MAT109 · Erich Prisner · Franklin College · 2007-2011

Election I

Note to the Teacher: ...

Prerequisites: Chapters 1, 2, and the Excel Introduction.

The president of the USA is elected by electors from all states. All votes from each state go to the most popular candidate in that state. If one week before the election a candidate knows that she is behind in a few states, and leading in others, what would be a good strategy for the remaining time? Concentrating on those where she is behind, or accept that they are lost and concentrate on others? Of course this decision will depend on several factors, including whether the state can still be turned, and also on the size of the state. California is more important than Montana, at least as far as votes are concerned. In this chapter we look at a simplified model with only three districts/states and analyze a few cases formally. The hope is that looking at simpler cases will allow us to extract some rules for general larger situations.

1. First Example

We start with a special game, out of a larger family of games that will be explained later.

ELECTION 1 or ELECTION(7,8,13|-1,-1,1|3,3): There are three districts, C, D, and E, and similar to the election of the president of the USA, the president of this area is elected by electoral votes. There are 7 electoral votes from district C, 8 electoral votes from district D, and 13 electoral votes from district E. Districts do not split electoral votes. Now there are two presidential candidates, Ann and Beth, and in the last phase of their campaigns they simultaneously decide how to allocate the 3 remaining resources each has. Each one of these resources goes to one district. Each district will vote for the candidate having put overall (not just during the last phase) more resources into the district, and will abstain in case of a tie. In districts C and D, Ann is presently 1 resource unit behind, and in district E, Ann has an advantage of 1 resource unit. How would Ann and Beth distribute the resources?

There are only three payoffs, win, loss, draw, and either one loses and the other wins, or both draw, so we should model this game as a zero-sum game with payoffs 1 (win) 0 (draw) and -1 (lose). Obviously we have a 2-player simultaneous game with many possible moves. Let us list all moves first:

A move means allocating x resources to district C, y resources to district to district D, and z resources to district E, obeying the resource restriction x+y+z=3. This is abbreviated as (x,y,z). Furthermore x, y, z must be nonnegative integers. Thus Ann and Beth have the 10 moves (0,0,3), (1,0,2), (0,1,2), (2,0,1), (1,1,1), (0,2,1), (3,0,0), (2,1,0), (1,2,0), and (0,3,0).

Calculating the outcome and therefore the payoffs for a given pair of moves is rather easy. Let's demonstrate this at the example where Ann plays (0,1,2) and Beth plays (1,0,2). Beth puts one more effort into district C, and had already the lead there before, so Beth wins district C. Ann puts one more resource into district D, in the last phase, but since Beth was leading there before by 1, we get a tie in district D. And for district E, both put the same effort in the last phase, but since Ann had an advantage of 1 before the last phase, the game phase, Ann wins district E. Therefore Ann gets 13 votes (from district E), and Beth gets 7 votes (from district C). District D is undecided and doesn't vote for either candidate. Thus Ann is elected president for that particular pair of moves, and gets a payoff of 1, while Beth's payoff is -1.

Class Activity: Before proceeding with the mathematical analysis, please spend five minutes to think how Beth should play.

Ann's payoffs for the 10 · 10 pairs of moves are given in the matrix below:

0,0,31,0,20,1,2 2,0,11,1,10,2,1 3,0,02,1,01,2,00,3,0
0,0,3 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1
1,0,2-1 -1 1 -1 -1 1 -1 -1 -1 1
0,1,2-1 1 -1 1 -1 -1 1 -1 -1 -1
2,0,1-1 -1 -1 -1 1 1 -1 -1 1 1
1,1,1-1 -1 -1 1 -1 1 1 -1 -1 1
0,2,1-1 1 -1 1 1 -1 1 1 -1 -1
3,0,0-1 -1 -1 -1 -1 -1 -1 1 1 1
2,1,0-1 -1 -1 -1 -1 -1 1 -1 1 1
1,2,0-1 -1 -1 1 -1 -1 1 1 -1 1
0,3,0-1 -1 -1 1 1 -1 1 1 1 -1

