderstanding?
$endgroup$
– lulu
Jan 23 at 13:21
$begingroup$
To be clear: my argument, if correct, would show that if $P_1(x)>P_2(x)$ for all $x>0$ then shooter $2$ should just shoot immediately to guarantee a $0$ outcome. Waiting at all leads to a negative expectation for the inferior shooter.
$endgroup$
– lulu
Jan 23 at 13:22
$begingroup$
To be clear: my argument, if correct, would show that if $P_1(x)>P_2(x)$ for all $x>0$ then shooter $2$ should just shoot immediately to guarantee a $0$ outcome. Waiting at all leads to a negative expectation for the inferior shooter.
$endgroup$
– lulu
Jan 23 at 13:22
$begingroup$
Hey, sorry for not being clear. I've updated the question. Both duelists only have 1 shot each. Also, if player 2 gets hit and misses his own shot, he gets -1 while player 1 gets +1. If player 2 shoots too early with low probability, player 1 can just wait until he has 100% chance to hit, and then shoot.
$endgroup$
– armara
Jan 23 at 13:28
$begingroup$
Hey, sorry for not being clear. I've updated the question. Both duelists only have 1 shot each. Also, if player 2 gets hit and misses his own shot, he gets -1 while player 1 gets +1. If player 2 shoots too early with low probability, player 1 can just wait until he has 100% chance to hit, and then shoot.
$endgroup$
– armara
Jan 23 at 13:28
1
1
$begingroup$
Ah, maybe. Here's an interesting thing: If I return fire, your expectation is $P_1(x)-P_2(x)$. If I don't, your expectation is $2P_1(x)-1$. These are equal when $P_1(x)-P_2(x)=2P_1(x)-1implies P_1(x)+P_2(x)=1$. So I think the trick is to show that equilibrium occurs when those two expectations coincide.
$endgroup$
– lulu
Jan 23 at 14:06
$begingroup$
Ah, maybe. Here's an interesting thing: If I return fire, your expectation is $P_1(x)-P_2(x)$. If I don't, your expectation is $2P_1(x)-1$. These are equal when $P_1(x)-P_2(x)=2P_1(x)-1implies P_1(x)+P_2(x)=1$. So I think the trick is to show that equilibrium occurs when those two expectations coincide.
$endgroup$
– lulu
Jan 23 at 14:06
1
1
$begingroup$
Well, glad you like it, but I am not persuaded that this really does provide an equilibrium. It seems likely, but I think further argument is required. Anyway I'll post it and edit as I think of more.
$endgroup$
– lulu
Jan 23 at 14:11
$begingroup$
Well, glad you like it, but I am not persuaded that this really does provide an equilibrium. It seems likely, but I think further argument is required. Anyway I'll post it and edit as I think of more.
$endgroup$
– lulu
Jan 23 at 14:11
|
show 9 more comments
1 Answer
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active
oldest
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$begingroup$
Note what happens if shooter $1$ decides to fire. This gives shooter $2$ a choice. Either he can return fire or he can wait. Of course, if he waits (and survives) he is guaranteed a payout of $1$ at distance $0$. If he returns fire then shooter $1's$ expected payout is $$(P_1(x))(1-P_2(x))-(1-P_1(x))(P_2(x))=P_1(x)-P_2(x)$$
If shooter $1$ holds off then shooter $1's$ expected payout is $$P_1(x)-(1-P_1(x))=2P_1(x)-1$$
If equilibrium occurs when these expectations coincide then we have $$P_1(x)-P_2(x)=2P_1(x)-1$$ which is equivalent to $$P_1(x)+P_2(x)=1$$
All that remains is to show that when the two expectations do not coincide, both players should wait.
Toward that end: Consider the difference of the two expectations $$F(x)=P_1(x)-P_2(x)-(2P_1(x)-1)=1-P_1(x)-P_2(x)$$
Then $F(0)=1$ and $F(x)$ is strictly decreasing.
Case I: $F(x)>0$. Then if shooter $1$ fires, shooter $2$ should hold off so shooter $1's$ expectation is $2P_1(x)-1$. That function is strictly increasing so shooter $1$ should wait.
