Vickrey auction: What is it, and why does the winner pay the second-highest price?

# What Is a Vickrey Auction, and How Does It Work?

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Sept. 9 2013 4:25 PM

# What Is a Vickrey Auction, and How Does It Work?

This question originally appeared on Quora.

Answer by Alon Amit, Ph.D. in Mathematics:

(Note: A Vickrey auction is one in which the winner pays the second-highest price, not the price they themselves bid.)

I'll try to answer at a naive, intuitive level.

When you end up paying the price you bid ("first price"), you have a strong incentive to lie about how much you're willing to pay. Suppose there's an item for sale that you'd be happy paying up to \$1,000 for if necessary, but of course you'd rather pay less. In a first-price auction, if you bid \$1,000 and you lose. Well, someone else was willing to pay more than you were willing to, so that's OK, but if you win, you know that nobody else offered that price and you'd be slapping yourself for not going for \$950 and saving a little. Or, who knows, maybe there are really few buyers and you later discover that the second person was only valuing the item at \$600? Damn, you could have walked away with it for \$610! You feel cheated.

Therefore, in a first-price auction, you're not bidding your true value, you're trying to guess what everyone else is going to bid and then put down a bid that's slightly higher than the next person's (unless you think someone is going to offer more than what you're willing to pay, in which case you just bow out).

In a second-price auction, there's no reason for you to do that. You can simply say exactly the maximum price you are willing to pay, and there's never any advantage for you in saying anything else:

• You don't want to post a higher bid since you might be forced to pay it, and you don't want to do that.
• You don't want to post a lower bid since you might lose the item for no good reason at all.
• You'll end up paying exactly what it takes to win the item: one dollar or one cent more than the next person's maximum bid.