If Beth plays (0,0,3) she wins, no matter what Ann does. The explanation is simple. If Ann invests also all 3 resources into district E, then Ann keeps E, but Beth keeps C and D where she was leading before, and since 7+8 > 13, Beth wins overall. If Ann puts two resources into district E, then there is a draw in E and also in the other district where Ann invests. But Beth keeps the third district and wins overall. Finally, if Ann invests less than 2 resources into district E, then Beth wins district E, but Ann cannot win (turn around) both districts C and D with the 3 resources she has, therefore Beth wins then as well.

2. Second Example

ELECTION 2 or ELECTION(7,8,13|-1,-1,1|4,4): We modify ELECTION 1 by allowing 4 resources for each player

We have even more moves here. Ann and Beth have the 1+2+3+4+5=15 moves (0,0,4), (1,0,3), (0,1,3), (2,0,2), (1,1,2), (0,2,2), (3,0,1), (2,1,1), (1,2,1), (0,3,1), (4,0,0), (3,1,0), (2,2,0), (1,3,0), and (0,4,0). I listed first the one move with z=4, then the two moves with z=3, and so on. Note also that Beth can no longer always win by puting all resources into district E, since Ann could respond by putting 2 resources into district C and D each, thereby winning both.

The 15 × 15 matrix is best generated by a computer. We can do it in this Excel sheet. The payoff matrix looks as follows:

0,0,41,0,30,1,3 2,0,21,1,20,2,2 3,0,12,1,11,2,10,3,1 4,0,03,1,02,2,0 1,3,00,4,0
0,0,4-1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1
1,0,3-1 -1 1 -1 -1 1 -1 -1 -1 1 -1 -1 -1 -1 1
0,1,3-1 1 -1 1 -1 -1 1 -1 -1 -1 1 -1 -1 -1 -1
2,0,2-1 -1 -1 -1 1 1 -1 -1 1 1 -1 -1 -1 1 1
1,1,2-1 -1 -1 1 -1 1 1 -1 -1 1 1 -1 -1 -1 1
0,2,2-1 1 -1 1 1 -1 1 1 -1 -1 1 1 -1 -1 -1
3,0,1-1 -1 -1 -1 -1 -1 -1 1 1 1 -1 -1 1 1 1
2,1,1-1 -1 -1 -1 -1 -1 1 -1 1 1 1 -1 -1 1 1
1,2,1-1 -1 -1 1 -1 -1 1 1 -1 1 1 1 -1 -1 1
0,3,1-1 -1 -1 1 1 -1 1 1 1 -1 1 1 1 -1 -1
4,0,0-1 -1 -1 -1 -1 -1 -1 -1 -1 -1 -1 1 1 1 1
3,1,0-1 -1 -1 -1 -1 -1 -1 -1 -1 -1 1 -1 1 1 1
2,2,01 -1 -1 -1 -1 -1 1 -1 -1 -1 1 1 -1 1 1
1,3,0-1 -1 -1 -1 -1 -1 1 1 -1 -1 1 1 1 -1 1
0,4,0-1 -1 -1 -1 -1 -1 1 1 1 -1 1 1 1 1 -1

This matrix does not have any pure Nash equilibrium. Ann's moves (0,0,4) and (0,1,3) are weakly dominated. For Beth, the list of weakly dominated moves is longer, namely (2,0,2), (0,2,2), (3,0,1), (0,3,1), (4,0,0), (3,1,0), (1,3,0), and (0,4,0). After eliminating all these moves, we get the following matrix:

