Auction and best response Consider an auction in which $k$ identical objects are sold to $
n&gt;k $ bidders. Each bidder $i$ needs only one object and has a valuation $v_{i}$  for
  the object. In the auction, simultaneously, every bidder $i$ bids $ b_{i} $ . The highest $k$
  bidders win. Each winner gets one object and pays the $k+1 ^{st}$ highest bidder (i.e.,
  the price $p$ is the highest bid among the bidders who do not get an object). (The
  ties are broken by a coin toss.) Each of the losing bidders gets a gift of value $w$
  for their participation. (The winners do not get a gift.) Show that the game has
  a dominant strategy equilibrium, and compute the equilibrium.


It has a solution on page 9 here, but I cannot understand it.

Can someone explain me the answer or provide one? Also, do the losing bidders pay their bid amount here? And what does this line mean: "Each winner gets one object and pays the $k+1 ^{st}$ highest bidder"?
Hint: I will talk through a simpler case, and let you extend it to the generic case.

Let's say there's only two bidders and only the top 1 bidders win. Whoever wins pays whatever the loser bid. This situation is basically a second price auction.

You, player 1, like the object up for auction with a value of $v_1$, but the consolation prize of $w$ is also nice. The solutions you gave say that the optimal bid is $b_1^* = v_1 - w$. Let's confirm this. 

If you win this auction with this bid, that means the other bidder bid less than you, $b_1^* &gt; b_2^*$. The total value of winning is $v_1 - b_2^*$. If you lose, then the other bidder bid higher than you, and you receive $w$. So the difference between winning and losing is

$$v_1 - b_2^* - w &gt; v_1 - b_1^* - w = v_1 - (v_1 - w) - w = 0$$

So hooray! There is value in winning compared to taking the consolation. :)

What happens if you unilaterally deviate from this proposed strategy (so your opponent doesn't change their bid) and bid higher up to  $b'_1 = v_1 + \epsilon - w$? Well, if you win when you were going to win anyway compared to the above case...you still get $v_1 - b_2^*$ as the total reward. If you lose when you were going to lose anyway, then you still get $w$. The interesting part is when you bid high enough to then win when you would have previously lost.

That implies we would have $b_1^* &lt; b_2^* &lt; b'_1$. You win and receive $v_1 - b_2^*$, so seems like nothing is different...or is it?

$$v_1 - b_2^* - w &lt; v_1 - b_1^* - w = v_1 - (v_1 - w) - w = 0$$

So now you've won...but you should've just taken the consolation prize. :(



You can do a similar thought experiment for bidding lower. If you bid lower and win, then what you pay doesn't change from bidding the proposed strategy. If you lose when you would've lost anyway, then nothing changes. If you lose when originally you would've won, then you're taking a relative loss again.

So there is no outcome that makes you better off by unilaterally deviating from the proposed strategy of bidding $b_1^*$. You can extend this to realize it is a dominant strategy for all players to follow this line of reasoning.

What about with three players with two winners? Three players with one winner? This question is really just a weird extension of a second price auction. If you understand that, that will help, but otherwise, try following the intuition above and re-reading your solutions given to you.