Question

Someone at a party pulls out a $100 bill and announces that he is going to auction it off. There are n = 10n=10 people at the party who are potential bidders. The owner of the $100 bill puts forth the following procedure: All bidders simultaneously submit a written bid. Only the highest bidders pay their bid (assuming that the highest bid is positive). If mm people submit the highest bid, then each receives 1/m1/m of the $100. Each person’s strategy set is \{0, 1, 2, . . . , 1000\}{0,1,2,...,1000} so bidding can go as high as $1,000

How many pure-strategy Nash equilibria does this game have?

Answer 1

If highest bid will be chosen as winning bid then there are m possible Nash equilibrium available.

If valuation of any participant is v_i 1000

Then revenue for " ith " participant is (v_i-b_i)>0

If any bid "b"<1000 then there is an opportunity to other bidder to choose bid=B=max(b,1000)

Hence by choosing bid B bidder can always make positive revenue.

Therefore it is always beneficial to choose highest bid =1000 because if you choose any bid less than 1000 then other bidder will choose marginally higher bid less than 1000.

Therefore we have 2 pure Nash equilibrium such as (m,m,m...10 times ) and m =1000

Hence each chooses 10000 as bid and this is the only Nash equilibrium we have in this game