Case II: $F(x)<0$ Then, if shooter $1$ is holding off, shooter $2's$ expectation is $2P_2(x)-1$ which is increasing. It follows that shooter $1's$ expectation must be decreasing in that region, so shooter $1$ should shoot when $F(x)=0$.
answered Jan 23 at 14:16
lulululu
42.9k25080
$endgroup$
add a comment |
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$begingroup$
Note what happens if shooter $1$ decides to fire. This gives shooter $2$ a choice. Either he can return fire or he can wait. Of course, if he waits (and survives) he is guaranteed a payout of $1$ at distance $0$. If he returns fire then shooter $1's$ expected payout is $$(P_1(x))(1-P_2(x))-(1-P_1(x))(P_2(x))=P_1(x)-P_2(x)$$
If shooter $1$ holds off then shooter $1's$ expected payout is $$P_1(x)-(1-P_1(x))=2P_1(x)-1$$
If equilibrium occurs when these expectations coincide then we have $$P_1(x)-P_2(x)=2P_1(x)-1$$ which is equivalent to $$P_1(x)+P_2(x)=1$$
All that remains is to show that when the two expectations do not coincide, both players should wait.
Toward that end: Consider the difference of the two expectations $$F(x)=P_1(x)-P_2(x)-(2P_1(x)-1)=1-P_1(x)-P_2(x)$$
Then $F(0)=1$ and $F(x)$ is strictly decreasing.
Case I: $F(x)>0$. Then if shooter $1$ fires, shooter $2$ should hold off so shooter $1's$ expectation is $2P_1(x)-1$. That function is strictly increasing so shooter $1$ should wait.
Case II: $F(x)<0$ Then, if shooter $1$ is holding off, shooter $2's$ expectation is $2P_2(x)-1$ which is increasing. It follows that shooter $1's$ expectation must be decreasing in that region, so shooter $1$ should shoot when $F(x)=0$.
answered Jan 23 at 14:16
lulululu
42.9k25080
$endgroup$
add a comment |
$begingroup$
Note what happens if shooter $1$ decides to fire. This gives shooter $2$ a choice. Either he can return fire or he can wait. Of course, if he waits (and survives) he is guaranteed a payout of $1$ at distance $0$. If he returns fire then shooter $1's$ expected payout is $$(P_1(x))(1-P_2(x))-(1-P_1(x))(P_2(x))=P_1(x)-P_2(x)$$
If shooter $1$ holds off then shooter $1's$ expected payout is $$P_1(x)-(1-P_1(x))=2P_1(x)-1$$
If equilibrium occurs when these expectations coincide then we have $$P_1(x)-P_2(x)=2P_1(x)-1$$ which is equivalent to $$P_1(x)+P_2(x)=1$$
All that remains is to show that when the two expectations do not coincide, both players should wait.
Toward that end: Consider the difference of the two expectations $$F(x)=P_1(x)-P_2(x)-(2P_1(x)-1)=1-P_1(x)-P_2(x)$$
Then $F(0)=1$ and $F(x)$ is strictly decreasing.
Case I: $F(x)>0$. Then if shooter $1$ fires, shooter $2$ should hold off so shooter $1's$ expectation is $2P_1(x)-1$. That function is strictly increasing so shooter $1$ should wait.
Case II: $F(x)<0$ Then, if shooter $1$ is holding off, shooter $2's$ expectation is $2P_2(x)-1$ which is increasing. It follows that shooter $1's$ expectation must be decreasing in that region, so shooter $1$ should shoot when $F(x)=0$.
answered Jan 23 at 14:16
lulululu
42.9k25080
$endgroup$
add a comment |
$begingroup$
Note what happens if shooter $1$ decides to fire. This gives shooter $2$ a choice. Either he can return fire or he can wait. Of course, if he waits (and survives) he is guaranteed a payout of $1$ at distance $0$. If he returns fire then shooter $1's$ expected payout is $$(P_1(x))(1-P_2(x))-(1-P_1(x))(P_2(x))=P_1(x)-P_2(x)$$
If shooter $1$ holds off then shooter $1's$ expected payout is $$P_1(x)-(1-P_1(x))=2P_1(x)-1$$
If equilibrium occurs when these expectations coincide then we have $$P_1(x)-P_2(x)=2P_1(x)-1$$ which is equivalent to $$P_1(x)+P_2(x)=1$$
All that remains is to show that when the two expectations do not coincide, both players should wait.
