game theory; pure and mixed strategies

Handout 7

ECO 444

Konrad Grabiszewski

Second-price Seal-bid Auction

There is an auction with one object on sale. There are N bidders interested in buying that object.

In the second-price seal-bid auction, each bidder writes down his bid and places it in an envelope.

Then, the envelopes are opened and everybody learns all bids. This implies that we are dealing

with a static game (simultaneous moves). The highest bidder wins the auction and pays a price

equal to the second-highest bid (hence, the name “second”-price).

Let vi denote a player i’s valuation of the object. I.e., this is a player’s utility from the object if

s/he does not have to pay for the object.

1 Normal-form representation

Normal-form representation of the second-price seal-bid auction; G = (N; S1, ..., SN ; u1, ..., uN ):

• N is the set of players;

• Si = [0,∞) is the set of player i’s strategies; a strategy si is a bid;

• ui is the utility function of player i; let ri = maxj 6=i sj denote the highest bid among all bids

except for the bid of player i;

ui =

 

0 if i does not with the auction

vi − ri if i wins the auction (1)

2 Main theorem

Our goal is to prove the following theorem.

Theorem

For each player, bidding own’s evaluation of the object is a weakly dominant strategy.

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3 Proof

To make our proof a little bit easier, we assume that (a) there are three bidders (i.e., N =

{Ann, Bob, Chris}) and (b) Ann’s valuation of the object is 100. However, you should try to

conduct the proof for the case in which you do not make these additional assumptions.

In our analysis, we focus on Ann and show that s∗A = 100 weakly dominates other strategies of

Ann. That is, for every strategy sA 6= 100, we want to show that the following is true:

a) uA(100, sB, sC) ≥ uA(sA, sB, sC) for each (sB, sC), and

b) there is at least one (s̃B, s̃C) such that uA(100, s̃B, s̃C) > uA(sA, s̃B, s̃C).

Since every player’s set of strategies is infinite, we are unable to represent our auction in a matrix

form. Hence, checking rows for weak dominance will not work. Thus, we need a trick. This trick is

to use rA. Instead of comparing uA(100, sB, sC) to uA(sA, sB, sC) for each pair of strategies (sB, sC),

we will conduct this comparison for different values of rA. We can do that because comparison of

Ann’s utilities for different values of rA is equivalent to comparison of Ann’s utilities for different

values of (sB, sC).

Our proof consists of two parts. First, in Part I, we show that s∗A = 100 weakly dominates any

strategy sA strictly smaller than 100. Second, in Part II, we show that s ∗ A = 100 weakly dominates

any strategy sA strictly bigger than 100.

3.1 Part I: sA < 100.

We compare s∗A = 100 to sA < 100. Recall that rA = max{sB, sC}.

3.1.1 rA > 100.

If Ann chooses s∗A = 100, then she does not win the auction; hence, uA(100, sB, sC) = 0. If Ann

chooses any sA < 100, then she also loses; hence, uA(sA, sB, sC) = 0. This implies that if rA > 100,

then bidding 100 and bidding less than 100 yield the same utility of zero.

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3.1.2 rA = 100.

If Ann chooses s∗A = 100, then she either wins or loses the auction. However, if she wins, then her

utility is 100 − rA = 0; hence, uA(100, sB, sC) = 0. If Ann chooses any sA < 100, then she also

loses; hence, uA(sA, sB, sC) = 0. This implies that if rA = 100, then bidding 100 and bidding less

than 100 yield the same utility of zero.

3.1.3 100 > rA > sA.

If Ann chooses s∗A = 100, then she wins the auction; hence, uA(100, sB, sC) = 100 − rA > 0. If

Ann chooses any sA < 100, then she does not win the auction; hence, uA(sA, sB, sC) = 0. This

implies that if 100 > rA > sA, then bidding 100 is strictly better than bidding less than 100.

3.1.4 rA = sA.

If Ann chooses s∗A = 100, then she wins the auction; hence, uA(100, sB, sC) = 100 − rA > 0. If

Ann chooses any sA < 100, then she either wins or loses the auction. Let m be the number of

people whose bids are equal sA. Note that m is at least 2 because we have Ann bidding sA and

we know that either Bob or Chris (or both) also bade sA. Hence, Ann wins the auction with

probability 1 m

and loses the auction with probability 1 − 1 m

. This implies that her (expected)

utility is 1 m

(100−rA) + ( 1 − 1

m

) 0 = 100−rA

m < 100−rA. This implies that if rA = sA, then bidding

100 is strictly better than bidding less than 100.

