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The Maximin moves are voting against a rise. There is no domination. There are seven pure Nash equilibria, 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.
Assume a simultaneous two-player game has the best response digraph shown to the right.
Display a possible payoff bimatrix. Can you find a possible zero-sum payoff bimatrix
generating this best response digraph?
| L | M | R | |
| U | -1, 1 | 1, -1 | 0, 1 |
| M | -1, 1 | 0, 0 | 1, -1 |
| D | 1, -1 | 1, -1 | -1, 1 |
| L | M | R | |
| U | 1,1 | 3,4 | 2,1 |
| M | 2,4 | 2,5 | 8,1 |
| D | 3,3 | 0,4 | 0,9 |
Maximin move is "M" for both Ann and Beth, guaranteeing a value of at least 2 for Ann and at least 4 for Beth.
For Ann there is no domination, weak or strong. For Beth, move "M" dominates (strongly, and therefore also weakly) move "L".
After removing option "L" for Beth, suddenly Ann's move "D" is strictly (and therefore also weakly) dominated by both "U" and "M". After eliminating this option as well, we arrive at the following bimatrix:
| M | R | |
| U | 3,4 | 2,1 |
| M | 2,5 | 8,1 |
| M | |
| U | 3,4 |
The best responses are U-->M, M-->M, and D-->R for Beth, and L-->D, M-->U, and R-->M for Ann.
Therefore the Nash equilibrium is "U" versus "M", since "M" is Beth's best response against Ann's "U", and "U" is Ann's best response against Beth's "M".
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| L | R | |
| U | 1, -1 | 2, -2 |
| D | 3, -3 | 4, -4 |
| 1 | 2 | 4 | 4 | |
| 1 | 0.5, 0.5 | 1, 0 | 1, 0 | 1, 0 |
| 2 | 0, 1 | 0.5, 0.5 | 1, 0 | 1, 0 |
| 3 | 0, 1 | 0, 1 | 0.5, 0.5 | 1, 0 |
| 4 | 0, 1 | 0, 1 | 0, 1 | 0.5, 0.5 |
There is no strict domination, so there is no dominant move equilibrium. However move "1" weakly dominates all others. Therefore the IEWD matrix is
| 1 | |
| 1 | 0.5, 0.5 |
The pair "1" versus "1" is the only Nash equilibrium of the game.
| 2 | 4 | 5 | |
| 2 | 2·(x/2+2), 2·(x/2+2) | 2·(x/2+4), 4·(x/2) | 2·(x/2+4), 5·(x/2) |
| 4 | 4·(x/2), 2·(x/2+4) | 4·(x/2+2), 4·(x/2+2) | 4·(x/2+4), 5·(x/2) |
| 5 | 5·(x/2), 2·(x/2+4) | 5·(x/2), 4·(x/2+4) | 5·(x/2+2), 5·(x/2+2) |
| friday | saturday | |
| friday | -10,-10 | 4,5 |
| saturday | 5,4 | -10,-10 |
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| Cooperate | Defect | |
| Cooperate | 1, 1 | 0, 3 |
| Defect | 3, 0 | 2, 2 |
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| Cooperate | Defect | |
| Cooperate | 2, 1 | 1, 3 |
| Defect | 3, 0 | 0, 2 |
Two cars are meeting at an intersection and want to proceed as indicated by the
arrows in the picture. Each player can proceed or yield. If both proceed, there is an accident.
A would have a payoff of -100 in this case, and B a payoff of -1000 (since B would be made responsible
for the accident, since A has the right of way). If one yields and the other proceeds, the
one yielding has a payoff of -5, and the other one of 5. If both yield,
it takes a little longer until they can proceed, so both have a payoff of -10.
Analyze this simultaneous game, draw the payoff bimatrix and find pure Nash equilibria.
| proceed | yield | |
| proceed | -100,-1000 | 5, -5 |
| yield | -5, 5 | -10, -10 |
Three cars are meeting at an intersection and want to proceed as indicated by the
arrows in the picture. Each player can proceed or yield. If two with intersecting paths proceed,
there is an accident. The one having the right of way has a payoff of -100 in this case,
the other one a payoff of -1000. If a car proceeds without causing an accident, the payoff
for that car is 5. If a car yields and the all others intersecting its path proceed, the
yielding car has a payoff of -5. If a car yields and a conflicting path car as well,
it takes a little longer until they can proceed, so both have a payoff of -10.
Analyze this simultaneous game, draw the payoff bimatrices and find all pure Nash equilibria.
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| 0 | 1 | 2 | 3 | 4 | 5 | |
| 0 | 0, 0 | 0, 1 | 0, 2 | 0, 3 | 0, 4 | 0, 5 |
| 1 | 1, 0 | 1, 1 | 1, 2 | 1, 3 | 1, 4 | 0, 0 |
| 2 | 2, 0 | 2, 1 | 2, 2 | 2, 3 | 0, 0 | 0, 0 |
| 3 | 3, 0 | 3, 1 | 3, 2 | 0, 0 | 0, 0 | 0, 0 |
| 4 | 4, 0 | 4, 1 | 0, 0 | 0, 0 | 0, 0 | 0, 0 |
| 5 | 5, 0 | 0, 0 | 0, 0 | 0, 0 | 0, 0 | 0, 0 |
Ann and Beth are running for president of the USB. There are three states. Similar to the president of the USA, this president is elected by votes from the three states, where states have different numbers of votes, and cannot split their votes. A state votes for that candidate that visited the state more often. In case of a tie, they abstain. We assume that the payoff for the winner is 1, and the payoff for the loser is -1. In case of a tie the payoffs are 0 for both.
It's three days before the election, and the two candidates have to determine now, simultaneously, which states to visit on each on the remaining three days. They have one visit per day, and they have to change states every day, so either they visit all three states, or they visit one twice and one once.
For parts (a) to (d) we also assume that the first state has 7 votes, the second state 8, and the third one 13.