Optimal Bid in a First-Price Auction Against a Uniform Opponent
A box has a random dollar value $V \sim \text{Binomial}(200,\, 1/2)$. You are in a one-shot first-price sealed-bid auction against a single opponent. Your opponent bids an integer drawn uniformly at random from $\{0, 1, \ldots, 200\}$. Ties are broken in your favor.
You must choose an integer bid $b \in \{0, 1, \ldots, 200\}$. If you win, you pay your bid $b$ and receive the box (profit $= V - b$). If you lose, your profit is $0$.
- Find the integer bid $b$ that maximizes your expected profit.
- Compute the maximum expected profit exactly.
Hints
- Write expected profit as win probability times expected surplus. What simplification does the independence of $V$ and the opponent's bid give you?
- The expected profit becomes $\frac{(b+1)(100 - b)}{201}$. This is a quadratic in $b$ -- find its peak.
- The continuous optimum is at $b = 49.5$. Check $b = 49$ and $b = 50$; both give $f(b) = 2550$.
Worked Solution
How to Think About It: You are bidding on a box whose value averages $E[V] = 100$. Your opponent is naive -- bidding uniformly at random, so you can exploit their randomness. The tension is classic first-price auction logic: bid higher to win more often, but every dollar you add to your bid comes straight out of your profit when you win. Your expected profit is the product of your win probability and your expected surplus. Since $V$ and the opponent's bid are independent, this simplifies to a clean one-variable optimization.
Quick Estimate: If you bid around half the expected value, say $b = 50$, you win about $51/201 \approx 25\%$ of the time and net
Approach: Write the expected profit as a function of $b$, then optimize the resulting quadratic.
Formal Solution:
You win whenever the opponent bids at most $b$ (ties go to you). Since the opponent bids uniformly on $\{0, 1, \ldots, 200\}$:
$P(\text{win}) = \frac{b + 1}{201}$
Conditional on winning, your profit is $V - b$. Since $V$ is independent of the opponent's bid:
$E[\text{profit}] = P(\text{win}) \cdot E[V - b] = \frac{(b+1)(100 - b)}{201}$
Define $f(b) = (b+1)(100 - b)$. Expanding:
$f(b) = -b^2 + 99b + 100$
This is a downward-opening parabola. The continuous maximum is at $b = 99/2 = 49.5$. Since $b$ must be an integer, check both neighbors:
- $f(49) = 50 \times 51 = 2550$
- $f(50) = 51 \times 50 = 2550$
Both $b = 49$ and $b = 50$ achieve the same maximum. The maximum expected profit is:
$E[\text{profit}] = \frac{2550}{201} = \frac{850}{67} \approx 12.687$
Answer: The optimal bid is $b = 49$ or $b = 50$ (both are equally optimal). The maximum expected profit is $\frac{850}{67} \approx 12.69$ dollars.
Intuition
This problem illustrates the fundamental first-price auction trade-off: you want to bid high enough to win, but every extra dollar you bid erodes your surplus. Against a uniform opponent, this trade-off is clean enough to optimize in closed form -- the expected profit is a quadratic function of your bid, and the optimal bid sits right around the midpoint between zero and the expected value of the box. That the optimum is near $E[V]/2$ is not a coincidence. In a first-price auction against a single uniform opponent, you are essentially splitting the expected surplus: half goes to your bid (which the "market" absorbs) and half is your profit.
In practice, this is the logic behind bid shading. Even though the box is worth