Question

1. Consider a sealed-bid auction in which the winning bidder pays the average of the two highest bids. As in the auction models considered in class, assume that players have valuations v1 > v2 > ... > vn, that ties are won by the tied player with the highest valuation, and that each player’s valuation is common knowledge.

   1. Is there any Nash equilibrium in which the two highest bids are different? If there is, give an example. If there is not, prove that no such equilibrium exists.

   2. Is there any Nash equilibrium in which a player other than the one with the highest valuation wins? If there is, give an example. If there is not, prove that no such equilibrium exists.

   3. Will bidding more than one’s own valuation be weakly dominated in this auction? Will bidding one’s own valuation exactly be weakly dominated? Will bidding less than one’s own valuation be weakly dominated?

   4. What is the highest possible winning bid in any Nash equilibrium of this game? What is the lowest possible winning bid in Nash equilibrium.

Answer 1

Consider the game problem here there are 2 bidders, “bidder-1” and “bidder-2”, both of them have same valuation for the good, which is “$6”, this is the common knowledge, => both the bidders know the other bidders valuation for the good.

Here both the bidder have “7” possible choice to bid, these are {$1, $2, $3, $4, $5, $6, $7}. Now, bidder-1 will bid more than bidde-2, then 1^st bidder will get the good and will pay the 2^nd highest bid, => the bid od the 2^nd bidder.

So, if “Bidder-1” will bid “$1” and “bidder-2” will bid {$2, $3, $4, $5, $6, $7}, that is more than “$1”, then “bidder-2’s” payoff is “5” and “bidder-1’s” payoff will be “$0”. Similarly, if “Bidder-1” will bid “$2” and “bidder-2” will bid {$3, $4, $5, $6, $7}, that is more than “$2”, then “bidder-2’s” payoff is “4” and “bidder-1’s” payoff will be “$0”. So, in this way we have to derive all this pay off matrix for both players.

Now, if both the bidder will bid the same amount, then the decision will be taken by flipping a fair coin, => each bidder have an equal probability of winning the good, that is “1/2”. So, here we have to find out the expected payoff. So, if both will bid “$1”, the, both have expected payoff, “(6-1)*(1/2) + 0*(1/2) = 5/2”. Similarly, if both will bid “$2”, the, both have expected payoff, “(6-2)*(1/2) + 0*(1/2) = 4/2”, in this way we have to find out all the expected payoffs.

So, the entire payoff matrix is given below.

In this game there don’t have any strictly dominated strategy for both the bidders, but there are “weakly” dominated strategies. If we compare “$1” with “$6” of “bidder-2”, then we will get all the payoff of “bidder-2” is either less or equal under “$1” then “$6”, => “$1” is weakly dominated by “$6”. Similarly, if we compare all these possible strategies with “$6”, we will get all these are weakly dominated by “$6”. So, we can eliminate all the possible choices only “$6” will remain.

So, now “bidder-1” have {$1, $2, $3, $4, $5, $6, $7} possible choices but “bidder-2” have only “$6”. Now, under this situation the “bidder-1” will also choose “$6”, because still there is an probability of winning the good.

So, the optimum solution is both the player will choose “$6”.