Pay-to-bid Auctions
In this blog post I will talk about the results of a paper which analyzes pay-to-bid auctions. This paper, called “Pay-to-bid Auctions” by Brennan C. Platt, Joseph Price and Henry Tappen, was one of the first to analyze this new and exciting kind of auction (you can find it here http://www.nber.org/papers/w15695). I’ll start out by setting up the problem and then I’ll discuss their calculation of the mixed strategy Nash Equilibrium for this game.
A pay-to-bid auction is set up as follows. The price of the item up for sale starts at zero dollars. A time of T seconds is put on the clock and the clock begins counting down to zero. Players may choose to bid on the item and bidding on the item raises the price of the item by S dollars. The twist is that a player must pay B dollars to the auction house every time he places a bid. As soon as a bid is made, the clock is reset to T seconds and begins counting down again. The auction ends when the clock ticks down to zero. The person who made the last bid then wins the auction.
To begin their analysis, the authors assume that all players value the item at the same price, V. This is reasonable since the items sold in these pay-to-bid auctions are just everyday items one can buy at the store (plasma TV’s, video games, etc.). We see then that V is just equal to the item’s retail price. The attraction of the pay to bid auction is that the final price paid for an item is often much less than the retail price V. The catch is that one also has to pay just to bid. Suppose that q bids are made before the item is sold and that the winner of the auction made q_{w} bids. The payoff for the winner is then P_{w} = V – ( qS + q_{w}B ). We see that if the winner makes too many bids, his payoff will become negative even though he has won the item. Meanwhile, the revenue for the auction house is R = q( B + S ) – V (they can make quite a bit of money on these auctions even though the item may sell for much less than V).
To find the Nash Equilibrium we need to examine the expected payoffs for each player on the qth bid. Since all players value the item the same, their expected payoff for each bid is the same, so we need only look at one player. Firstly, a player can choose not to bid, in which case his expected payoff is P_{not bid}(q) = 0. If a player chooses to bid, then his expected payoff is P_{bid}(q) = ( V-qS )( 1 – m(q+1) ) – b, where m(q+1) is the probability that anyone will place the q+1th bid. Then the probability that no one makes the q+1th bid is just ( 1-m(q+1) ), which is the also the probability that the player who made the qth bid wins the item. Also note that for q > Q = ( V – B )/S, P_{bid}(q) < 0. This means that it is in no one’s interest to ever place a Q+1th bid. Therefore m(q) = 0 for all q > Q, P_{bid}(Q) = 0 and P_{bid}(q) < 0 for all q > Q.
We are now in a position to find the Nash Equilibrium for this game. Setting P_{not bid}(q) = P_{bid}(q), we find that there is a Nash Equilibrium at m(q) = 1 – B/( V – (q-1)S ), 1 < q <= Q and m(q) = 0, q > Q. The interesting thing to notice about m(q) is that it is a function of q. The probability that someone will place the qth bid approaches 1 ( for q <= Q ) as q gets very large (the second term is inversely proportional to q) and then suddenly drops to zero at q = Q + 1. This makes sense since one might try to maximize one’s payoff by bidding only once towards the end of the auction, near q = Q.
Once they have the function m(q), the authors then go on to calculate the expected revenue for the auction house (to do this you need to calculate the mean value of q using the probability distribution defined by m(q), which is not so easy, see their appendix). One could also use m(q) to calculate the expected payoff for the winner of the auction, but to do this you would also need to know how many bids the winning player made, q_{w}.
We can see from just this analysis that pay-to-bid auctions are an interesting thing to study. There is also much more to be studied about these auctions, as the authors of this paper found that their model did not work very well for video games and other “hot” items. They author’s also go on to analyze a model in which bidders arrive at the auction at random times (with a frequency given by a Poisson distribution) instead of all at once. To see one of these pay-to-bid auctions in action, I recommend checking out www.bidstick.com.