Zero Profit at Every Bid, Because Winning Says the Firm Was Cheap

Acceptance restricts the value to below your bid, where a uniform variable averages half of it, and doubling half your bid returns exactly your bid. The expected profit is therefore identically zero at every bid up to 100 and 100 minus b above it, so there is no optimal bid to find. With a general multiplier the profit is b squared times k minus 2, over 200, making doubling the exact break-even multiple, and the article shows a value distribution starting at 50 where the same bidder profits.

A company is worth $V$ , somewhere between 0 and 100 with every value equally likely. In your hands it would be worth $2V$ , because you can run it better. You make an offer, and the offer is accepted only if it is at least the true value. What do you bid?

Anything you like. The expected profit is exactly zero at every bid from 0 to 100, and negative above 100. There is no optimum to find, which is a more interesting answer than a number would have been.

Where the apparent twenty comes from

The reasoning that produces a bid of 80 is clean, and every step of it is arithmetically correct. The value averages 50 before you learn anything, since:

\mathbb{E}[V] = \int_0^{100} \frac{v}{100}\,dv = 50

(1)

Twice fifty is a hundred, so the company is worth 100 to you on average. Bid 80 and pocket 20. The subtraction $2 \times 50 - 80 = 20$ is not where this goes wrong.

What goes wrong is which average was used. Equation (1) is the average over all values $V$ can take. You never own the company at all values of $V$ . You own it only in the cases where your offer was accepted, and those cases are not a fair sample.

Winning is information

The event you are averaging over

A bid $B$ is accepted exactly when $B \ge V$ . So conditional on winning, $V$ is uniform on $[0, B]$ rather than on $[0, 100]$ . Acceptance is not a neutral administrative step. It is the seller telling you that the company was worth less than you offered.

The conditional mean follows immediately:

\mathbb{E}[V \mid V \le B] = \frac{1}{B}\int_0^{B} v\,dv = \frac{B}{2}

(2)

At $B = 80$ , winning means the value averaged 40, not 50. Double it and you get exactly 80, which is precisely what you paid. The doubling and the conditioning cancel, and they cancel exactly rather than approximately.

Fig. 1 — Winning restricts the value to the shaded half of the line, and the average of a uniform variable over that region is the midpoint, B/2.

Zero at every bid in the range

The cancellation in (2) has nothing to do with 80, so it is worth doing in general. If the offer is rejected the profit is zero, and if it is accepted the profit is $2V - B$ , so:

\mathbb{E}[\Pi(b)] = \int_0^{b} \frac{2v - b}{100}\,dv = \frac{b^2}{100} - \frac{b^2}{100} = 0 \qquad (0 \le b \le 100)

(3)

Flat. Identically zero across the whole range, which means the winner's curse does not merely reduce the apparent edge, it removes all of it at every level simultaneously. There is no bid to look for and no first-order condition to solve, because the objective has no slope.

Above the range the integral saturates, since acceptance becomes certain:

\mathbb{E}[\Pi(b)] = \int_0^{100} \frac{2v - b}{100}\,dv = 100 - b \qquad (b > 100)

(4)

So overbidding costs exactly the overpayment. A bid of 125 loses 25 on average, and there is nothing subtle left in that region: you are certain to win and certain to have paid above what the company is worth to you.

Fig. 2 — The whole objective function. Nothing to maximise inside the range, and a strict penalty outside it.

Why doubling is exactly the wrong multiple

Replace the doubling with a general multiplier $k$ , so the company is worth $kV$ to you. The same integral gives:

\mathbb{E}[\Pi(b)] = \int_0^{b} \frac{kv - b}{100}\,dv = \frac{b^2 (k-2)}{200}

(5)

The sign is carried entirely by $k - 2$ . At $k = 2$ the profit vanishes for every bid, which is the case in front of us. At $k > 2$ profit is positive and increasing in the bid, so you should bid as high as the range allows: at $k = 3$ and $b = 100$ the expected profit is 50. At $k < 2$ every bid loses money and the right move is not to play.

Doubling sits exactly on the knife edge, which is what makes the problem feel like a trick. The factor of two in the payoff is cancelled by the factor of two hiding in equation (2), where the conditional mean is half the bid rather than the bid. Being worth twice as much as the seller is precisely enough to break even against the information you gave away by winning, and not a penny more.

What the uniform assumption is doing

The exact zero is a property of pairing a doubling with a value that is uniform on an interval starting at zero. That is what makes $\mathbb{E}[V \mid V \le B]$ equal to $B/2$ exactly. Change the distribution and the answer changes, which is worth seeing once so that the zero is not mistaken for a law of nature.

Suppose instead the value is uniform on $[50, 100]$ , so the seller is known to have something reasonably good. Winning still tells you the value is below your bid, but it can only drag the conditional mean down to the midpoint of $[50, b]$ , which is well above $b/2$ . Working the integral out:

\mathbb{E}[\Pi(b)] = \int_{50}^{b} \frac{2v-b}{50}\,dv = b - 50 \qquad (50 \le b \le 100)

(6)

Positive throughout, and increasing, so here you should bid the maximum and expect 50. The curse is still operating, it is simply not strong enough to eat the whole doubling, because the floor on the value limits how bad the news of winning can be. The general lesson survives and the number does not: whatever distribution you assume, the quantity to compute is the mean of $V$ over the acceptance region, never over everything.

One reading has to be pinned down or the answer moves for the wrong reason. The doubling applies to the value of the company, not to whatever you paid for it. If it applied to the bid, the problem would be a different and much easier one, and every claim above would be false.

Sources and further reading

The effect the whole problem is about — Winner's curse
The step in equation (2) — Conditional expectation
The distribution assumed for the value — Continuous uniform distribution
What bidders do about it in practice — Bid shading

The expected profit was computed twice by different routes. Summing over twenty thousand equally likely midpoints in exact fractions returns exactly zero at bids of 10, 25, 50, 75, 80 and 100, and exactly $100 - b$ at 110, 125 and 150. A hundred thousand simulated draws at six bid levels agreed within 0.15 inside the range. A sweep of the multiplier on a hundredth-wide grid found the first profitable multiple at 2.01, confirming that the break-even multiple in (5) is exactly two.