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Alice and Bob would both like to own the same manuscript. The manuscript is worth 5 million to Alice and worth 3 million to Bob. The present owner of the manuscript proposes the following method of sale, known as a “second-price auction”. Alice and Bob will each simultaneously write down a “bid” for the manuscript. Let $b_A$ be Alice's bid, and let $b_B$ be Bob's bid. The manuscript will go to the person whose bid is highest, and that person will have to pay the other person's bid. If the bids are tied, then a fair coin will be tossed to decide who gets the manuscript, and that person will have to pay the tied bid.

Everything above is common knowledge to the players (in particular, they know each other's valuations). There are no other bidders. Negative bids are not allowed.

a) Write down each player’s payoff function in this game (payoff as a function of bids). Assume the payoff is in millions of USD.

b) What are Alice's best responses to $b_B = 2$? What are Alice's best responses to $b_B = 10$?

c) Are there any equilibria in which Bob wins the manuscript? Explain very briefly.


For a)

I regard payoff function as their expected payoff:

For Alice: $P(b_A \ge b_B)[5-b_B]$

For Bob: $P(b_B\ge b_A)[3-b_A]$

For b)

In either case, the best response for alice is to bid her true value. The reason for this is that when $b_B > 5$, this could make the probability of winning as small as possible; and when $b_B < 5$, this could make the probability of winning as large as possible.

Am I correct?

For c)

I do not know how to answer this one, but seems there is not any equilibrium for this case..

thanks for helping!

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  • $\begingroup$ thanks for correcting them! $\endgroup$
    – Steve
    Commented Dec 8, 2012 at 17:16

1 Answer 1

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I think, a) requires the payoff functions given actual bid values, not (yet) probabilities - or rathe with expected values only playing a role with repset to the tie-breakon coin toss. That is for Alice, the payoff is $$f_A(b_A,b_B)=\begin{cases}5-b_A&b_A> b_B\\ \frac{5-b_A}2&b_A= b_B\\ 0&b_A< b_B\end{cases}$$ and similarly $$f_B(b_A,b_B)=\begin{cases}0&b_A> b_B\\ \frac{3-b_B}2&b_A= b_B \\3-b_B&b_A<b_B\end{cases}$$

If $b_B=2$, then $b_A=2+\epsilon$ yields $f_A(b_A,b_B)=3-\epsilon$ for Alice (but $b_B=0$ yields only $\frac32$).

If $b_B=10$, then any response $b_A\le 5$ yields $f(b_A,b_B)=0$.

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  • $\begingroup$ should it be 5-bB (5-bB)/2 and 3-bA (3-bA)/2 $\endgroup$
    – Steve
    Commented Dec 8, 2012 at 11:05
  • $\begingroup$ since the winner only needs to pay the second highest bid for the prize $\endgroup$
    – Steve
    Commented Dec 8, 2012 at 11:05
  • $\begingroup$ So do you mean that the best response for alice when bB = 2 is to bid an amount greater than 2 and less than 5, since any bid between this amount would give alice a payoff of 3? And when bB = 10, the best response for alice is to bid between 0 and 5, because any bid within this amount would not bring her any payoff right? $\endgroup$
    – Steve
    Commented Dec 8, 2012 at 11:08
  • $\begingroup$ Hagen von eitzen: Could you explain it more detail about the function for bA > bB? i do not understand why it is alice value minus 'her' bid but not bob's bid since the winner only needs to pay the second highest bid? $\endgroup$
    – Steve
    Commented Dec 8, 2012 at 17:15
  • $\begingroup$ i do not think this is the answer, but still thanks Hagen! $\endgroup$
    – Steve
    Commented Dec 8, 2012 at 17:21

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