<p>The given solution is as follows:</p>
<p>Suppose <span class="math-container">$a_1 &lt; a_3$</span>. Straightforward argument shows that the
set of actions that constitute pure best response for player 2 is <span class="math-container">$\{1, . . . , a_1\}$</span>.
When <span class="math-container">$a_1 &gt; a_3$</span>, the set is <span class="math-container">$\{a_1, . . . , 10\}$</span> and when <span class="math-container">$a_1 = a_3$</span>, then the set is
<span class="math-container">$\{1, . . . , 10\}$</span>.</p>
<p>Suppose we set <span class="math-container">$a_1=5$</span> and <span class="math-container">$a_3=3$</span> as in the case when <span class="math-container">$a_1&gt;a_3$</span>, as described in the comment as well
<span class="math-container">$$u_2(5, a_2, 3)=-|5-a_2|+|3-a_2|$$</span> then player 2 would be better playing from <span class="math-container">$\{a_1,....10\}$</span> and vice versa.</p>

CC BY-SA 4.0
2022-01-26T09:46:38.093Z
2022-01-26T09:46:38.093Z
Maybeline Lee
<p>Consider the following strategic game with complete information played
by three players. Each player <span class="math-container">$i ∈ {1, 2, 3}$</span> chooses her action from <span class="math-container">$A =
\{1, 2, . . . , 10\}$</span>. Utility functions, mapping each action profile <span class="math-container">$(a_1, a_2, a_3) ∈ A^3$</span>
into utils, of the three players are as follows:
<span class="math-container">$$u_1(a_1, a_2, a_3)=-|a_3-a_1|+|a_2-a_1|$$</span>
<span class="math-container">$$u_2(a_1, a_2, a_3)=-|a_1-a_2|+|a_3-a_2|$$</span>
<span class="math-container">$$u_3(a_1, a_2, a_3)=-|a_2-a_3|+|a_1-a_3|$$</span></p>
<p>The given solution is as follows:</p>
<p>Suppose <span class="math-container">$a_1 &lt; a_3$</span>. Straightforward argument shows that the
set of actions that constitute pure best response for player 2 is <span class="math-container">$\{1, . . . , a_1\}$</span>.
When <span class="math-container">$a_1 &gt; a_3$</span>, the set is <span class="math-container">$\{a_1, . . . , 10\}$</span> and when <span class="math-container">$a_1 = a_3$</span>, then the set is
<span class="math-container">$\{1, . . . , 10\}$</span>.</p>
<p>This solution is too short for me to understand how to start solving it. I understand the conditions when one is &gt;,&lt; or = but I do not seem to follow how the BR is calculated</p>

CC BY-SA 4.0
2022-01-24T15:52:00.977Z
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?
Giskard
CC BY-SA 4.0
2022-01-24T16:11:18.360Z
Ah I see. For the second case when $a_1>a_3$ then 2 would play 8 and up? if we assume $a_1=8$ and $a_3=5$
Maybeline Lee
CC BY-SA 4.0
2022-01-24T16:27:58.617Z
"*When $a_1 > a_3$, the set is $\{a_1, . . . , 10\}$*", so yes.
Giskard
CC BY-SA 4.0
2022-01-24T18:26:38.297Z
I’m voting to close this question because it is a specific self-study problem and was resolved in the comments.
Giskard
CC BY-SA 4.0
2022-01-25T16:26:45.253Z
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?
Walrasian Auctioneer
CC BY-SA 4.0
2022-01-26T09:54:42.807Z
@WalrasianAuctioneer See [answer](https://economics.stackexchange.com/a/50224/1601).
Giskard
CC BY-SA 4.0
2022-01-26T14:49:33.763Z
@Giskard I am slightly lost for words.
Walrasian Auctioneer
<game-theory><nash-equilibrium><dominant-strategy>
Strategic game with complete informaation
Maybeline Lee
# Strategic game with complete informaation

(Asked by: username_0 on 2022-01-24)

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

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

> **Comments:**
> **username_1**: 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?  
> **username_0**: Ah I see. For the second case when $a_1>a_3$ then 2 would play 8 and up? if we assume $a_1=8$ and $a_3=5$  
> **username_1**: "*When $a_1 > a_3$, the set is $\{a_1, . . . , 10\}$*", so yes.  
> **username_1**: I’m voting to close this question because it is a specific self-study problem and was resolved in the comments.  
> **username_2**: 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?  
> **username_1**: @WalrasianAuctioneer See [answer](https://economics.stackexchange.com/a/50224/1601).  
> **username_2**: @username_1 I am slightly lost for words.  

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## 1 answers

### Answer by username_0 on 2022-01-26. Score: 1

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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