Toward that end: Consider the difference of the two expectations $$F(x)=P_1(x)-P_2(x)-(2P_1(x)-1)=1-P_1(x)-P_2(x)$$
Then $F(0)=1$ and $F(x)$ is strictly decreasing.
Case I: $F(x)>0$. Then if shooter $1$ fires, shooter $2$ should hold off so shooter $1's$ expectation is $2P_1(x)-1$. That function is strictly increasing so shooter $1$ should wait.
Case II: $F(x)<0$ Then, if shooter $1$ is holding off, shooter $2's$ expectation is $2P_2(x)-1$ which is increasing. It follows that shooter $1's$ expectation must be decreasing in that region, so shooter $1$ should shoot when $F(x)=0$.
answered Jan 23 at 14:16
lulululu
42.9k25080
$endgroup$
Note what happens if shooter $1$ decides to fire. This gives shooter $2$ a choice. Either he can return fire or he can wait. Of course, if he waits (and survives) he is guaranteed a payout of $1$ at distance $0$. If he returns fire then shooter $1's$ expected payout is $$(P_1(x))(1-P_2(x))-(1-P_1(x))(P_2(x))=P_1(x)-P_2(x)$$
If shooter $1$ holds off then shooter $1's$ expected payout is $$P_1(x)-(1-P_1(x))=2P_1(x)-1$$
If equilibrium occurs when these expectations coincide then we have $$P_1(x)-P_2(x)=2P_1(x)-1$$ which is equivalent to $$P_1(x)+P_2(x)=1$$
All that remains is to show that when the two expectations do not coincide, both players should wait.
Toward that end: Consider the difference of the two expectations $$F(x)=P_1(x)-P_2(x)-(2P_1(x)-1)=1-P_1(x)-P_2(x)$$
Then $F(0)=1$ and $F(x)$ is strictly decreasing.
Case I: $F(x)>0$. Then if shooter $1$ fires, shooter $2$ should hold off so shooter $1's$ expectation is $2P_1(x)-1$. That function is strictly increasing so shooter $1$ should wait.
Case II: $F(x)<0$ Then, if shooter $1$ is holding off, shooter $2's$ expectation is $2P_2(x)-1$ which is increasing. It follows that shooter $1's$ expectation must be decreasing in that region, so shooter $1$ should shoot when $F(x)=0$.
answered Jan 23 at 14:16
lulululu
42.9k25080
edited Jan 23 at 14:40
answered Jan 23 at 14:16
lulululu
42.9k25080
answered Jan 23 at 14:16
lulululu
42.9k25080
answered Jan 23 at 14:16
lulululu
42.9k25080
42.9k25080
add a comment |
add a comment |
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$begingroup$
Interesting. Why would the inferior shooter wait to allow the superior shooter to get to their preferred distance? If that would be positive expectation for the superior shooter, then the inferior one would be better off shooting at distance $0$, and thereby guaranteeing a $0$ outcome. Or am I misunderstanding?
$endgroup$
– lulu
Jan 23 at 13:21
$begingroup$
To be clear: my argument, if correct, would show that if $P_1(x)>P_2(x)$ for all $x>0$ then shooter $2$ should just shoot immediately to guarantee a $0$ outcome. Waiting at all leads to a negative expectation for the inferior shooter.
$endgroup$
– lulu
Jan 23 at 13:22
$begingroup$
Hey, sorry for not being clear. I've updated the question. Both duelists only have 1 shot each. Also, if player 2 gets hit and misses his own shot, he gets -1 while player 1 gets +1. If player 2 shoots too early with low probability, player 1 can just wait until he has 100% chance to hit, and then shoot.
$endgroup$
– armara
Jan 23 at 13:28
1
$begingroup$
Ah, maybe. Here's an interesting thing: If I return fire, your expectation is $P_1(x)-P_2(x)$. If I don't, your expectation is $2P_1(x)-1$. These are equal when $P_1(x)-P_2(x)=2P_1(x)-1implies P_1(x)+P_2(x)=1$. So I think the trick is to show that equilibrium occurs when those two expectations coincide.
$endgroup$
– lulu
Jan 23 at 14:06
1
$begingroup$
Well, glad you like it, but I am not persuaded that this really does provide an equilibrium. It seems likely, but I think further argument is required. Anyway I'll post it and edit as I think of more.
$endgroup$
– lulu
Jan 23 at 14:11