MAT109 · Erich Prisner · Franklin College · 2007-2011

# 2-Auction

Prerequisites: Chapters 1, 2, 3,

2-Auction (v1,v2,mA,mB) Two paintings, the first one worth \$v1, the second one worth \$v2, are auctioned, and there are only two bidders, Ann and Beth. The first painting is auctioned first, then the second one. Starting bid is always \$1000, and it is increased by \$1000 in each round. The two bidders alternate bidding. Initially Ann owns \$mA and Beth owns \$mB.

After one passes, the other one gets the painting, but then the one passed has the first offer for \$1000 for the new painting. There is one exception, however: If Ann passes in the first round for \$1000, then Beth can also bid for \$1000. If she passes too, then nobody gets the first painting, and Ann gets the first offer for \$1000 for the second painting.

Note that players can bid more than they have. As long as they don't have to pay it in the end, it doesn't matter. But if they have to pay more than they have at the end (when both paintings have been auctioned), then the player is bancrupt and gets a payoff of -100000. Note also that each player knows how much cash each player has.

Ann's and Beth's goal is to maximize the difference between the value in paintings and the money the paid for it. The player's goal is not to be better than the other player---they don't care about each other.

Of course, we don't have just one game here, but a whole family of games. Different worth of the first and second painting, and diferent starting money for Ann and Beth results in a different game.

Assumptions for what follows: For simplicity we will assume that each painting is worth one of the values \$2500, \$3500, \$4500, \$5500, \$6500, \$7500. We will also concentrate on the case mA = mB = 4000, where both players start with \$4000.

AppletAuction3.html, AppletAuction4.html, AppletAuction5.html, AppletAuction55.html, AppletAuction6.html, In this game applet you can and should simulate the game, but unfortunately you have to play both roles. The payoff is the sum of money left after the auction and value of the pictures obtained. The two players are either slightly friendly to each other, or slightly hostile. Note that in both cases the players don't care what the other gets as payoff, they just want to maximize their own payoff. But in the end, if a player sees that he or she will not get a certain painting, he or she may think "Ok, I see you will get the painting, I will not make it more difficult to you" in the slightly friendly version, or "Ok, you will get it, but I make you pay a high price for it" in the slightly hostile version. Theoretically there would also be a neutral version, but this neutral version is more difficult to analyze. In my opinion the slightly hostile version is more plausible in a game like this.

Since this is a sequential game without randomness and with perfect information, we can analyze it easily provided we have the extensive form. Unfortunately this extensive form is rather large. The figure to the right doesn't show the full game tree but just the "pruned" one for the auction of the first painting. The nine vertices that look like end vertices are not really end vertices, but rather the beginning of the auction of the second painting. In the five vertices labeled by "Ann", Ann has the first move for the second painting (bidding \$1000 or not), and owns presently still all her money, since she wasn't successfull obtaining the first painting. At these five vertices except the second one, Beth owns now the first painting, and has less money than before. At the four vertices below labeld by "Beth" the situation is just the other way---Ann owns the first painting but has less money left, and Beth has still all money and decides whether or not to bid \$1000 for the second painting. These nine branch subtrees that are cut at these nine vertices are almost as large as the "trunk", therefore the whole game tree has around 150 vertices. We implicitely assume, probably quite reasonably, since none of the paintings is worth more than \$8000, that Beth will not bid \$8000 for the first painting). Without such assumptions, the tree would be infinite and the game could not be analyzed.

Our strategy is to try to assign values, expected payoffs for Ann and Beth, at the nine cut vertices. If we can do this, we can analyze the whole game simply by performing backward induction on the game tree trunk. To obtain these values we need to look at auctions with only one painting.

## 1. One Painting

If there is only one painting, worth \$v2, and Ann has \$mA and Beth \$mB, and Ann starts, then it is obvious how to play: Everybody could bid as long as the bid is lower or equal to the painting worth v2 and also to the money, either mA, or mB, she has. So it is rather clear who gets the painting, and the other will stop bidding immediately (in the friendly version) or bid as long as the other would bid higher---even bidding over her own money---in the hostile version. This game is abbreviated by A(v2,mA,mB). Stretching our terminology a little, the same symbol will be used for the pair of payoffs that will result from this game. Of course the version where Beth starts the bidding, called B(v2,mA,mB), has a similar analysis.

