Say we have an uniform price auction where 1 really rare stamp is for sale. Only 5 randomly selected stamp collectors are invited to bid and the stamp ends up selling for 1 million dollars. Now pretend that auction never happened.

If 6 randomly selected stamp collectors were invited instead of 5 in the previous example (again, original auction never happened), what expected percent increase in revenue should the stamp sell for - 10% more, 15% more? Similarly, what if only 4 bidders were allowed to the auction?

In short, I'm trying to determine how increasing or decreasing bidder participation by 1 impacts the ultimate selling price on a percentage basis. I'm sure lots of variables are at play, so am not sure if any economic theory can be used to address this.

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    $\begingroup$ I think you are asking an interesting question. I'd say it could have both an empirical and theoretic solution. 1st point to make is that not enough is known about the distribution of potential bidders WRT marginal benefit. 2nd point is that any answer you will get will be probabilistic. There is an x% chance of an increase in price. I don't think based on the information you have provided you would be able to make a statement about % increase in price changes. $\endgroup$ – Jamzy May 14 '15 at 1:16

It is a known result in Mechanism Design that both first-price as well as second-price auctions yield the same expected revenue, under certain conditions (like independence of valuations, private information, etc). Consult Jehle, G. A., & Reny, P. J. (2011). Advanced Microeconomic Theory (3d ed.), ch.9 on Auctions and Mechanism Design, for a very accessible exposition of the basics, as well as for the full set of conditions that must hold.

The answer to your question is bound to be distribution-specific. Assume as an example, that, from the point of view of the seller, all (unknown to him) valuations of the bidders for the object for sale, come from a Uniform $U(0,1)$ distribution: this implies that we have normalized the value of the object for sale and that we express the possible valuations of it as a percentage of its possible maximum value, which is assumed common for all bidders. I.e. What we say here is :"we don't know how much each bidder actually values the object, but we do know that the maximum possible valuation by a bidder will be some $V>0$, and this $V$ is common to all bidders". We do not assert that there exists some bidder that actually values the object at $V$.

In such a setup, the Expected Revenue of the Seller is

$$ER(N) = \frac {N-1}{N+1} \tag {1}$$

where $N$ is the number of bidders. Again, this essentially expresses expected revenue as a fraction of the object (unidentified) maximum value.

Now you can play around with $N$ to see how the expected revenue changes, in absolute, relative and percentage terms. It sure isn't changing linearly, but it is everywhere increasing in $N$, approaching unity (i.e. the object maximum valuation).

This should be intuitive in the framework adopted: the more the bidders are, the more probable becomes to obtain higher and higher valuations of the object, and so also to sell it at a higher value.

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  • $\begingroup$ Alecos, thanks so much for your insight! A few questions: 1) If I plug 5 in for N in the ER(N) function, we get 67%. 6 yields 71%. Am I making a fair statement in saying that increasing the number of bidders from 5 to 6 should theoretically yield a 4% increase in expected seller revenue? 2) Taking this one step further, I see you've noted that the 71% can be interpreted as the expected revenue of the objects maximum value. But I was also under the impression that the value of an object is whatever someone is willing to pay for it. Can you expand on this? Thanks again! $\endgroup$ – Jeff May 14 '15 at 16:27
  • $\begingroup$ 1) No. The percentage increase of revenue would be $(0.71/0.67) -1 \approx 6\%$ 2) 71% indicates "expected revenue is 71% of the objects possible maximum value", where we have assumed that "possible maximum value" is a finite number and common to all bidders. This is a consequence of the assumption that valuations are modeled as coming from a distribution that has a specific upper bound: again, the results will be distribution-specific. $\endgroup$ – Alecos Papadopoulos May 14 '15 at 16:47
  • $\begingroup$ 1) Rookie move on my end - agreed and thanks. And so the ER(N) function you provided only applies if the bidders are uninformally distributed and have equal probabilities of winning the bid? How would the function change if it were normally distributed? 2) Thanks! $\endgroup$ – Jeff May 14 '15 at 17:12

I am not sure I have the knowledge to provide you with a complete answer, but I can provide you with some insight.

First assume a pool of potential bidders. They are normally distributed. Each bidder has their own marginal benefit for this stamp. Call this $\alpha$. Each bidder is a utility maximiser with the utility function


A bidder will continue bidding until they assess that they will be happier with the money in their pocket.

5 bidders are pulled from the pool and given the opportunity to participate in the auction. The result of the auction is that the price the stamp sold for is


The price of the stamp is equal to the marginal benefit of the second highest bidder + an arbitarily small number. This is the price at which no party will raise the bid.

If there were a second auction with the same first 5 bidders+ an additional bidder, there is a possibility that this new bidder has a low marginal benefit of this product and has no impact on the price at an auction.

The condition for auction 2 to sell higher than auction one is


The marginal benefit of the entrant has to be higher than the marginal benefit of the second highest bidder.

The potential price with the addition of a new bidder is bounded between $[\alpha_2,\alpha_1]$. The price will not go any higher than what the second highest bidder is prepared to pay for it.

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  • $\begingroup$ Jamzy, thanks for your perspective! Definitely helpful. $\endgroup$ – Jeff May 14 '15 at 16:34

As Alecos noted in his answer, and standard auction results in the same revenue on average—the revenue equivalence theorem. To keep things simple, suppose that the assumptions necessary for this theorem to apply are satisfied (i.e that bidders are risk neutral, etc.) I assume also that values are private and that each bidder's value is a draw from the same probability distribution $F$, with density $f$.

We know from basic auction theory that the revenue from a second price auction is just the value of the second highest bidder. The revenue equivalence theorem therefore implies that if $v^{(i)}$ is the willingness to pay of the $i^\text{th}$ smallest value bidder and there are $n$ bidders then the expected revenue of essentially any auction is $E(v^{(n-1)})$. The probability density of the $n-1^\text{th}$ order statistic is $$nf(v)\frac{(n-1)!}{(n-2)!}F(v)^{n-2}(1-F(v)).$$ Thus, the expected revenue when there are $n$ bidders is $$E(r_n)=\int_{-\infty}^{\infty}nf(v)\frac{(n-1)!}{(n-2)!}F(v)^{n-2}(1-F(v))v\,dv.$$ Similarly, when there are $n+1$ bidders, expected revenue is $$E(r_{n+1})=\int_{-\infty}^{\infty}(n+1)f(v)\frac{(n)!}{(n-1)!}F(v)^{n-1}(1-F(v))v\,dv.$$ The percentage gain from the extra bidder is $$\frac{E(r_{n+1})-E(r_n)}{E(r_n)}.$$ If you know the distribution, $F$ of bidder's values then you can put it into these equations and calculate the gain for any $n$.

Some caveats

  • This analysis works if bidders are smart enough to bid in accordance with the equilibrium of the auction. If they choose their bids heuristically then there will be an error in the forecast revenue gain.

  • This gives the expected revenue gain. The actual revenue gain will depend upon the actual values of the bidders.

  • Obviously, the above is only accurate to the extent that the assumptions underlying it are satisfied.

  • In particular, note that the private values assumption probably does not hold for stamp collectors where there is likely to be a strong common values component.

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