Google adwords is a well known application of algorithmic game theory. It has a system wherein buyers of ad placements bid to put their ads up on various sellers websites, then pay the second bid below them.

If this system were turned on its head, and sellers of ad placements (websites, in other words) bid to be the cheapest space (relative to their quality) for buyers of ads and were paid the second cheapest price instead, would this result in the same prices being paid for ad placements?

If these two scenarios are not equivalent, where is the breakdown?

In general, is there any duality between these problems?


Generally speaking, auctions can go both ways in a dual manner: you can sell something via an auction among buyers or you can buy something via a procurement auction among possible providers (sellers). The asymmetry usually comes from where the competition lies: the auctioneer holds a "monopoly" while the bidders compete with each other.

In the example of ad auctions we usually view the advertisers as having unlimited demand for "impressions" and thus there is no effective competition between websites which each can be viewed as having a monopoly over the impressions it delivers. More realistic models may take into account some limits on the demand of advertisers for impressions (e.g max impressions or budget limits) in which case taking competing websites into account too would take us out of the usual auction model and into a two-sided market scenario.

Generally speaking there is no reason for the prices of a buyer-led auction to be equal to those from a seller-led one. Consider even the case of one buyer and one seller: certainly the side running the "auction" (which in this case is just setting a transaction price for the other to take or leave) can set it at the others' value in order to optimize for himself. A related example would be the stable-marriage problem where a "men-propose algorithm" produces a man-optimal matching, while a "women-propose algorithm" produce a women-optimal one.

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    $\begingroup$ Moreover, in the stable matching case, the man-optimal matching is woman-pessimal and vice versa. $\endgroup$ – mhum Mar 8 '11 at 6:48

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