I am interest in finding the optimal strategy for a single bidder (the BNE), when I have a third-price auction with N bidders and two identical goods. The bidders have iid valuations: U(0,1)

I have been told by my teacher that it is a weakly dominant strategy for the bidders to bid their valuations, but I cannot see why this would be?
(and I know from the literature that it is optimal to bid slightly above your true valuation if only one item is auctioned off, i.e. to bid $b_i=\frac{n-1}{n-2}v_i$)

The utility function for bidder i looks like this:
$ u_i(b_i, b_{-i}) = \left\{ \begin{array}{lr} v_i-\max_{(2)}\{b_{-i}\} & \text{if } b_i>\max_{(1)}\{b_{-i}\}>\max_{(2)}\{b_{-i}\} \\ v_i-\max_{(2)}\{b_{-i}\} & \text{if } \max_{(1)}\{b_{-i}\}>b_i>\max_{(2)}\{b_{-i}\} \\ 0 & \text{if } b_i<\max_{(2)}\{b_{-i}\} \end{array} \right. $
since in this auction, the highest and second highest bidder wins a good each, and both pay the third highest bid, because the bidders gain equal value from either good.

If anyone knows why it supposedly is a weakly dominant strategy to bid your true value in this type of auction, or where I might find a good explanation, I would really appreciate any help I can get!

  • $\begingroup$ Should we assume that $n \geq 3$? Also: I guess that, if this is indeed dominant, then it should be dominant for any value distribution (not just uniform)? $\endgroup$
    – afreelunch
    Nov 2, 2021 at 9:49
  • $\begingroup$ Yes, sorry for not clarifying that I assume $n\geq 3$. And indeed, I would think that it should hold for any continuous distribution with increasing probability. The reason why I asked specifically for case with a uniform value distribution is because that is the distribution I assumed in my thesis. $\endgroup$ Nov 2, 2021 at 10:57

1 Answer 1


Consider bidder $i$. Let $p = \max_{(2)}\{b_{-i}\}$.

By bidding $v_i$, she wins if $v_i > p$ and not if $v_i < p$ (and indifferent if $x_i = p$).

Suppose, however, she bids $z_i < v_i$.

(i) If $v_i > z_i > p$, she still wins the auction and still gets $v_i - p$.

(ii) If $p > v_i > z_i$, she still loses the auction and still gets $0$.

(iii) However, if $v_i > p > z_i$, she now loses instead of winning.

Thus, bidding less than $v_i$ can never improve $i$'s payoffs.

Can you finish the argument for the case $z_i > v_i$?

  • $\begingroup$ Thank you! It makes good sense. However, do you happen to know if there is a way to solve this as an equation? Or at least write up the expected utility given that each bidder bids her true value, sinse I am also interested in comparing bidder's expected utility in a third-price auction with her expected utility in a second-price auction? $\endgroup$ Nov 2, 2021 at 8:44
  • $\begingroup$ I dont see the benefit of attempting to write this as an equation? This is enough to show this is indeed the equilibrium. As for expected utilities, I would advice you to start a new question instead asking for that! $\endgroup$ Nov 2, 2021 at 11:45

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