Here's the problem that relates to a whole class of problems to which I am trying to figure out a general solution.

Given two players 1 and 2 who can select a number from the interval $[0, 1]$, define payoffs from a game as follows:

$\frac{s_i+s_j}{2}$ for $i<j$, $1-\frac{s_i+s_j}{2}$ for $i>j$ and $\frac{1}{2}$ to each player when $s_i=s_j$.

So the goal is to find such a combination of $s_i$ and $s_j$ that they are both best responses to each other, that is, there is no profitable deviation for either player. I have considered several cases, for example if $s_i<s_j$, there is always a profitable deviation for player $i$ where he/she can pick a higher number as long as it is arbitrarily smaller than $s_j$.

Symmetrically, the same argument applies when $s_i>s_j$. Then player $j$ can select a number arbitrarily smaller than $s_i$, thus increasing his/her payoff. That leaves us only the possibility of $s_i=s_j$ to investigate, and even then I managed to find only one set of responses ($\frac{1}{2}$ for each) that results into a Nash equilibrium.

Otherwise, say both players choose $0.4$. Then any player who chooses $0.5$ in response will get a profit of $1-\frac{.9}{2}$, exceeding the profits of $\frac{1}{2}$, thus there is a profitable deviation.

The question I have is how can I mathematically prove/show that ($\frac{1}{2}, \frac{1}{2}$) is the only Nash equilibrium (or are there more?). I managed to find those numbers by inspection, and would really like a more rigorous way of doing that since it would apply to all similar problems.

  • $\begingroup$ Do you mean "player i wins $(s_i + s_j)/2$ for $s_i < s_j$, and $1 - (s_i + s_j)/2$ for $s_i > s_j$" ? $\endgroup$
    – mercio
    Apr 18 '11 at 20:33
  • 1
    $\begingroup$ Line breaks go a long way :) $\endgroup$
    – Justin
    Apr 18 '11 at 21:08

First, I want to clarify your payoff functions. The way you defined does not make sense to me. $i<j$ and $i>j$ should be $s_i<s_j$ and $s_i>s_j$ ? I guess you mean whoever chooses a strictly smaller number gets $\frac{s_1+s_2}2$ and the other one gets $1-\frac{s_1+s_2}2$; and if they choose the same number, both get $1/2$.

Second, I think you almost have a rigorous proof. Let me try to summerize. Suppose $(s_1,s_2)$ is a Nash equilibrium, then using proof by contradiction, we can show that $s_1=s_2$. Suppose not, either $s_1<s_2$ or $s_2<s_1$. In the first case, Player 1 benifits from deviating from this strategy by choosing any $s_1'\in (s_1,s_2)$. The same for the second case.

Now the only type of Nash equlibrium must be $(s,s)$. Consider different cases: If $s>1/2$, either player can get a higher payoff by choosing $s'\in (1-s,s)$. If $s<1/2$, either player can get a higher payoff by choosing $s'\in (s, 1-s)$. If $s=1/2$, neither player benifits from deviating because increasing $s$ to $s'>s$ results in a payoff of $1-\frac{s+s'}2<1/2$ and decreasing $s$ to $s'<s$ results in $\frac{s+s'}2<1/2$.

  • $\begingroup$ Sorry, that was my typo, and thanks for the answer, it looks rigorous! $\endgroup$ Apr 18 '11 at 20:35

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