0,0,41,0,30,1,3 1,1,22,1,11,2,1 2,2,0
1,0,3-1 -1 1 -1 -1 -1 -1
2,0,2-1 -1 -1 1 -1 1 -1
1,1,2-1 -1 -1 -1 -1 -1 -1
0,2,2-1 1 -1 1 1 -1 -1
3,0,1-1 -1 -1 -1 1 1 1
2,1,1-1 -1 -1 -1 -1 1 -1
1,2,1-1 -1 -1 -1 1 -1 -1
0,3,1-1 -1 -1 1 1 1 1
4,0,0-1 -1 -1 -1 -1 -1 1
3,1,0-1 -1 -1 -1 -1 -1 1
2,2,01 -1 -1 -1 -1 -1 -1
1,3,0-1 -1 -1 -1 1 -1 1
0,4,0-1 -1 -1 -1 1 1 1

Now Ann's moves (2,0,2), (1,1,2), (3,0,1), (2,1,1), (1,2,1), (4,0,0), (3,1,0), (1,3,0), and (0,4,0) and Beth's moves (1,1,2) and (2,1,1) are weakly dominated. We eliminate them and get:

0,0,41,0,30,1,3 1,2,12,2,0
1,0,3-1 -1 1 -1 -1
0,2,2-1 1 -1 -1 -1
0,3,1-1 -1 -1 1 1
2,2,01 -1 -1 -1 -1

Since in this reduced submatrix, Beth's moves (1,2,1) and (2,2,0) have identical payoffs against all of Ann's moves, it suffices to take one of them, let's say (1,2,1). We get the following matrix, which now doesn't have any weakly dominated moves. It is the result of the IEWD procedure.

0,0,41,0,30,1,3 1,2,1
1,0,3-1 -1 1 -1
0,2,2-1 1 -1 -1
0,3,1-1 -1 -1 1
2,2,01 -1 -1 -1

What would Ann and Beth choose? Is any of Ann's four options better or even different to the others? The resulting 4 × 4 matrix is completely symmetric in the sense that each one of Ann's remaining options beats exactly one of Beth's remaining options, and also each one of Beth's moves beats exactly three of Ann's remaining options. This may imply that Ann and Beth each chooses any of their fours options. You may also agree that Beth has an advantage in this game ELECTION(7,8,13|-1,-1,1|4,4). Therefore, having an advantage of 1 in the small districts C and D is better than an advantage of 1 in the large district E. We finish this example in the second part, where we have the tool of mixed strategies available.

0,0,41,0,30,1,3 0,2,21,2,12,2,0
1,0,3-1 -1 1 1 -1 -1
0,2,2-1 1 -1 -1 -1 -1
0,3,1-1 -1 -1 -1 1 1
2,2,01 -1 -1 -1 -1 -1
is another 25% versus 25%,25%,12.5%,12.5%,12.5%,12.5% mixed Nash equilibrium.

3. The General Model

The general formulation of these games may be as follows:

ELECTION(c,d,e|c1,d1,e1|a,b): There are three districts, C, D, and E, and the president is elected by electoral votes. There are c electoral votes from district C, d electoral votes from district D, and e electoral votes from district E. Districts do not split electoral votes, and they vote for the candidate having put the most resources into the district, and abstain in case of a tie. The two presidential candidates, Ann and Beth, simultaneously decide how to allocate Ann's remaining a resources and Beth's remaining b resources over the districts. Presently Ann leads in diestricts C, D, and E by c1, d1, and e1 resources. How would Ann and Beth distribute their resources?

4. Third Example

ELECTION(7,9,13|-1,-1,1|4,4) is like ELECTION 2, except that district D has 9 votes instead of 8.

Of course, we can just do the whole procedure once again, but what will change? Will there change anything?

