# what is the Nash equilibrium in a Third price auction?

Consider an auction in which the winner is the highest bidder, and he pays the third highest bid. Suppose there are three players and each player $$i\in\{1,2,3\}$$ has a valuation $$v_i$$ and bids $$b_i$$. Without loss of generality, assume $$v_1>v_2>v_3$$. In case of a tie, the bidder with the lowest index $$i$$ wins and pays $$\min_{i}(b_i)$$. The payoff of the winner $$i$$ is given by $$v_i-p_i$$, where $$p_i$$ is the value he will have to pay. The other bidders get a payoff of zero.

(a) Show that truth-telling (i.e. $$b_i=v_i$$ $$\forall i$$) is in general not a Nash equilibrium of this auction.

For this I have said:

Truth telling is not a dominant strategy with this auction. In order to explain this I will need you to suppose that I value an item at \$100, and there are 2 other bids, \$200 and \$10. I should bid \$201 and pay only \\$10 for the item (note that the bid is higher than my private valuation). Consider three bidders. Suppose bidders 1 and 2 submit bids , respectively, and bidder 3's value is such that . If bidder 3 bids truthfully, her payoff is 0, because bidder 2 will win the object. However, if bidder 3 overbids, so that , then she would win the auction and get a positive payoff This shows that truthful bidding is not a dominant strategy: there exists a situation where playing truthful bidding is not a best response.

In the third price auction, the expected utility of for the bidder is: ( is your own bid, is the bid of the third place.)

(b) Show that everyone bidding the highest valuation (i.e. $$b_i=v_1$$ $$\forall i$$) is a Nash equilibrium of this auction.

In (a) you are not asked whether bidding the valuation is a dominant strategy. Instead, you need to show that $$(b_1,b_2,b_3)=(v_1,v_2,v_3)$$ is not a Nash equilibrium. To do this, you just need to show that a player has a strictly profitable deviation. So, suppose $$(b_1,b_2,b_3)=(v_1,v_2,v_3)$$. Player 2's pay off is zero. If they increase their bid to something more than $$v_1$$, then they win the auction, pay the third highest bid $$v_3$$, and get a payoff of $$v_2-v_3>0$$. This is a strictly profitable deviation, so all players bidding their valuation is not a Nash equilibrium.
For (b), to show $$(b_1,b_2,b_3)=(v_1,v_1,v_1)$$ is a Nash equilibrium, you need to show that no player has a strictly profitable deviation from bidding $$v_1$$ (when the other players are bidding $$v_1$$). When players all bid $$v_1$$, they all get zero payoff (players 2 and 3 because they do not win the auction, and player 1 because they win the auction and pay $$v_1$$).
• If player 1 raises their bid, they still win the auction and pay $$v_1$$ and get zero payoff. If player 1 lowers their bid, they lose the auction and get zero payoff.
• If player 2 or 3 raise their bid, they win the auction and pay $$v_1(>v_2>v_3)$$ so make a loss. If player 2 or 3 lowers their bid, they lose the auction and get zero payoff.
Thus, no player has a strictly profitable deviation, and hence it is a Nash equilibrium for all players to bid the highest valuation $$v_1$$.