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Bid cost: the eighth invitation is the first that loses money

A bid costs 20,000 to prepare and 150,000 if it wins. One chance in seven still pays 1,429; one in eight loses 1,250. Seven holds and the eighth invitation is the first that loses money. The page names no number to invite.

Live result

Invitation 8 is the first loss; 7 still pays.
  • Expected profit at 7$1,428.57
  • Expected profit at 8−$1,250.00
At your count of 7, preparing a bid pays.
Invitation 8 is the first loss; 7 still pays.At your count of 7, preparing a bid pays.

Givens

Try a scenario

Bid preparation cost
20,000.00
$
From 0.01 to 10,000,000, in one currency.
Expected margin if won
150,000.00
$
After costs, from 0.01 to 1,000,000,000 in the same currency.
Count invited
7
A whole count from 1 to 30.
Chance of winning
one in the count
Preparation cost
paid win or lose
Margin
same whoever wins

Expected profit from bidding at each count invited

Expected profitLast paying countFirst losing count
Live chart
Expected profit from bidding at each count invitedExpected profit at counts 1 to 20, with 7 the last paying count and 8 the first without profit.58.8k130kExpected profit, $0Last paying count7 · $1,428.57First losing count8 · −$1,250.0011020Count invited
Profit falls as the invited count grows; a loss carries a minus sign in the same currency.

Count invited

invited
invited
invited
invited

It does not check the values in between.

How the answer sits across the values sampledA rail of the 20 values sampled, from 1 invited to 20 invited in steps of 1 invited. Across the values sampled, the verdict on preparing a bid stays a profit from 1 invited, the lowest value sampled to 7 invited. The nearest value where it changes is 8 invited, where it becomes a loss.first change8 invited7 invitedon the page

Across the values sampled, the verdict on preparing a bid stays a profit from 1 invited, the lowest value sampled to 7 invited. The nearest value where it changes is 8 invited, where it becomes a loss.

Across the values sampled, the verdict on preparing a bid stays a profit from 1 invited, the lowest value sampled to 7 invited. The nearest value where it changes is 8 invited, where it becomes a loss.
The answera profit
Values sampled1 invited to 20 invited, 20 values, step 1 invited
Holds from1 invited, lowest sampled
Holds to7 invited
Nearest value that changes it8 invited
The answer therea loss

What an unchanged region does and does not prove is set out in a short lesson.

Every admissible integer in this range was checked. Values outside the range remain unchecked.

The same count, read from two seats

For the bidder, each extra invitation spreads the same margin across another chance while preparation still costs the same.

For the buyer, a larger invitation list can leave fewer bidders expecting a profit.

Where this stops

The page calculates expected profit from your values, names no number to invite and predicts nobody's decision. It shows what a mechanism does under stated rules and says nothing about any public procurement regulation.

Here is the maths

▾
Scale
Linear on both axes, with zero drawn
Basis
One equal chance of winning among those invited, independent of who else bids
Assumptions
One margin and one preparation cost, the same for every count read
Model version
1.0.0

The formula, in words

The margin is shared across equal chances of winning. Subtract the preparation cost, paid regardless of the outcome, to find expected profit from bidding.

E(n)=mn−cE(n)=\frac{m}{n}-cE(n)=nm​−c

where E(n)E(n)E(n) is the expected profit from bidding, mmm the margin if the bid wins, nnn the count invited, and ccc the cost of preparing the bid.

The count that stops it

The last paying count keeps expected profit above zero:

n∗=max⁡{ n≥1: mn>c }n^{*}=\max\left\{\,n\ge 1:\ \frac{m}{n}>c\,\right\}n∗=max{n≥1: nm​>c}

The next count is n∗+1n^{*}+1n∗+1. An exact whole ratio breaks even there; the last paying count is one below it.

A worked example

The extra invitation turns a profit into a loss with preparation costing 20,000 and winning worth 150,000.

The ratio marks where profit reaches zero:

150,00020,000=7.50\frac{150{,}000}{20{,}000}=7.5020,000150,000​=7.50

The last whole count below it still pays:

n∗=7n^{*}=7n∗=7

Profit at that count, in the same currency:

E(7)=150,0007−20,000=1,428.57E(7)=\frac{150{,}000}{7}-20{,}000=1{,}428.57E(7)=7150,000​−20,000=1,428.57

At the next count, profit becomes a loss:

E(8)=150,0008−20,000=−1,250.00E(8)=\frac{150{,}000}{8}-20{,}000=-1{,}250.00E(8)=8150,000​−20,000=−1,250.00
Show the full maths