We only need to investigate the following cases, since only they will occur in our initial two-rounds game:

• A(v2,4000,4000): Beth gets the last (reasonable) bid of 4000. That means that if v2 > 4000, then Beth will get the painting and the payoffs are 0 for Ann and v2-4000 for Beth (in the hostile version). For v2=3500 .... and for v2=2500 ....
• A(v2,4000,3000): For v2>3000, Ann gets the painting for 3000, thus Ann's and Beth's payoffs in that case are v2-3000 and 0. If v2=2500 then Beth gets the painting for 2000 and Ann's and Beth's payoffs are 0 and v2-2000.
• A(v2,4000,2000): The same analysis as in the previous case: If v2>3000, the payoffs are (v2-3000, 0). If v2=2500 the payoffs are (0, v2-2000).
• A(v2,4000,1000): Ann bids 1000 and gets the painting. The payoff for Ann equals v2-1000, and for Beth it is 0.
• A(v2,4000,0): The same as in the previous case: Ann bids 1000 and gets the painting. The payoff for Ann equals v2-1000, and for Beth it is 0.
• B(v2,4000,4000): The same analysis as in A(v2,4000,4000), with just the roles of Ann and Beth reversed. Thus if v2 > 4000, ......... . If v2=3500 ............... . If v2=2500 ...........
• B(v2,3000,4000): The same analysis as in A(v2,4000,3000), with the roles of Ann and Beth reversed. If v2>3000, the payoffs are (0,v2-3000). If v2=2500 the payoffs are (v2-2000,0).
• B(v2,2000,4000): The same analysis as in the previous case: If v2>3000, the payoffs are (0,v2-3000). If v2=2500 the payoffs are (v2-2000,0).
• B(v2,1000,4000): Beth bids 1000 and gets the painting. The payoff for Ann is 0, and that for Beth equals v2-1000.
• B(v2,0,4000): The same as in the previous case: Beth bids 1000 and gets the painting. The payoff for Ann is 0, and that for Beth equals v2-1000.

## 3. Formulating the game as a simultaneous game

If the two-painting auction starts, every player would have made up her mind of how high to bid for the first painting. Assume the players, since they are rational, stick to this pregame decision, which will later be called a pure strategy.

To be more precise, Ann has only the choice whether to stop before \$1000, or before \$3000, or before \$5000, or before \$7000, or not to stop and even bid \$7000 if necessary. Similar, Beth has the choice to stop before \$1000, before \$2000, before \$4000, before \$6000, or never.

 Ann/Beth stop before \$1000 \$2000 \$4000 \$6000 never \$1000 A(v2,mA,mB) A(v2,mA,mB-1000)+(0,v1-1000) A(v2,mA,mB-1000)+(0,v1-1000) A(v2,mA,mB-1000)+(0,v1-1000) A(v2,mA,mB-1000)+(0,v1-1000) \$3000 B(v2,mA-1000,mB)+(v1-1000,0) B(v2,mA-1000,mB)+(v1-1000,0) A(v2,mA,mB-2000)+(0,v1-2000) A(v2,mA,mB-2000)+(0,v1-2000) A(v2,mA,mB-2000)+(0,v1-2000) \$5000 B(v2,mA-1000,mB)+(v1-1000,0) B(v2,mA-1000,mB)+(v1-1000,0) B(v2,mA-3000,mB)+(v1-3000,0) A(v2,mA,mB-4000)+(0,v1-4000) A(v2,mA,mB-4000)+(0,v1-4000) \$7000 B(v2,mA-1000,mB)+(v1-1000,0) B(v2,mA-1000,mB)+(v1-1000,0) B(v2,mA-3000,mB)+(v1-3000,0) B(v2,mA-5000,mB)+(v1-5000,0) A(v2,mA,mB-6000)+(0,v1-6000) never B(v2,mA-1000,mB)+(v1-1000,0) B(v2,mA-1000,mB)+(v1-1000,0) B(v2,mA-3000,mB)+(v1-3000,0) B(v2,mA-5000,mB)+(v1-5000,0) B(v2,mA-7000,mB)+(v1-7000,0)