Winner's curse: the lowest bid understood the job least

Six contractors price the same scope from the same information and one comes in well below the rest. Predict what happens to the winning bid's distance below true cost as more bidders are invited, then move the count from 2 to 20 and watch it rise. At an estimating spread of 8% the lowest of five bids is expected 9.3% under and the lowest of six is 10.1%, which is the first count past a 10% contingency. No bid is labelled and no number of bidders to invite is named.

Live result

At 5 centered estimates, the expected shortfall is 9.3%. Move bidder count once to reveal how selecting from more estimates changes it.

The belief

Inviting another bidder makes the lowest observed price more informative about the job's true cost.

Every bidder prices the same scope from the same information. Estimating errors are independent, centered on the same true cost, and normally distributed with the spread you type. A bidder with different information, scope, or a systematic cost advantage is outside this model.

Try a scenario

The typed statistical frame

Estimating spread%

The standard deviation of the centered estimating error as a percentage of true cost.

Typed contingency%

A comparison line only. Crossing it is not a label on a bid or invitation advice.

Bidder count and the selected minimum

Same scope and information, independent centered normal errors, one typed spread.

Bidder count5bidders

This is the one dominant control. It moves in whole counts from 2 through 20.

First count strictly above contingency

This is arithmetic on a typed spread, typed contingency, and bidder count under one same scope and same information model. It does not label a bid, evaluate a tender, estimate true cost, name how many bidders to invite, or advise combining awards.

Here is the maths

The formula, in words

Write each estimate as true cost plus a centered normal error. Selecting the lowest of $n$ estimates gives the same expected distance below zero as the expected maximum of $n$ standard normal values lies above zero. Multiply that distance $m(n)$ by the typed estimating spread. The crossing occurs only when this expected shortfall is strictly greater than the typed contingency.

The expected distance below zero:

m(n)=\int_{-\infty}^{\infty}z\,n\,\phi(z)\,\Phi(z)^{n-1}\,dz

The expected shortfall:

s=\sigma\times m(n)

where

s

is expected shortfall percent and

\sigma

estimating spread percent.

crossing only when expected shortfall percent is greater than contingency percent

A worked example

At 5 bidders, $m(5)$ is 1.162964. With an 8% spread, the expected shortfall is $1.162964\times8$ , which is 9.303716%, displayed as 9.3%.
At 6 bidders, $m(6)$ is 1.267206. With the same spread, the expected shortfall is 10.137651%, displayed as 10.1%.
Scanning whole counts from 2 upward finds no earlier value strictly above a 10% contingency, so 6 is the first crossing. The sixth estimate did not change true cost in the model.

What this does not model

It cannot separate a genuinely lower supplier cost from a larger estimating error in a real tender. It does not model dependent errors, different information, different scopes, systematic advantages, technical acceptability, commercial terms, or award structure.

Source class

The mechanism is standard order statistic mathematics for the minimum of independent normal estimates. Its application class is public common value bidding literature describing the selection effect. No definitions list, taxonomy, or method table is reproduced.

Calculation version

1.0.0

This lesson describes the expected minimum of independent centered normal estimates under the assumptions printed on the page. It never accepts bid labels, actual bid prices, supplier identities, a bidder count presented as advice, invitation advice, or an award choice. A statistical crossing is not a finding about any real tender.