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.
I have no idea how to answer this one... someone please help
 A: 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 $v_1$.
