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I have come across this problem while trying to understand game theory. The problem presents two players (hunters) that want to hunt in one of two regions available. Region $A$ has $12$ rabbits and region $B$ has $18$ rabbits. If the two hunters go to different regions they will acquire all the rabbits in their respective hunting ground, but if they go to the same region they'll acquire half the rabbits each, assuming they have the same skills. I have found the mixed strategy that they need to follow, which will be the same for both: they will go to region $A$ with probability $p=\frac{3}{5}$ or to region $B$ with probability $1-p=\frac{2}{5}$.

The problem emerges when I try to further the problem by adding a third hunter in the game. I would assume that an obvious answer would come up, like one region would give the best result when chosen for all hunters. However, what I'm seeing by creating the payoff tables is that if the third hunter chooses region $A$ then it's in everyone's benefit if the other two hunters choose region $B$, thus having the result array be $(H_1,H_2,H_3)=(9,9,12)$. But if hunter 3 chooses region $B$, the best outcome is when the other two hunters don't choose the same region between themselves.

I am unsure as to how I can interpret this or if there's another way to find the answer to this more complicated question. Can I apply a method to this problem and find mixed strategies for the three hunters? I am not necessarily looking for the answer directly, but for someone to point out a method or way of thinking for me to find the answer alone.

Edit: Trying again after the suggestions, I did this and hopefully someone can verify if my solution is correct:

Suppose that $p$ is the probability with which hunter 1 goes to region $A$ and $(1-p)$ if he goes to region $B$. That same probability applies to the other hunters. There are two cases: the other two go to the same region or to different regions. Therefore, we have the two expected payoffs:

• $V(H_1 | H_2=H_3=A)=4\cdot p+18\cdot (1-p)=18-14\cdot p$
• $V(H_1 | H_2=A, H_3=B)=6\cdot 9+9\cdot (1-p)=9-3\cdotp$
• $V(H_1 | H_2=H_3=B)= 12\cdot p+6\cdot (1-p)=6+6\cdot p$

The best payoff comes when the above expectations are equal. Let's take the first two: $$18-14\cdot p=9-3\cdot p \Rightarrow p=\frac{9}{11}$$ but if we take the last two: $$9-3\cdot p=6+6\cdot p \Rightarrow p=\frac{1}{3}$$ and if we the the 1st and 3rd: $$18-14\cdot p=6+6\cdot p \Rightarrow p=\frac{3}{5}$$

So what does this mean for the problem? Surely, it doesn't have a sadle point, but what's the interpretation of this?

I have come across this problem while trying to understand game theory. The problem presents two players (hunters) that want to hunt in one of two regions available. Region $A$ has $12$ rabbits and region $B$ has $18$ rabbits. If the two hunters go to different regions they will acquire all the rabbits in their respective hunting ground, but if they go to the same region they'll acquire half the rabbits each, assuming they have the same skills. I have found the mixed strategy that they need to follow, which will be the same for both: they will go to region $A$ with probability $p=\frac{3}{5}$ or to region $B$ with probability $1-p=\frac{2}{5}$.

The problem emerges when I try to further the problem by adding a third hunter in the game. I would assume that an obvious answer would come up, like one region would give the best result when chosen for all hunters. However, what I'm seeing by creating the payoff tables is that if the third hunter chooses region $A$ then it's in everyone's benefit if the other two hunters choose region $B$, thus having the result array be $(H_1,H_2,H_3)=(9,9,12)$. But if hunter 3 chooses region $B$, the best outcome is when the other two hunters don't choose the same region between themselves.

I am unsure as to how I can interpret this or if there's another way to find the answer to this more complicated question. Can I apply a method to this problem and find mixed strategies for the three hunters? I am not necessarily looking for the answer directly, but for someone to point out a method or way of thinking for me to find the answer alone.

I have come across this problem while trying to understand game theory. The problem presents two players (hunters) that want to hunt in one of two regions available. Region $A$ has $12$ rabbits and region $B$ has $18$ rabbits. If the two hunters go to different regions they will acquire all the rabbits in their respective hunting ground, but if they go to the same region they'll acquire half the rabbits each, assuming they have the same skills. I have found the mixed strategy that they need to follow, which will be the same for both: they will go to region $A$ with probability $p=\frac{3}{5}$ or to region $B$ with probability $1-p=\frac{2}{5}$.