3.1.5 rA < sA.

If Ann chooses s∗A = 100, then she wins the auction; hence, uA(100, sB, sC) = 100 − rA. If Ann

chooses any sA < 100, then she also wins; hence, uA(sA, sB, sC) = 100 − rA. This implies that if

rA < sA, then bidding 100 and bidding less than 100 yield the same utility of 100 − rA.

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3.1.6 Part I: summary.

We summarize our analysis and show that (a) uA(100, sB, sC) ≥ uA(sA, sB, sC) for each value of

rA, and (b) there are values of rA such that uA(100, sB, sC) > uA(sA, sB, sC).

value of rA uA(100, sB, sC) uA(sA, sB, sC)

rA > 100 0 0

rA = 100 0 0

100 > rA > sA 100 − rA > 0 0

rA = sA 100 − rA > 0 100−rAm < 100 − rA rA < sA 100 − rA > 0 100 − rA > 0

3.2 Part II: sA > 100.

We compare s∗A = 100 to sA > 100. Recall that rA = max{sB, sC}.

3.2.1 rA > sA.

If Ann chooses s∗A = 100, then she does not win the auction; hence, uA(100, sB, sC) = 0. If Ann

chooses any sA > 100, then she also loses; hence, uA(sA, sB, sC) = 0. This implies that if rA > sA,

then bidding 100 and bidding more than 100 yield the same utility of zero.

3.2.2 rA = sA.

If Ann chooses s∗A = 100, then she does not win the auction; hence, uA(100, sB, sC) = 0. If Ann

chooses any sA > 100, then she either wins or loses the auction. Let m be the number of people

whose bids are equal sA. This implies that her (expected) utility is 1 m

(100 − rA) + ( 1 − 1

m

) 0 =

100−rA m

< 0. This implies that if rA = sA, then bidding 100 is strictly better than bidding more

than 100.

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3.2.3 sA > rA > 100.

If Ann chooses s∗A = 100, then she does not win the auction; hence, uA(100, sB, sC) = 0. If Ann

chooses any sA > 100, then she wins the auction; hence, uA(sA, sB, sC) = 100 − rA < 0. This

implies that if sA > rA > 100, then bidding 100 is strictly better than bidding more than 100.

3.2.4 rA = 100.

If Ann chooses s∗A = 100, then she either wins or loses the auction. However, if she wins, then her

utility is 100 − rA = 0; hence, uA(100, sB, sC) = 0. If Ann chooses any sA > 100, then she wins

the auction; hence, uA(sA, sB, sC) = 100 − rA = 0. This implies that if rA = 100, then bidding

100 and bidding more than 100 yield the same utility of zero.

3.2.5 rA < 100.

If Ann chooses s∗A = 100, then she wins the auction; hence, uA(100, sB, sC) = 100 − rA. If Ann

chooses any sA > 100, then she also wins the auction; hence, uA(sA, sB, sC) = 100 − rA. This

implies that if rA = 100, then bidding 100 and bidding more than 100 yield the same utility of

100 − rA.

3.2.6 Part II: summary.

We summarize our analysis and show that (a) uA(100, sB, sC) ≥ uA(sA, sB, sC) for each value of

rA, and (b) there are values of rA such that uA(100, sB, sC) > uA(sA, sB, sC).

value of rA uA(100, sB, sC) uA(sA, sB, sC)

rA > sA 0 0

rA = sA 0 100−rA

m < 0

sA > rA > 100 0 100 − rA < 0

rA = 100 0 0

rA < 100 100 − rA 100 − rA

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4 Additional readings

If you are interested in learning more about auctions, then I recommend Vijay Krishna. Auction

Theory. Academic Press, 2nd edition, 2009.

5 Misc

The second-price seal-bid auction has been proposed by William Vickrey who was awarded the

Nobel Prize in Economics in 1996. Interesting and irrelevant fact: As you might know, the Nobel

prize is not awarded posthumously; you have to be alive on the day they announce the prize.

William Vickrey passed away three days after the announcement of his Nobel prize. According

to Wikipedia, “there are only three other cases where a Nobel Prize has been presented posthu-

mously.”

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