To answer this question, let me reformulate who wins in the games ELECTION(c,d,e|c1,d1,e1|a,b):

  1. If all three districts are stuck in a tie, then the game ends in a tie.
  2. If two districts show a tie, the third districts decides the president.
  3. If one district ties, then the player winning the larger district becomes president. If both these districts have equal size and are won by different players, then there is an overall tie. But both in ELECTION(7,8,13|c1,d1,e1|a,b) as in ELECTION(7,9,13|c1,d1,e1|a,b) no districts have the same size.
  4. If no district is stuck in a tie, we have to add the numbers won and compare. Since 7+8>13, and 7+9>13, in ELECTION(7,8,13|c1,d1,e1|a,b) and in ELECTION(7,9,13|c1,d1,e1|a,b) the player winning (at least) two districts wins the presidency.
Now obviously the sizes of the districts only matter in the third and fourth case, and since ELECTION(7,9,13|-1,-1,1|4,4) behaves just like ELECTION(7,8,13|-1,-1,1|4,4), both are played exactly the same way. Although the descriptions are different, one could say that the games are the same---we have the same moves and the same payoffs. Mathematicians would call such games isomorphic.

5. The Eight Cases

Let us look again at the four cases above. In case 3, it is only important whether some of c, d, or e have the same size. And the exact description of case 4 above is as follows:

  1. Assume c ≤ d ≤ e, and no district is stuck in a tie. If c+d > e, then the player winning (at least) two districts wins the presidency. If c+d = e, then the player winning (at least) two districts wins the presidency, except in the case where one wins C and D and the other district E, in which case we get an overall tie. Finally, if c+d < e, then the player winning district E wins the presidency.

Combining these properties on repeated sizes and c+d versus e, we get eight cases to consider.

All examples discussed above belong to the group "7-8-13".

6. Voting Power Indices (optional)

Both players will treat district D the same, no matter whether it has 8 or 9 (or 10 or 11 or 12) votes, provided district C has 7 votes and district E has 13. Thus the number of votes is not proportional to the importance of the district. There is some connection to so-called "voting power indices", but not an immediate one. These indices apply to situations where voters have different numbers of votes. However, the power of a voter is usually not proportional to the number of votes. Take the obvious example of three voters with 4, 8, and 13 votes, where the two voters with the small numbers of votes have no power, and the other one has all. But two very important assumptions not valid for our model are that no voter abstains--- voters can only vote for or against a proposal, or if they can abstain, these abstentions count the same as votes against--- and that no tie is possible---the proposal is accepted if it gets more than half of the votes, and otherwise rejected.

Now the Banzhaf-Penrose Index is calculated as follows: A set of voters is called a winning coalition if together they have enough votes to pass a proposal. In the example where A, B, C, and D have 4,3,2,and 1 vote, and where 6 votes are needed for accepting a proposal, winning coalitions are {A,B}, {A,C}, {B,C,D}, and every set containing any of these three, namely {A,B,C}, {A,B,D}, {A,C,D}, and {A,B,C,D}. A swing voter is any member in a winning coalition whose removal makes the group non-winning. We list all seven winning coalitions again, this time underlining all swing voters in them. We get {A,B}, {A,C}, {B,C,D}, {A,B,C}, {A,B,D}, {A,C,D}, and {A,B,C,D}. Now the A occurs 5 times among these 12 underlined swing voters, B and C both occur 3 times, and D occurs only once. Therefore the Banzhaf-Penrose Indices of A, B, C, and D are 5/12, 3/12, 3/12, and 1/12.

In the Shubik-Shapley Index we look at all ordered lists, as A,B,C,D or C,D,A,B where each voter occurs exactly once. For each such list we include the voters in the list until we get a winning coalition. The last voter added, changing a still losing coalition into a winning one, is called "pivotal" for that sequence. Let us list all 24 such sequences, underlining the pivotal element in each case. We get
A,B,C,D; A,B,D,C; A,C,B,D; A,C,D,B; A,D,B,C; A,D,C,B;
B,A,C,D; B,A,D,C; B,C,A,D; B,C,D,A; B,D,A,C; B,D,C,A;
C,A,B,D; C,A,D,B; C,B,A,D; C,B,D,A; C,D,A,B; C,D,B,A;
D,A,B,C; D,A,C,B; D,B,A,C; D,B,C,A; D,C,A,B; D,C,B,A.
For each voter, the number of sequences where that voter is pivotal divided by the number of all such sequences equals its Shubik-Shapley Index. So we get 10/24, 6/24, 6/24, and 2/24 for A, B, C, respectively D.

