Optimal Bidding with Resale in a First-Price Auction

Game Theory · Hard · Free problem

You are bidding in a first-price sealed-bid auction against $n - 1$ other bidders. Each bidder (including you) independently draws a private valuation $V$ from a $\text{Uniform}(0, M)$ distribution. However, the winner can resell the item on a secondary market for twice their private valuation, so your effective value for the item is

V$.

In a first-price auction, the highest bidder wins and pays their own bid.

  1. Derive the symmetric Bayesian Nash equilibrium bid function $b(v)$ as a function of your valuation $v$, the number of bidders $n$, and the upper bound $M$.
  1. Compute the explicit bid function for $n = 2$ and $n = 10$.
  1. Explain qualitatively how the resale option and the number of bidders each affect optimal bidding behavior.

Hints

  1. The resale option changes your effective value for the item. What is the item really worth to you if you can sell it for twice your private valuation?
  2. Once you account for the resale value, this reduces to a standard first-price auction. Recall the equilibrium bid in a first-price auction with $n$ i.i.d. $U(0, M)$ bidders: $b(v) = \frac{n-1}{n} v$.
  3. Substitute $w = 2v$ as the effective value into the standard formula. The boundary condition is $\beta(0) = 0$ -- the bidder with the lowest possible valuation bids zero.

Worked Solution

How to Think About It: The resale option doubles your effective value for the item -- if you value it at $v$, you can resell for v$, so winning is worth v$ to you. This means the auction is really a standard first-price auction, just with effective values V$ instead of $V$. The classic result for first-price auctions with i.i.d. uniform values is that you shade your bid below your true value by a factor of

/n$ -- you bid $\frac{n-1}{n}$ of your true value. Here, "true value" is
v$ because of resale. So expect the answer to look like $b(v) = \frac{n-1}{n} \cdot 2v$.

Quick Estimate: Take $n = 2$, $M = 1$, and suppose your valuation is $v = 0.5$. Your effective value is (0.5) = 1.0$. With two bidders, you should shade by

/n = 1/2$, so bid around $\frac{1}{2} \cdot 1.0 = 0.5$. That means $b(0.5) = 0.5 = v$. For $n = 10$ with $v = 0.5$: effective value
.0$, shade factor $9/10$, so bid $0.9$. That is
.8v$. These match the formula we expect.

Approach: Derive the equilibrium bid function using the standard symmetric equilibrium technique for first-price auctions with i.i.d. valuations.

Formal Solution:

Each bidder has private valuation $V_i \sim U(0, M)$ and effective value $W_i = 2V_i$. Since $V_i \sim U(0, M)$, we have $W_i \sim U(0, 2M)$. This transforms the problem into a standard first-price auction with i.i.d. $U(0, 2M)$ effective values.

In a symmetric equilibrium, each bidder uses the same strictly increasing bid function $\beta(w)$ of their effective value $w = 2v$. A bidder with effective value $w$ who bids $b$ wins if all $n - 1$ opponents bid less. Since $\beta$ is increasing, this happens when all opponents have effective values below $\beta^{-1}(b)$:

$P(\text{win} \mid b) = \left(\frac{\beta^{-1}(b)}{2M}\right)^{n-1}$

The bidder maximizes expected payoff:

$\max_b \; (w - b) \cdot \left(\frac{\beta^{-1}(b)}{2M}\right)^{n-1}$

In a symmetric equilibrium, $b = \beta(w)$. Taking the first-order condition and using the standard envelope-theorem argument (or solving the resulting ODE with boundary condition $\beta(0) = 0$), the equilibrium bid function for $U(0, 2M)$ values is:

$\beta(w) = \frac{n-1}{n} \cdot w$

Substituting back $w = 2v$:

$b(v) = \frac{n-1}{n} \cdot 2v = \frac{2(n-1)}{n} \cdot v$

This is valid for $v \in [0, M]$.

Explicit cases:

  • $n = 2$: $b(v) = \frac{2}{2} v = v$. You bid your raw private valuation. Your profit margin is the entire resale premium $v$.
  • $n = 10$: $b(v) = \frac{18}{10} v = 1.8v$. You bid 90% of your effective value
v$. Competition squeezes your margin down to $0.2v$.

Qualitative effects:

  • *Resale option:* The resale doubles your effective value, so you bid more aggressively than you would without resale. Without resale, the equilibrium bid would be $b(v) = \frac{n-1}{n} v$.
  • *More bidders:* As $n$ grows, $\frac{n-1}{n} \to 1$, so $b(v) \to 2v$. Competition drives bids toward the full effective value, and the winner's informational rent (the gap between value and bid) shrinks to zero.

Answer: The symmetric equilibrium bid function is $b(v) = \frac{2(n-1)}{n} v$. For $n = 2$, $b(v) = v$; for $n = 10$, $b(v) = 1.8v$. The resale option doubles the effective value, shifting all bids up, while more bidders drive the shading factor $\frac{n-1}{n}$ toward

$.

Intuition

The key insight is that a resale option is economically equivalent to simply having a higher valuation. Once you recognize that the item is worth v$ to you instead of $v$, the problem collapses into a textbook first-price auction -- the resale mechanics are a red herring that tests whether you can identify the true payoff structure before diving into game theory.

This pattern shows up constantly in trading: whenever you can transform or repackage an asset (resell, hedge, convert), your bid should reflect the transformed value, not the raw one. The other critical lesson is the $\frac{n-1}{n}$ shading factor. In practice, the number of competing bidders (or the intensity of competition) is the dominant driver of how aggressively you should bid. With two bidders you capture half the surplus; with ten you keep only 10%. Understanding this trade-off between competition and margin is the core of any auction or market-making strategy.

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