I'm trying to find the correct formula to calculate de expected revenue of a reverse auction, properly, a second-price sealed-bid procurement auction to identify the lowest cost supplier.

Prices are private, independent, and uniformly distributed between 50 to 100. no bidder will go below 50. And the number of bidders are 10

We know that it does not matter whether we use a first-price or a second price auction - the expected final price is the same.

The formula E(p)=(n-1)/(n+1) stems from the fact that it is the valuation of the (n-1)th bidder that determines the expected price.

Now, this is a reverse auction, so the expected price stems from the ordering of bids. In a reverse auction, the price is not dependent on the value of the (n-1)th bidder, but on the second bidder. So we get that instead, E(p)=2/(n+1).

  • $\begingroup$ By your formula, $E(p) = \frac{2}{10+1} = 0.1818$, but no one's price is supposed to fall below $50$? Also I am very unsure what your question is. $\endgroup$
    – Kitsune Cavalry
    Dec 18, 2015 at 2:48

1 Answer 1


Hint: In second price sealed bid, it is optimal to bid your own valuation for the good. So in a reverse auction, it is optimal for each person to bid their lowest possible price they are willing to supply at. Intuitively, if one individual deviates from that strategy, it doesn't increase their chance of winning the auction or increase their chance at getting a higher priced contract, since the price they supply is based on other contractors' bids.

So the expected revenue is going to be the average value of the second lowest bidder amongst ten bidders. Try to set up an integral to describe the distribution and go from there.

For a normal second price auction (not reverse):

$F(x)$ is the probability that the price will be less than or equal to some given $x$, and $f(x)$ is the corresponding density function.

$$F(x) = \frac{x-50}{50}; \quad f(x) = F'(x) = \frac{1}{50}$$

Probability that second highest bid is $x$, with $n$ bidders:

$$h(x,n) = n(n−1)f(x)(1−F(x))F(x)^{n−2}$$

Expected value of $x$ (the price):

$$\int^{100}_{50} x h(x, n) dx$$

So you can easily apply a similar logic to your reverse auction. Set up $F(x)$ so that it represents the probability that the price will be greater than or equal to some given $x$.

  • $\begingroup$ Something is fishy with the $F(x)$, because $F(100) = 2$. Then again this is a numbers question, so a few errors in the answer don't hurt. $\endgroup$
    – Giskard
    Dec 18, 2015 at 8:05
  • $\begingroup$ I think you may be right, but it is a while since I did auction theory, and I'm not sure how to fix it. $\endgroup$
    – Kitsune Cavalry
    Dec 18, 2015 at 17:40
  • $\begingroup$ @denesp Ah! I think I figured it out. Will edit the answer. $\endgroup$
    – Kitsune Cavalry
    Dec 18, 2015 at 22:58

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