The problem emerges when I try to further the problem by adding a third hunter in the game. I would assume that an obvious answer would come up, like one region would give the best result when chosen for all hunters. However, what I'm seeing by creating the payoff tables is that if the third hunter chooses region $A$ then it's in everyone's benefit if the other two hunters choose region $B$, thus having the result array be $(H_1,H_2,H_3)=(9,9,12)$. But if hunter 3 chooses region $B$, the best outcome is when the other two hunters don't choose the same region between themselves.

I am unsure as to how I can interpret this or if there's another way to find the answer to this more complicated question. Can I apply a method to this problem and find mixed strategies for the three hunters? I am not necessarily looking for the answer directly, but for someone to point out a method or way of thinking for me to find the answer alone.

Edit: Trying again after the suggestions, I did this and hopefully someone can verify if my solution is correct:

Suppose that $p$ is the probability with which hunter 1 goes to region $A$ and $(1-p)$ if he goes to region $B$. That same probability applies to the other hunters. There are two cases: the other two go to the same region or to different regions. Therefore, we have the two expected payoffs:

• $V(H_1 | H_2=H_3=A)=4\cdot p+18\cdot (1-p)=18-14\cdot p$
• $V(H_1 | H_2=A, H_3=B)=6\cdot 9+9\cdot (1-p)=9-3\cdotp$

The best payoff comes when the above expectations are equal, so we get: $$18-14\cdot p=9-3\cdot p\Rightarrow p=\frac{9}{11}$$ Therefore, the strategy that all three hunters must follow is $\bigg\{\bigg(A,\frac{9}{11}\bigg), \bigg(B,\frac{2}{11}\bigg) \bigg\}$. If that is correct, I find it interesting that it would be to all their benefits to go to region A, since if they all go they'll just get $4$ rabbits each. They could at least all go to region $B$ and get $9$ each. If we're looking for a strategy that gives none of the other hunters motive to change region, how does this not give motive to any hunter to go to region $B$?

I have come across this problem while trying to understand game theory. The problem presents two players (hunters) that want to hunt in one of two regions available. Region $A$ has $12$ rabbits and region $B$ has $18$ rabbits. If the two hunters go to different regions they will acquire all the rabbits in their respective hunting ground, but if they go to the same region they'll acquire half the rabbits each, assuming they have the same skills. I have found the mixed strategy that they need to follow, which will be the same for both: they will go to region $A$ with probability $p=\frac{3}{5}$ or to region $B$ with probability $1-p=\frac{2}{5}$.

The problem emerges when I try to further the problem by adding a third hunter in the game. I would assume that an obvious answer would come up, like one region would give the best result when chosen for all hunters. However, what I'm seeing by creating the payoff tables is that if the third hunter chooses region $A$ then it's in everyone's benefit if the other two hunters choose region $B$, thus having the result array be $(H_1,H_2,H_3)=(9,9,12)$. But if hunter 3 chooses region $B$, the best outcome is when the other two hunters don't choose the same region between themselves.

I am unsure as to how I can interpret this or if there's another way to find the answer to this more complicated question. Can I apply a method to this problem and find mixed strategies for the three hunters? I am not necessarily looking for the answer directly, but for someone to point out a method or way of thinking for me to find the answer alone.

Edit: Trying again after the suggestions, I did this and hopefully someone can verify if my solution is correct:

Suppose that $p$ is the probability with which hunter 1 goes to region $A$ and $(1-p)$ if he goes to region $B$. That same probability applies to the other hunters. There are two cases: the other two go to the same region or to different regions. Therefore, we have the two expected payoffs:

• $V(H_1 | H_2=H_3=A)=4\cdot p+18\cdot (1-p)=18-14\cdot p$
• $V(H_1 | H_2=A, H_3=B)=6\cdot 9+9\cdot (1-p)=9-3\cdotp$
• $V(H_1 | H_2=H_3=B)= 12\cdot p+6\cdot (1-p)=6+6\cdot p$

The best payoff comes when the above expectations are equal. Let's take the first two: $$18-14\cdot p=9-3\cdot p \Rightarrow p=\frac{9}{11}$$ but if we take the last two: $$9-3\cdot p=6+6\cdot p \Rightarrow p=\frac{1}{3}$$ and if we the the 1st and 3rd: $$18-14\cdot p=6+6\cdot p \Rightarrow p=\frac{3}{5}$$

So what does this mean for the problem? Surely, it doesn't have a sadle point, but what's the interpretation of this?