The six orderings of C, D, and E are CDE, CED, DCE, DEC, ECD, and EDC. The pivotal district---pushing the coalition from losing to winning, are given for each of the six cases:

C,D,EC,E,DD,C,E D,E,CE,C,DE,D,C
13-13-13DECECD
8-8-15DECECD
8-8-16D,EEC,EEE,CE,D
8-8-17EEEEEE
8-13-13DECECD
7-8-13DECECD
5-8-13D,EEC,EEE,CE,D
4-8-13EEEEEE
CDE
13-13-132/62/62/6
8-8-152/62/62/6
8-8-161/61/64/6
8-8-17001
8-13-132/62/62/6
7-8-132/62/62/6
5-8-131/61/64/6
4-8-13001

2. The Version where candidates have to move every day

The case 7-8-13

The payoff matrix for this zero-sum game looks as follows:
7-8-13 c,d,e c,d,c c,e,c d,c,d d,e,d e,c,e e,d,e
c,d,e 0 1 1 1 -1 -1 -1
c,d,c -1 0 -1 -1 -1 1 -1
c,e,c -1 1 0 1 -1 -1 -1
d,c,d -1 1 -1 0 -1 -1 1
d,e,d 1 1 1 1 0 -1 -1
e,c,e 1 -1 1 1 1 0 -1
e,d,e 1 1 1 -1 1 1 0

All the moves are maxmin moves. Moves cde, cdc, and cec are weakly dominated. There are no Nash Equilibria in pure strategies, but there must be some in mixed strategies. These look as follows: ..................

Mixed strategy: Choose 1/3 of each of dcd, ece, and ede.

The case 5-8-13

The payoff matrix looks as follows: It differs to the 7-8-13 matrix only in four places.
5-8-13 c,d,e c,d,c c,e,c d,c,d d,e,d e,c,e e,d,e
c,d,e 0 1 1 1 -1 -1 -1
c,d,c -1 0 -1 -1 -1 0 -1
c,e,c -1 1 0 1 -1 -1 -1
d,c,d -1 1 -1 0 -1 -1 0
d,e,d 1 1 1 1 0 -1 -1
e,c,e 1 0 1 1 1 0 -1
e,d,e 1 1 1 0 1 1 0

Analysis: Maximin is ede. The moves cde, cdc, cec, and dcd are weakly dominated. ede versus ede is the pure Nash equilibrium.

The case 4-8-13

The payoff matrix looks as follows:
4-8-13 c,d,e c,d,c c,e,c d,c,d d,e,d e,c,e e,d,e
c,d,e 0 1 1 1 -1 -1 -1
c,d,c -1 0 -1 -1 -1 -1 -1
c,e,c -1 1 0 1 -1 -1 -1
d,c,d -1 1 -1 0 -1 -1 -1
d,e,d 1 1 1 1 0 -1 -1
e,c,e 1 1 1 1 1 0 -1
e,d,e 1 1 1 1 1 1 0

Analysis: Every move except ede is weakly dominated. There is one pure Nash equilibrium, ede versus ede.

The case 8-13-13

The payoff matrix looks as follows:
8-8-13 c,d,e c,d,c c,e,c d,c,d d,e,d e,c,e e,d,e
c,d,e 0 1 1 0 -1 0 -1
c,d,c -1 0 0 -1 -1 1 -1
c,e,c -1 0 0 1 -1 -1 -1
d,c,d 0 1 -1 0 -1 0 1
d,e,d 1 1 1 1 0 -1 0
e,c,e 0 -1 1 0 1 0 -1
e,d,e 1 1 1 -1 0 1 0

Analysis: cdc and cec are weakly dominated. There are no pure Nash equilibria. The mix seems to be a little unclear.

