# Third and average price auction

Third price auction: the winner is the highest bidder but this time instead of paying the second highest bid, he would pay the third highest bid. -assume there are at least 3 bidders. -

Average price auction: the winner is the highest bidder but pays the average of his and the second highest bid.

For these,

How to show whether truthtelling is a dominant strategy or not? And how to show whether truthtelling is a nash equlibrium?

• Somehow, I feel, there is not enough context. What are players' payoffs? Commented May 12, 2014 at 21:46
• Not given :( Dear @mathse
– 1190
Commented May 12, 2014 at 21:49
• Then, I can't answer it. Maybe this makes sense to someone else, but I don't know. Commented May 12, 2014 at 21:50
• Okay thank you so much:) @mathse
– 1190
Commented May 12, 2014 at 21:53

Similarly, truth-telling is not dominant in the average-price auction. Suppose I value the item at $100 and the bid is £10. If I bid my true value I will pay £55, but if I bid £12 I will pay only £11. Both examples show that truth-telling is not a Nash equilibrium, because you could suppose that the other bids are at their truthful values. The logic is the same as the second price auction. Suppose you value the object at$v$and the two highest other bids are$a,b$with$a \gt b$. If your value is greater than$a$, bidding above your value doesn't hurt, but it doesn't change anything. You get the object and pay$b$, making a profit of$v-b$. If you bid low, you might drop below$a$and forego the profit. If your value is between$a$and$b$, by bidding high you might get the object, paying$b$and making a profit (at the expense of the$a\$ bidder). As long as you value the object more than the second highest other bid, you should bid an arbitrarily large amount. This clearly does not encourage truthful bidding.