First-Price Auction¶

Definition¶

In a first-price sealed-bid auction, each of $n$ bidders submits a bid $b_i\ge0$ simultaneously and without seeing the others' bids. The highest bid wins (ties are broken at random) and the winner pays her own bid; losers pay nothing. In a private-value auction with valuation $\theta_i$ , bidder $i$ 's payoff is $\theta_i-b_i$ if she wins and $0$ otherwise. A strategy is a bidding function $b_i(\theta_i)$ .

Symmetric equilibrium (uniform case). If $\theta_1,\dots,\theta_n$ are i.i.d. with the uniform distribution on $[0,1]$ , the symmetric Bayesian Nash equilibrium with an increasing bidding function is

b(\theta)=\frac{n-1}{n}\,\theta.

Interpretation¶

Raising the bid has two opposing effects: it raises the probability of winning and lowers the surplus $\theta_i-b_i$ obtained when winning. Bidding one's value guarantees a payoff of $0$ , so in equilibrium every bidder shades her bid below her value. The amount of shading, $\theta/n$ in the uniform case, falls as competition increases.

Expected payoff of a bidder with value 0.9 facing two rivals, plotted against her bid: it peaks at 0.243 at the bid 0.6 and falls to 0 when she bids her value.

The figure uses the $n=3$ numbers of the Examples: a bidder with $\theta=0.9$ whose two rivals bid $\tfrac{2}{3}\theta_j$ earns the expected payoff $\mathbb{E}[u]=(0.9-b)(1.5b)^2$ for $b\le\tfrac{2}{3}$ $\frac{2}{3}$ , and $0.9-b$ for higher bids, which win for sure. Low bids rarely win and high bids leave little surplus, so the curve peaks at the equilibrium bid $b(0.9)=0.6$ with payoff $0.243$ . The deviations to $0.5$ and $0.65$ (orange) give $0.225$ and $0.238$ , and bidding the value gives $0$ .

Move the bid $b$ : the expected payoff peaks at $b(\theta)=\tfrac{n-1}{n}\theta$ $\frac{n - 1}{n} θ$ , here $0.6$ with payoff $0.243$ . Add bidders and the optimal shading $\theta/n$ shrinks.

Unlike the second-price auction, the first-price auction has no dominant strategy: the best bid depends on what the rivals bid, so a bidder needs beliefs about the rivals' values and strategies. This is why the equilibrium is computed as a BNE rather than by a dominance argument.

A useful reading of the equilibrium bid: a bidder bids the expected highest rival value, conditional on that value being below her own. In the uniform case the highest of $n-1$ rival values, given that all are below $\theta$ , has expectation $\frac{n-1}{n}\theta$ , which is exactly $b(\theta)$ .

Key Results¶

General formula for independent private values (IPV). For i.i.d. values with a continuous cumulative distribution function (CDF) $F$ on $[0,\overline\theta]$ , the symmetric increasing BNE is $b(\theta)=\theta-\dfrac{\int_0^\theta F(x)^{n-1}\,\mathrm{d}x}{F(\theta)^{n-1}}$ for $F(\theta)>0$ $0$ ; for $\operatorname{Unif}(0,1)$ values this gives $\frac{n-1}{n}\theta$ .
Win probability. Under a symmetric increasing strategy, a bidder of type $\theta$ wins if and only if she has the highest value, with probability $F(\theta)^{n-1}$ ( $=$ $=\theta^{n-1}$ for $\operatorname{Unif}(0,1)$ ).
Expected revenue with $\operatorname{Unif}(0,1)$ values: the winner is the highest order statistic $\theta_{(n)}$ with $\mathbb{E}[\theta_{(n)}]=\frac{n}{n+1}$ $\frac{n}{n + 1}$ , so revenue is $\frac{n-1}{n}\cdot\frac{n}{n+1}=\frac{n-1}{n+1}$ $\frac{n - 1}{n + 1}$ , the same as in the second-price auction.
Interim expected surplus of type $\theta$ (uniform case): $\left(\theta-\tfrac{n-1}{n}\theta\right)\theta^{n-1}=\theta^n/n$ $θ^{n} / n$ .
Any bid above one's value is weakly dominated by bidding the value, but bidding the value itself is not a best response to the equilibrium bids of the others when $\theta>0$ $0$ .
The Dutch auction is strategically equivalent to the first-price auction.

Examples¶

Deriving $b(\theta)$ for $\operatorname{Unif}(0,1)$ values. Suppose the rivals use $b(\theta)=\beta\theta$ with $\beta>0$ $0$ . A bidder with value $\theta$ who bids $b\le\beta$ beats a rival $j$ when $\beta\theta_j<b$ $b$ , which has probability $b/\beta$ , so her expected payoff $\mathbb{E}[u]$ is

\mathbb{E}[u]=(\theta-b)\left(\frac{b}{\beta}\right)^{n-1}.

The first-order condition $-b^{n-1}+(n-1)(\theta-b)b^{n-2}=0$ gives $b=\frac{n-1}{n}\theta$ $\frac{n - 1}{n} θ$ , independent of $\beta$ . Consistency requires $\beta=\frac{n-1}{n}$ $\frac{n - 1}{n}$ .

Numbers with $n=3$ $3$ . A bidder with $\theta=0.9$ bids $0.6$ , wins with probability $0.9^2=0.81$ and expects $0.3\times0.81=0.243$ $0.243$ . Deviating to $0.5$ wins only if both rivals have $\theta_j<0.75$ $0.75$ : payoff $0.4\times0.5625=0.225$ $0.225$ . Deviating to $0.65$ wins if $\theta_j<0.975$ $0.975$ : payoff $0.25\times0.950625\approx0.238$ $0.238$ . Both deviations do worse. Expected revenue is $\frac{2}{4}=0.5$ $0.5$ .

Two bidders. With $n=2$ the bid is $\theta/2$ and expected revenue is $\frac{1}{2}\mathbb{E}[\max(\theta_1,\theta_2)]=\frac{1}{2}\cdot\frac{2}{3}=\frac{1}{3}$ $\frac{1}{3}$ .

Prerequisites¶

Sources¶

Game Theory · Week 6

Aliases: first-price sealed-bid auction, First-Price Auction Bidding, FPA, FPSB auction Keywords: bid shading, b(v)=(n-1)v/n, symmetric bidding function, expected revenue (n-1)/(n+1), pay-your-bid auction Symbols: beta, integral, theta

Last updated: 17 October 2026 21:53