The payoff matrix looks as follows:
8-8-13 all three s-l-s l1-s-l1 l1-l2-l1
all three 0 1 0 -1
s-l-s -1 0 0 -1
l1-s-l1 0 0 0 0
l1-l2-l1 1 1 0 0

Many mixed Nash equilibria, I guess.

The case 8-8-15

The payoff matrix looks as follows:
8-8-15 c,d,e c,d,c c,e,c d,c,d d,e,d e,c,e e,d,e
c,d,e 0 1 0 1 0 -1 -1
c,d,c -1 0 -1 0 -1 1 -1
c,e,c 0 1 0 1 0 -1 -1
d,c,d -1 0 -1 0 -1 -1 1
d,e,d 0 1 0 1 0 -1 -1
e,c,e 1 -1 1 1 1 0 0
e,d,e 1 1 1 -1 1 0 0

Analysis: cde, cec, and ded are weakly dominated.

8-8-15 all three s-s-s s1-l-s1 l-s-l
all three 0 1 0 -1
s-s-s -1 0 -1 0
s1-l-s1 0 1 0 -1
l-s-l 1 0 1 0

Pure Nash equilibrium: l-s-l versus l-s-l (large-small-large).

The case 8-8-16

The payoff matrix looks as follows:
6-8-16 c,d,e c,d,c c,e,c d,c,d d,e,d e,c,e e,d,e
c,d,e 0 1 0 1 0 -1 -1
c,d,c -1 0 -1 0 -1 0 -1
c,e,c 0 1 0 1 0 -1 -1
d,c,d -1 0 -1 0 -1 -1 0
d,e,d 0 1 0 1 0 -1 -1
e,c,e 1 0 1 1 1 0 0
e,d,e 1 1 1 0 1 0 0

Analysis: Nash equilibrium: Any of ece or ede versus any of ece or ede.

The case 8-8-17

The payoff matrix looks as follows:
8-8-17 c,d,e c,d,c c,e,c d,c,d d,e,d e,c,e e,d,e
c,d,e 0 1 0 1 0 -1 -1
c,d,c -1 0 -1 0 -1 -1 -1
c,e,c 0 1 0 1 0 -1 -1
d,c,d -1 0 -1 0 -1 -1 -1
d,e,d 0 1 0 1 0 -1 -1
e,c,e 1 1 1 1 1 0 0
e,d,e 1 1 1 1 1 0 0

Analysis: Nash equilibrium: Any of ece or ede versus any of ece or ede.

The case 13-13-13

The payoff matrix looks as follows:
13-13-13 c,d,e c,d,c c,e,c d,c,d d,e,d e,c,e e,d,e
c,d,e 0 0 0 0 0 0 0
c,d,c 0 0 0 0 -1 1 0
c,e,c 0 0 0 1 0 0 -1
d,c,d 0 0 -1 0 0 0 1
d,e,d 0 1 0 0 0 -1 0
e,c,e 0 -1 0 0 1 0 0
e,d,e 0 0 1 -1 0 0 0

Analysis: ..................

3. The General Version

Excel sheet

The case 7-8-13

Here is the normal form of the game for the different pure strategies. It is a zero-sum game, thus only the payoff for player 1 is shown.
(0,0,3) (0,1,2) (0,2,1) (0,3,0) (1,0,2) (1,1,1) (1,2,0) (2,0,1) (2,1,0) (3,0,0)
(0,0,3) 0 1 1 1 1 -1 -1 1 -1 1
(0,1,2) -1 0 1 1 1 1 -1 1 1 1
(0,2,1) -1 -1 0 1 -1 1 1 1 1 1
(0,3,0) -1 -1 -1 0 -1 -1 1 -1 1 1
(1,0,2) -1 -1 1 1 0 1 1 1 -1 1
(1,1,1) 1 -1 -1 1 -1 0 1 1 1 1
(1,2,0) 1 1 -1 -1 -1 -1 0 -1 1 1
(2,0,1) -1 -1 -1 1 -1 -1 1 0 1 1
(2,1,0) 1 -1 -1 -1 1 -1 -1 -1 0 1
(3,0,0) -1 -1 -1 -1 -1 -1 -1 -1 -1 0