Weight both outcomes before counting entrants

Winning leaves the margin less preparation cost; losing leaves only the preparation cost to pay. Weight each result by its chance:
E(n)=1n(m−c)+(1−1n)(−c)=mn−cE(n)=\frac1n(m-c)+\left(1-\frac1n\right)(-c)=\frac mn-cE(n)=n1​(m−c)+(1−n1​)(−c)=nm​−c
This uses the same cost whether the bid wins or loses.
Preparation pays only when expected profit is strictly positive. Both the cost and the invited count are positive, so rearranging keeps the inequality’s direction:
E(n)>0  ⟺  mn>c  ⟺  n<mcE(n)>0\iff\frac mn>c\iff n<\frac mcE(n)>0⟺nm​>c⟺n<cm​
The last paying count is therefore the largest whole number strictly below the margin divided by cost:
n∗=max⁡ ⁣(0,⌈mc⌉−1)n^*=\max\!\left(0,\left\lceil\frac mc\right\rceil-1\right)n∗=max(0,⌈cm​⌉−1)
where n∗n^{*}n∗ is the last paying count. Rounding the ratio up and subtracting one handles both whole and fractional ratios. Zero means that no positive invited count pays.
Break even is a different condition:
E(n)=0  ⟺  n=mcE(n)=0\iff n=\frac mcE(n)=0⟺n=cm​
It occurs at an invited count only when the ratio is a whole number. At that count there is no profit, so it is excluded from the paying counts.
The worked example has a fractional threshold, so seven pays and the next whole count loses:
150,00020,000=7.5n∗=⌈7.5⌉−1=7\frac{150{,}000}{20{,}000}=7.5\qquad n^*=\lceil7.5\rceil-1=720,000150,000​=7.5n∗=⌈7.5⌉−1=7150,00020,000=7.5n∗=⌈7.5⌉−1=7\begin{gathered}\frac{150{,}000}{20{,}000}=7.5 \\ n^*=\lceil7.5\rceil-1=7\end{gathered}20,000150,000​=7.5n∗=⌈7.5⌉−1=7​
With the same preparation cost and a margin of 140,000, the threshold is whole; six is the last paying count and seven breaks even:
140,00020,000=7n∗=⌈7⌉−1=6E(7)=0\frac{140{,}000}{20{,}000}=7\qquad n^*=\lceil7\rceil-1=6\qquad E(7)=020,000140,000​=7n∗=⌈7⌉−1=6E(7)=0140,00020,000=7n∗=⌈7⌉−1=6E(7)=0\begin{gathered}\frac{140{,}000}{20{,}000}=7 \\ n^*=\lceil7\rceil-1=6 \\ E(7)=0\end{gathered}20,000140,000​=7n∗=⌈7⌉−1=6E(7)=0​

The same count, in the other lesson

One more bidder thins expected profit here and deepens the gap between the lowest estimate and true cost in the linked lesson. More competition can discourage entry while making the lowest bid more likely to underestimate cost.

Why the lowest bid tends to be an underestimate

What this does not model

No unequal bidders or changing preparation costs. No learning, reserve price or bid subsidy. No prediction of entry.

What the sweep checks

Only the count changes, in whole steps; each verdict uses the same margin and preparation cost as the finding.

Source basis

Nicola Dimitri, Gustavo Piga and Giancarlo Spagnolo, Handbook of Procurement: costly preparation changes the participation a buyer can expect; Paul Milgrom, Putting Auction Theory to Work: entry costs meet diminishing gains from another entrant. Paul Klemperer, Auctions: Theory and Practice: entry responds to incentives and can lower common value auction revenue.

Calculation version

1.0.0

The whole record prints. Your browser can save it as a PDF instead of printing it.

Your margin, preparation cost and count determine the figures; the page neither assesses bidders nor recommends inviting, bidding or staying away.

Related

  • Winner's curse: the lowest bid understood the job leastSix 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.Interactive lesson · 4 min
  • Supplier fixed cost: the lowest line prices give the highest billAward each line to its cheapest quote and the total is expected to be the lowest one available. Type two items, two suppliers and one cost that attaches to each supplier engaged rather than to a line, and it stops being true. At 50 a supplier the line by line award totals 310 while the whole basket at one supplier totals 270. The crossing sits at 10, below which the split award really is cheaper, and no supplier is ranked and no line is awarded here.Calculator · 3 min

Bid cost and the count invited

Bid cost: the eighth invitation is the first that loses money

Values typed

Bid preparation cost
$20,000.00
Expected margin if won
$150,000.00
Count invited
7

What the count gives

Last count that pays
7
First count that does not
8
Expected profit at the last paying count
$1,428.57
Expected profit at the next count
−$1,250.00
Verdict at the count typed
bidding pays

Record fields

Calculation version
1.0.0
Boundary
The page calculates expected profit from your values, names no number to invite and predicts nobody's decision. It shows what a mechanism does under stated rules and says nothing about any public procurement regulation.
Source basis
Nicola Dimitri, Gustavo Piga and Giancarlo Spagnolo, Handbook of Procurement, Paul Milgrom, Putting Auction Theory to Work, Paul Klemperer, Auctions: Theory and Practice

Your margin, preparation cost and count determine the figures; the page neither assesses bidders nor recommends inviting, bidding or staying away.

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