I have come across this problem while trying to understand game theory. The problem presents two players (hunters) that want to hunt in one of two regions available. Region $A$ has $12$ rabbits and region $B$ has $18$ rabbits. If the two hunters go to different regions they will acquire all the rabbits in their respective hunting ground, but if they go to the same region they'll acquire half the rabbits each, assuming they have the same skills. I have found the mixed strategy that they need to follow, which will be the same for both: they will go to region $A$ with probability $p=\frac{3}{5}$ or to region $B$ with probability $1-p=\frac{2}{5}$.

The problem emerges when I try to further the problem by adding a third hunter in the game. I would assume that an obvious answer would come up, like one region would give the best result when chosen for all hunters. However, what I'm seeing by creating the payoff tables is that if the third hunter chooses region $A$ then it's in everyone's benefit if the other two hunters choose region $B$, thus having the result array be $(H_1,H_2,H_3)=(9,9,12)$. But if hunter 3 chooses region $B$, the best outcome is when the other two hunters don't choose the same region between themselves.

I am unsure as to how I can interpret this or if there's another way to find the answer to this more complicated question. Can I apply a method to this problem and find mixed strategies for the three hunters? I am not necessarily looking for the answer directly, but for someone to point out a method or way of thinking for me to find the answer alone.

I have come across this problem while trying to understand game theory. The problem presents two players (hunters) that want to hunt in one of two regions available. Region $A$ has $12$ rabbits and region $B$ has $18$ rabbits. If the two hunters go to different regions they will acquire all the rabbits in their respective hunting ground, but if they go to the same region they'll acquire half the rabbits each, assuming they have the same skills. I have found the mixed strategy that they need to follow, which will be the same for both: they will go to region $A$ with probability $p=\frac{3}{5}$ or to region $B$ with probability $1-p=\frac{2}{5}$.

The problem emerges when I try to further the problem by adding a third hunter in the game. I would assume that an obvious answer would come up, like one region would give the best result when chosen for all hunters. However, what I'm seeing by creating the payoff tables is that if the third hunter chooses region $A$ then it's in everyone's benefit if the other two hunters choose region $B$, thus having the result array be $(H_1,H_2,H_3)=(9,9,12)$. But if hunter 3 chooses region $B$, the best outcome is when the other two hunters don't choose the same region between themselves.

I am unsure as to how I can interpret this or if there's another way to find the answer to this more complicated question. Can I apply a method to this problem and find mixed strategies for the three hunters? I am not necessarily looking for the answer directly, but for someone to point out a method or way of thinking for me to find the answer alone.

Edit: Trying again after the suggestions, I did this and hopefully someone can verify if my solution is correct:

Suppose that $p$ is the probability with which hunter 1 goes to region $A$ and $(1-p)$ if he goes to region $B$. That same probability applies to the other hunters. There are two cases: the other two go to the same region or to different regions. Therefore, we have the two expected payoffs:

• $V(H_1 | H_2=H_3=A)=4\cdot p+18\cdot (1-p)=18-14\cdot p$
• $V(H_1 | H_2=A, H_3=B)=6\cdot 9+9\cdot (1-p)=9-3\cdotp$

The best payoff comes when the above expectations are equal, so we get: $$18-14\cdot p=9-3\cdot p\Rightarrow p=\frac{9}{11}$$ Therefore, the strategy that all three hunters must follow is $\bigg\{\bigg(A,\frac{9}{11}\bigg), \bigg(B,\frac{2}{11}\bigg) \bigg\}$. If that is correct, I find it interesting that it would be to all their benefits to go to region A, since if they all go they'll just get $4$ rabbits each. They could at least all go to region $B$ and get $9$ each. If we're looking for a strategy that gives none of the other hunters motive to change region, how does this not give motive to any hunter to go to region $B$?