23.4  Suppose that there are N bidders; each bidder is one of two types, and , and each type is equally likely.  Show that truthful bidding is a dominant strategy for the type.  Repeat for the type.

A type bidder could bid  + 1 and be certain that he will win against any other type.  But there is a risk in doing so. As long as no other  type bidder plays this same strategy and they all bid truthfully then the one bidder that applies the strategy will only pay and have zero net gain.  But suppose some other players recognize this strategy and some don't.  Those who have seen the strategy will find the auction being decided by chance among the j players who have seen the strategy.  The winner will pay , the next lowest bid, and have zero net gain 1/j % of the time.  As long as j < then number of type bidders, this is a good strategy.  But all types will see the opportunity and they will all want to bid a little above .  When they all see the opportunity the winner will be decided by chance and will pay , the next highest bid, for a net gain of - .  But this would be the same net gain if they all bid truthfully to begin with, so they will.

The type person will never bid below because he will never win if there is at least one other type in the pool of bidders.  Not knowing whether there is another type bidder, the person will not risk losing the auction.  Even if he is the only type person, he will never bid since then his fate is left to chance when he could have won the auction with certainty.

23.5 What is the probability of winning for a type bidder?

In general we need to find P(win|# of other players)xP(# of other players) for #=0,1,2,3,...N-1,

Let p(0) be the probability that there are no other type people. Since there are a total of N bidders and one of them is known to be there are N-1 others to be decided.  A given bidder is assigned to be either or with equal probability (1/2). So, the probability that all N-1 of the remaining bidders are is (1/2)^N-1. In this case, since bidding truthfully is dominant, she will win with certainty.

Let p(1) be the probability that there is exactly 1 other  player. The probability of exactly 1 other player is (1/2)(1/2)^N-1.  But there are N-1 ways to do this.

To find out p(j) we need to use the binomial theorem, which is . 

Putting the pieces together

P(win) =

To find the expected payoff just multiply this by - .

23.6 Repeat the question for a type player.

There is only one way a type player can win;  That is when every other player is also a type.  This probability is (1/2)^N-1.  The probability of a win when every player is of the same type is (1/N), and so the total probability is (1/N)(1/2)^N-1 .

Since this is a second price auction, everyone is of the same type, and players bid truthfully even a winner has zero net gain.  SO the expected winnings for a type player is zero.