<p>$(d)$ cannot be an answer because of the following.  If 1 and 2 are located as in $(d)$, then 3 wants to be located (i) either at $\frac{1}{3}-\varepsilon$ or (ii) at $\frac{2}{3}+\varepsilon$ because doing so yields almost $\frac{1}{3}$ while choosing $\frac{1}{2}$ yields $\frac{1}{2}\times\frac{1}{3}$.  So, 1 receives $\frac{1}{2}\times\frac{1}{3}$ if 3 chooses (i), or 2 receives $\frac{1}{2}\times\frac{1}{3}$ if 3 chooses (ii).  Then, 1 and 2 does not want to act according to $(d)$.</p>

<p>$(b)$ is the right answer. The reasoning is: if 1 and 2 are located as in the answer, 3 is indifferent of choosing anything in $\left[\frac{1}{4}, \frac{3}{4}\right]$ all of which yield $\frac{1}{4}$ and any other location yields a lower payoff.  In other words, each of 1 and 2 can guarantee himself at least $\frac{1}{4}$.   </p>

<p>Now, the question is: Does any of 1 and 2 want to deviate from $(b$)?  Suppose when both 1 and 2 played according to $(b)$, 3 chooses $\frac{1}{2}$ so that each of 1 and 2 obtains $\frac{3}{8}$.  Now, suppose that 1 deviates to a location $\frac{1}{4}-x$, where $x \in (0,1/4)$.  Then, he obtains $\frac{1}{4}-x + \frac{\frac{1}{2}-\left(\frac{1}{4}-x\right)}{2} =\frac{3}{8}-\frac{x}{2}$ which is clearly lower than $\frac{3}{8}$ as $x&gt;0$.  Thus, 1 does not want to deviate.  By a similar argument, 3 does not want to deviate too.  Hence, $(b)$ is an equilibrium</p>

CC BY-SA 4.0
2018-06-08T12:56:58.840Z
thanks a lot for such a clear and detailed explanation! I always find myself im doubts when I’m solving location games in game theory. Is backward induction not a good approach when solving these questions?
divya garg
CC BY-SA 4.0
2018-06-07T18:42:36.867Z
2018-06-07T18:42:36.867Z
Green.H
<p><a href="https://i.sstatic.net/OO3Mb.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/OO3Mb.jpg" alt="enter image description here"></a></p>

<p>I could solve Question 25 (simultaneous location game between 2 firms), but I'm confused between options b) and d) for Question 26 (sequential game involving a third firm). I have attached both questions since Question 25 provides context to Question 26. 
Basically, I got my answer as d) for Question 26, using the method of backward induction since it's a sequential game. While the answer key for this question says b) is the correct option. 
My method: Assuming that firm 3 locates at 1/2, firm 1 and 2's location described in option d) gives them both the maximum share than in any other option. Equilibrium should be such that firms 1 &amp; 2 are at an advantage since they are first movers in the game. </p>

<p>Could anyone please confirm as to which should be the correct option for Question 26? Any solution and views are welcome.</p>

<game-theory><nash-equilibrium>
Location game between 3 firms
divya garg
# Location game between 3 firms

(Asked by: username_0 on 2018-06-07)

[![enter image description here](https://i.sstatic.net/OO3Mb.jpg)](https://i.sstatic.net/OO3Mb.jpg)

I could solve Question 25 (simultaneous location game between 2 firms), but I'm confused between options b) and d) for Question 26 (sequential game involving a third firm). I have attached both questions since Question 25 provides context to Question 26.
Basically, I got my answer as d) for Question 26, using the method of backward induction since it's a sequential game. While the answer key for this question says b) is the correct option.
My method: Assuming that firm 3 locates at 1/2, firm 1 and 2's location described in option d) gives them both the maximum share than in any other option. Equilibrium should be such that firms 1 & 2 are at an advantage since they are first movers in the game.

Could anyone please confirm as to which should be the correct option for Question 26? Any solution and views are welcome.

**Tags:** game-theory, nash-equilibrium

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## 1 answers

### Answer by username_1 on 2018-06-07. Score: 0 | Accepted answer

$(d)$ cannot be an answer because of the following. If 1 and 2 are located as in $(d)$, then 3 wants to be located (i) either at $\frac{1}{3}-\varepsilon$ or (ii) at $\frac{2}{3}+\varepsilon$ because doing so yields almost $\frac{1}{3}$ while choosing $\frac{1}{2}$ yields $\frac{1}{2}\times\frac{1}{3}$. So, 1 receives $\frac{1}{2}\times\frac{1}{3}$ if 3 chooses (i), or 2 receives $\frac{1}{2}\times\frac{1}{3}$ if 3 chooses (ii). Then, 1 and 2 does not want to act according to $(d)$.

$(b)$ is the right answer. The reasoning is: if 1 and 2 are located as in the answer, 3 is indifferent of choosing anything in $\left[\frac{1}{4}, \frac{3}{4}\right]$ all of which yield $\frac{1}{4}$ and any other location yields a lower payoff. In other words, each of 1 and 2 can guarantee himself at least $\frac{1}{4}$.

Now, the question is: Does any of 1 and 2 want to deviate from $(b$)? Suppose when both 1 and 2 played according to $(b)$, 3 chooses $\frac{1}{2}$ so that each of 1 and 2 obtains $\frac{3}{8}$. Now, suppose that 1 deviates to a location $\frac{1}{4}-x$, where $x \in (0,1/4)$. Then, he obtains $\frac{1}{4}-x + \frac{\frac{1}{2}-\left(\frac{1}{4}-x\right)}{2} =\frac{3}{8}-\frac{x}{2}$ which is clearly lower than $\frac{3}{8}$ as $x>0$. Thus, 1 does not want to deviate. By a similar argument, 3 does not want to deviate too. Hence, $(b)$ is an equilibrium

> **Comments:**
> **username_0**: thanks a lot for such a clear and detailed explanation! I always find myself im doubts when I’m solving location games in game theory. Is backward induction not a good approach when solving these questions?  

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