Moves 4, 8, and 10 are dominated, according to that Gambit file. Therefore we arrive at the following reduced game:
(0,0,3) (0,1,2) (0,2,1) (1,0,2) (1,1,1) (1,2,0) (2,1,0)
(0,0,3) 0 1 1 1 -1 -1 -1
(0,1,2) -1 0 1 1 1 -1 1
(0,2,1) -1 -1 0 -1 1 1 1
(1,0,2) -1 -1 1 0 1 1 -1
(1,1,1) 1 -1 -1 -1 0 1 1
(1,2,0) 1 1 -1 -1 -1 0 1
(2,1,0) 1 -1 -1 1 -1 -1 0
The optimal mixed strategy is to use the three moves (0,0,3), (0,1,2), (1,0,2), (1,1,1), (1,2,0) with equal probability 1/5.

The case 5-8-13

Here is the normal form of the game for the different pure strategies. It is a zero-sum game, thus only the payoff for player 1 is shown.
(0,0,3) (0,1,2) (0,2,1) (0,3,0) (1,0,2) (1,1,1) (1,2,0) (2,0,1) (2,1,0) (3,0,0)
(0,0,3) 0 1 1 1 1 0 0 1 0 1
(0,1,2) -1 0 1 1 1 1 0 1 1 1
(0,2,1) -1 -1 0 1 -1 1 1 1 1 1
(0,3,0) -1 -1 -1 0 -1 -1 1 -1 1 1
(1,0,2) -1 -1 1 1 0 1 1 1 0 1
(1,1,1) 0 -1 -1 1 -1 0 1 1 1 1
(1,2,0) 0 0 -1 -1 -1 -1 0 -1 1 1
(2,0,1) -1 -1 -1 1 -1 -1 1 0 1 1
(2,1,0) 0 -1 -1 -1 0 -1 -1 -1 0 1
(3,0,0) -1 -1 -1 -1 -1 -1 -1 -1 -1 0

Moves (0,3,0), (2,0,1), (2,1,0), and (3,0,0) are dominated. Eliminating them, we arrive at the following submatrix:
(0,0,3) (0,1,2) (0,2,1) (1,0,2) (1,1,1) (1,2,0)
(0,0,3) 0 1 1 1 0 0
(0,1,2) -1 0 1 1 1 0
(0,2,1) -1 -1 0 -1 1 1
(1,0,2) -1 -1 1 0 0 1
(1,1,1) 0 -1 -1 -1 0 1
(1,2,0) 0 0 -1 -1 -1 0
Again there are weakly dominated moves, going on eliminating them, we arrive at the IEWD matrix consisting just of move (0,0,3).

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Exercises

  1. Use Excel sheets Election.xls and SIMzero.xls to find the IEWD matrix for the game ELECTION(8,8,15|1,-1,0|3,3).
  2. Use Excel sheets Election.xls and SIMzero.xls to analyze the game ELECTION(5,8,13|0,0,0|3,3).
  3. Use Excel sheets Election.xls and SIMzero.xls to find the IEWD matrix for the game ELECTION(5,8,13|0,0,0|4,3).
  4. Find the Banzhaf-Penrose Index and the Shubik-Shapley Index of four parties in a parliament, where party A has 44 seats, party B has 33 seats, party C has 15 seats, and party D has 8 seats.
  5. Find the Banzhaf-Penrose Index and the Shubik-Shapley Index of four parties in a parliament, where party A has 45 seats, party B has 30 seats, party C has 21 seats, and party D has 4 seats.