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Suppose a single unit of a good is to be sold at an auction. There are two bidders who may have different valuations of the object to be sold. The set of types T i = {vH, vL}, where i = 1, 2. The vL-type has a low valuation of the good whereas the vH-type has a high valuation. The probabilities of being a high or a low type are common knowledge. The seller wants to maximise expected revenue. Now suppose the probability that any player is of the vH-type is 0.7, and the probability that any player is of the vL-type is 0.3. In a second-price auction, calculate the probability that each type of player wins, their expected payment and their expected surplus.

The answer given is: In a second-price auction, truthful bidding is a dominant strategy for each type of player. The probabilities of winning for each type of player is: ρ(vH)=13/20 and ρ(vL)=3/20. How did they calculate ρ(vH)? Thanks in advance.

1 Answers 1

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Consider a vH type. She has a 0.7 chance of facing another vH type, in which case, since they both truthfully reveal their valuation, our vH person has a 0.5 chance of winning--that is, if they have a tie bid, they flip a coin or some such way to decide. So 0.7 time 0.5 is 0.35 chance. There is a 0.3 chance that the vH type faces an vL type, and then the vH type wins with certainty. So total chance of winning is 0.65, which is 13/20.

vL type: 0.7 chance of losing for sure, and 0.3 times 0.5 chance of winning (by facing another vL type) or 0.15, or 15/100, or 3/20.

Of course, this answer is formally incomplete because I did not prove truthful bidding is a dominant strategy.