MATH SOLVE

4 months ago

Q:
# Show that a sealed-bid third-price auction for three or more bidders is not strategy proof: A) In such an auction, the bidder with the highest bid wins, B) But pays a price equal to the third-highest bid.

Accepted Solution

A:

Explanation:Lets suppose that you are willing to pay m for an object. Note that you may bet more than m, because you arent actually going to pay for that amount.Imagine that you are on this situation: current bet is k1, previous bet was k2, and k2 ≤ m. If you make a bet and win, then you will pay less than or equal to m, so you are good. You have two options: you need to make sure that you bet for a number big enough no one is willing to pay you bet for something that you will to payIf you bet for a number k3, with k3 bigger than m and k2, and then someone else bets for k4, with k4 > k3, then you cant keep betting, because if you win you will have to pay at least k3, which is higher than the number you are willing to pay. This means that if you bet for an amount bigger than m, then you are better betting on a gigant number M, billions of times bigger than what anyone would pay. So that you scare everyone away. This strategy seems to work, right? However, there is a minor problem.If another person bets 2M and wins, then he has to pay only k2. And anyone trying to bet higher than those 2M has to be willing to pay M, which shoudnt be happening. This means that the strategy could fail if anyone is willing to pay k2. Even though this strategy failed, this gives us another strategy: Keep betting below the amount m you are willing to pay. If you win, you will have to pay at most m, so you are ok. Immediately after the current bet k is higher than (or practically iqual to) m, go and bet for MIf no one is willing to pay more than the number k, then you win. If someone bets, then you lose, but that person was willing to pay more than you.If a person doesnt follow this strategy, then there could be 2 possibilities:The person bets for a higher number than his 'm' (M) too fast, and he can lose against someone using this strategy even by having a lower m valueThe person doesnt bet when the current bet k beats his m value. Someone else bets instead of him for a value j, with j > k, and he cant keep betting. This person could lose against someone using this strategy, even when his m value is lower (he would bet a big number M after the current k value)So, this strategy is optimal, however, there is a problem. What happen if 2 (or more) persons are using it?If the current value k of the bet beats the m value of 2 persons using this strategy at the same time, both of them will rush to bet M. The fastest one is the winner.If three persons are using this strategy with m values m1 < m2 < m3, then each player will be carefull not to bet higher than his respective m value. Whenever the current bet beats the number m1, then that person will bet M and afterwards, the other 2 persons will rush to bet an even higher amount, such as 2M. Thus, the one who bets first is the winner, and he or she will have to pay a number slightly bigger than m1, despite having another player willing to pay way more than m1.This proves that such kind of auction is not strategy proof.I hope this answer helped you!