3. Sequential LEGISLATORS' VOTEWhat happens if all three have to vote at the same time? Remember that we describe the game not by one bimatrix but a system of trimatrices (matrices with three entries in each cell), one for each one of player A's moves. Note that all payoffs are given in hundreds in this chapter.
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In the worst case, a player is better off voting against a rise, since she can lose $1000 in the "voting for" case, but cannot lose in the "voting against" case. Therefore the maximin moves are voting against a rise.
None of the moves dominates another, strongly or weakly.
Note that the "best response" concept doesn't apply to three or more player games.
Let us now describe how to find all pure Nash equilibria. Thereby we give a procedure that can be used for general simultaneous games of more than two players. First we look at player A. Among all outcomes with fixed moves of all other players, we mark those where A's payoff is maximal. We underline this payoff. For instance, there are two outcomes where B votes for a raise and C votes for a raise, and the one most favorable for A (the "best response" to that situation) is where A votes against a raise. If B votes for a raise and C against, then A should vote for a rise. The same in the symmetric case where B votes against and C for a raise. If both B and C vote against a rise, then A should also vote against.
The same analysis could be done for B and C. When we look at B's "best responses", we underline the second entry of those cells in each of the four columns maximizing this second entry. When doing it for C, we underline the third entry of those cells in each of the four rows maximizing this third entry.
Finally, the pure Nash equilibira are those cells where all three entries are underlined. Our game has three pure Nash equilibria, and the corresponding cells are shaded. One is when all three vote against a raise, and the other three are when two of them vote for a raise and one votes against.
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Again the maximin moves are voting against a rise. Again there is no domination.
The pure Nash equilibria are found using the same method as described above. There are seven of them, namely when two of them vote for a raise and one opposes, or when one of them votes for the raise and two abstain, or when all three vote against it. The corresponding cells are shaded again. Two cells miss the Nash criterion of all three entries being underlined barely---the three cases where two players are opposed to a raise and one abstains. For the two players opposed this is a best response to the situation, but for the third player it is not, this player should deviate and also say no.
Let me also note that the "Gambit" program (see the main page here) [McKelvey, McLennan, Turocy 2007] finds 20 Nash equilibria in mixed strategies. We will talk about them later.
3. Sequential LEGISLATORS' VOTEWhat happens if A has to vote first, then B, then C. (This is a variant of a game described in [Kaminski].) The backwards induction solution is: A votes against raise, getting a payoff of 20, whereas B and C both vote for a raise and get 10 each. The first player has an advantage. He or she is allowed to keep his or her face, whereas the other two have to get the raise passed. If the order in which the three legislators have to declare their opinion in public is random, each one can expect the average of the three expectations in the game, namely a payoff of 13.3 hundreds, more precisely. of $1333.33.
The evening before the three legislators have to vote about the salary rise, sequentially and with an order chosen randomly and not yet known, two legislators are sitting in the hotel bar. Then suddenly a journalist calls, telling them that at the election district of the third legislator, Jim, large crowds are demonstrating against a salary rise for the legislators. That means that Jim's loss of face if voting for a raise would have to be valued even more than $ 1000 worth. How would you react to these news? Get another beer and celebrate Jim's bad luck? And is it bad luck for Jim? May it even affect you?
The answer to this question is the same as for almost all questions: "It depends." Namely, it depends how much Jim's loss of face is now worth. If he values it less than $ 2000, then he carries the whole cost of the changed situation. If he, however, now values it more than $ 2000, then he will never vote for a raise. If the other two know this, they both have to vote for a raise, even when voting first. That means that in that situation the expected payoff of the other two legislators sank from $1333 to $1000, and Jim's expected payoff raised from $ 1333 to $2000. He is getting the raise for free, provided the others just know about his changed payoff.
| ... considers his loss of face worth -1500$ | ... considers his loss of face worth -2500$ | |
| The first (red) player ... |
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| The second (blue) player ... |
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| The third (green) player ... |
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What happens if this other player is more scrupulous than the other two, but only slightly more? Let's assume his face is worth $1500. Then the first game tree, where Mister Proper has to move first, doesn't change. If Mister Proper moves second, the payoffs are now 20, 5, 10. If Mister Proper moves last, the payoffs are 20, 10, 5. Nothing changes for the other two, but our scrupulous guy is worse off than before, due to the damage with his face,
The conclusion is that being scrupulous hurts first the guy itself, but if it exceeds a certain value, then only the other two legislators have to pay.
Again, abstaining would carry a loss of face worth $200.