Consider there are two players in a simultaneous game. Each of them chooses and pays integer multiples of one dollar to play the game. The player with the highest bid wins 100 dollar and the loser gets nothing, though both players lose the amount they bid (e.g., if player 1 bids $b_h$ and player 2 bids $b_l$ with $b_h> b_l$, then player 1 gets $(100–b_h)$ and player 2 gets $(–b_l)$ atthe end). Construct a mixed strategy equilibrium (and prove that it is an equilibrium) in which every integer bid bϵ[0,99] has a positive probability.

My answer:

There is no pure strategy Nash equilibrium.

And when I look at the mixed strategy Nash equilibrium, I guess that $P_s = 1/100$ is Nash equilibrium, where $P_s$ is the probability of action of bidding $s= \{0,1,...,99\}$. Because the mixed strategy in this case implies that players are indifferent in choosing the action of bidding.

But, I cannot show my answer in a formal way properly. How can I show its proof?

  • $\begingroup$ What if they both bid the same? $\endgroup$
    – YJT
    Apr 10, 2021 at 14:37
  • $\begingroup$ The question gives no information about that @YJT $\endgroup$
    – 1190
    Apr 10, 2021 at 15:20

1 Answer 1


Suppose Player 1 chooses uniformly a bid in the set $\{0,\ldots,99\}$ and denote it by $X$. When Player 2 bids $i$, his payoff is $100\Pr(X< i) -i +50\Pr(X=i)$, where I assumed that when both bid the same, they split the $100$. Since $\Pr(X<i)=\tfrac{i}{100}$, the payoff is $50\Pr(X=i)=0.5$ it is independent of $i$. Thus, Player 2 is indifferent among all the bids. This is true when player 2 uses this strategy too, so when both chose a number uniformly, neither has a profitable deviation and it's an equilibrium.

Note that the $50$ doesn't really matter. Even if they don't split the $100$ when bid the same but each gets $100$, the indifference holds. However, there might be other pure equilibria in this case (each strategy profile where they bid the same is a pure equilibrium in this case).


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