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Consider the following strategic game with complete information played by three players. Each player $i ∈ {1, 2, 3}$ chooses her action from $A = \{1, 2, . . . , 10\}$. Utility functions, mapping each action profile $(a_1, a_2, a_3) ∈ A^3$ into utils, of the three players are as follows: $$u_1(a_1, a_2, a_3)=-|a_3-a_1|+|a_2-a_1|$$ $$u_2(a_1, a_2, a_3)=-|a_1-a_2|+|a_3-a_2|$$ $$u_3(a_1, a_2, a_3)=-|a_2-a_3|+|a_1-a_3|$$

The given solution is as follows:

Suppose $a_1 < a_3$. Straightforward argument shows that the set of actions that constitute pure best response for player 2 is $\{1, . . . , a_1\}$. When $a_1 > a_3$, the set is $\{a_1, . . . , 10\}$ and when $a_1 = a_3$, then the set is $\{1, . . . , 10\}$.

This solution is too short for me to understand how to start solving it. I understand the conditions when one is >,< or = but I do not seem to follow how the BR is calculated

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    $\begingroup$ So you don't understand "Suppose $a_1 < a_3$. Straightforward argument shows that the set of actions that constitute pure best response for player 2 is $\{1, . . . , a_1\}$."? Have you tried substituting numbers to see if that makes it easier for you to understand? E.g., do you see why if $a_1 = 5, a_3 = 8$, and Player $2$'s payoff function is $$u_2(5, a_2, 8)=-|5-a_2|+|8-a_2|$$ they would never play numbers larger than 5? $\endgroup$
    – Giskard
    Commented Jan 24, 2022 at 15:52
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    $\begingroup$ "When $a_1 > a_3$, the set is $\{a_1, . . . , 10\}$", so yes. $\endgroup$
    – Giskard
    Commented Jan 24, 2022 at 16:27
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    $\begingroup$ I’m voting to close this question because it is a specific self-study problem and was resolved in the comments. $\endgroup$
    – Giskard
    Commented Jan 24, 2022 at 18:26
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    $\begingroup$ If the OP has solved the question, they could answer it themselves so we have a completed Q and A instead of closing off the question? $\endgroup$
    – Walrasian Auctioneer
    Commented Jan 25, 2022 at 16:26
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    $\begingroup$ @Giskard I am slightly lost for words. $\endgroup$
    – Walrasian Auctioneer
    Commented Jan 26, 2022 at 14:49

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The given solution is as follows:

Suppose $a_1 < a_3$. Straightforward argument shows that the set of actions that constitute pure best response for player 2 is $\{1, . . . , a_1\}$. When $a_1 > a_3$, the set is $\{a_1, . . . , 10\}$ and when $a_1 = a_3$, then the set is $\{1, . . . , 10\}$.

Suppose we set $a_1=5$ and $a_3=3$ as in the case when $a_1>a_3$, as described in the comment as well $$u_2(5, a_2, 3)=-|5-a_2|+|3-a_2|$$ then player 2 would be better playing from $\{a_1,....10\}$ and vice versa.

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