Exercise 4.5 in Vijay Krishna's Auction Theory:

Suppose that bidder 1 always has value $X_1=2$, while bidder 2's value, $X_2$ is equally likely to be $0$ or $2$. Find equilibrium bidding strategies in a first-price auction. (Note that since values are discrete, the equilibrium will be in mixed strategies.)

  • I don't understand why it is that since values are discrete, the equilibrium will be in mixed strategies. Is there some rule/principle to this?

  • How do I set up for the problem? Since equilibrium are in mixed strategies, I can suppose that player 1 bids $b_1,\ldots,b_k$ with probabilities $p_1,\ldots,p_k$ respectively. But that seems too general to make any progress.


2 Answers 2


The suggested pure strategies do not create an equilibrium, because if $X_2=X_1=2$, there is a tie. Whichever tie-breaking rule is employed, one of the players would want to deviate to $\beta_i=1+\epsilon$ to avoid the tie and win with certainty. If tie breaking is always in favor of the second player, player 1 deviates to $0+\epsilon$.

Discrete values in this setting do suggest a mixed strategy equilibrium (non-degenerate), because the probability of a tie in valuations is non-zero. This is why the pure strategy equilibrium from the continuous case is missing here.

The more general case of this problem is in Doni & Menicucci (2013).


Any pure strategy Nash equilibrium is implicitly a mixed-strategies Nash equilibrium. Since the valuations vary, it's a good indicator we want to consider mixed-strategies. The fact that the problem tells us this is a stronger indicator, though I'm sure not the axiomatic justification you are seeking. :-)

Consider player $1$. We have player $1$'s expected profit: $\mathbb{E}[\Pi_{1}(b)] = (2-b) Pr[b \geq \beta_{2}(v_{2})]$, where $b$ is player $1$'s bid, $\beta_{2}$ is player $2$'s bidding strategy, and $v_{2}$ is player $2$'s valuation. We can assume $\beta_{2}(0) = 0$ (because if $\beta_{2}(0) > 0$, player $2$ can improve upon this by decreasing his bid). Since we are only considering two potential valuations for player $2$, we can assume $\beta_{2}(v) = av$, for some constant $a \in \mathbb{R}_{++}$. (That is, given the two points $(0, 0)$ and $(2, \beta_{2}(2))$, we just draw a line between them).

Observe that $Pr[b \geq av] = Pr[v \leq \frac{b}{a}] = \frac{b}{2a}$, with the last inequality since we have a 50-50 chance on the valuation of player $2$.

Now for a Nash equilibrium, player $1$ seeks to maximize his expected value. This is given by the following optimization problem:

$$\max_{b} (2-b) \cdot (\frac{b}{2a})$$

This yields the first order conditions:

$\frac{1}{2a} \cdot (2 - 2b) = 0$, and we obtain that $b = 1$ is our only solution for player $1$. This answer should be reasonably intuitive.

Now player $2$ only wins if his valuation is $2$. So he players $\beta_{2}(2) = 1$ and $\beta_{2}(0) = 0$.

  • 1
    $\begingroup$ This part $$ Pr[v \leq \frac{b}{a}] = \frac{b}{2a} $$ seems incorrect. For any given $b$ the probability could take three values: 0, $\frac{1}{2}$, 1. It is not continuous in $a$. $\endgroup$
    – Giskard
    Apr 26, 2016 at 21:00

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