Consider a seller who must sell a single private value good. There are two potential buyers, each with a valuation that can take on one of three values,θi∈{0,1,2}, each value occurring with an equal probability of 1/3.The players’ values are independently drawn. The seller will offer the good using a second-price sealed-bid auction, but he can set a “reserve price” of r≥0 that modifies the rules of the auction as follows. If both bids are below r then neither bidder obtains the good and it is destroyed. or above r then the regular auction rules prevail. If only one bid is at or above r then that bidder obtains the good and pays r to the seller.

What is the optimal reserve price in a second price sealed bid auction?

  • $\begingroup$ This $r$ that you describe is not really a traditional reserve price since it is not just the minimum price at which a transaction will occur, but rather it's the price at which any transaction will occur. So this is quite a peculiar auction. It's a good question though. $\endgroup$
    – Shane
    Commented Nov 20, 2014 at 18:27
  • $\begingroup$ @Shane: Quote: "If both bids are (...) above r then the regular auction rules prevail". Looks like a normal reserve price for an SPSB Auction to me... $\endgroup$
    – miwe
    Commented Nov 21, 2014 at 12:08
  • $\begingroup$ @miweiss I believe the question was edited as it used to say that the transaction takes place at the reserve price. $\endgroup$
    – Shane
    Commented Nov 23, 2014 at 5:02

1 Answer 1


First, we want to see which strategies the bidders will play: From a real bidders perspective, you can look at the reserve price as if it were the bid of a third bidder, regarding both allocation and payment rule. Therefore, as we know that the sealed bid second price auction is truthful, we can expect the bidders to bid their true value.

The probabilities of the bidders value profiles ($v_i, v_2$) are:

Propbabilities for each value profile

You can count the profiles if it helps you to understand the following case distinction:

  1. Assume we set $r > 2$, then the expected revenue is 0 (as item is always destroyed).
  2. Assume we set $2 \geq r > 1$, then the expected revenue is $$ E_{rev} = \frac{1}{9} * 2 + \frac{4}{9} * r + \frac{4}{9} * 0 $$ which is highest at r = 2 and takes the value $1.11$.
  3. Assume we set $1 \geq r \geq 0$ , then the expected revenue is $$ E_{rev} = \frac{1}{9} * 2 + \frac{3}{9} * 1 + \frac{4}{9} * r + \frac{1}{9} * 0$$ which is highest at r = 1 and takes the value $1$.

Therefore, the expected revenue is maximized with $r = 2$.

Note that I assumed your "above r" is actually "greater or equal than r", otherwise, the case, where the highest bid equals the reserve price, would not be defined.

Best, miweiss


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .