If we take the maximum in each row and then Javier is the minimum maximum, Javier is v+. If we take the minimum in each column, and Raoul is the maximum of those, then Raoul is the maximum minimum, or v−. Thus, Javier is richer.

Another way to think of this is that the poorest rich guy is wealthier than the richest poor guy. Common sense.

Since v+ = 2, v− = − 2, there are no pure optimal strategies.

since min { x, 0 } ≤ 0, no matter what x is. Similarly,

Thus, v− = v+ = 1 and there is a pure saddle at row 2, column 1.

For matrix B, we have v− = max { 1, min { x, 0 } } = 1, v+ = min { max { 2, x }, 1 } = 1, and there is a pure saddle at row 1, column 2, no matter what x is.

and

Thus, v = 0 with a pure saddle point at (n, n).

Without loss of generality, we may as well assume that the saddle is at row 1, column 1, v− = v+ = a11. Since (1, 1) is a saddle, we have

In particular, a21 ≤ a11. If it is also true that a22 ≤ a12 and a23 ≤ a13 then row 1 dominates row 2 and we are done. Thus, we need only suppose that a22 > a12. Then, from the saddle point inequalities,

But then a12 ≥ a11 and a22 > a21 says that column 2 is dominated by column 1.

Now consider the 3 × 3 matrix

Then v− = v+ = 1 and there is a saddle at row 2, column 3, but no row or column dominates another.

and the lower value is

since if y chooses first, y can choose y = x. Thus, v+ = 0, v− = 0.

This proves x f(x, y) is linear. Similarly, y f(x, y) is also linear.

Then

Similarly,

and then

Since min max ≥ max min, we have from

that we must have equality throughout.

The result for this problem is 3y + 2(1 − y) = y + 4(1 − y), which implies that y* = and v(A) = .

The payoff of 0.7 to Curly if he puts the gold bar at the office and the thief hits the office is obtained by calculating (+ 1)× 0.85 + (−1) × 0.15 = 0.7. The two lines for the expected payoffs of player I cross where E((x, 1 − x), 1) = E((x, 1 − x), 2), which becomes − x + (1 − x) = x + 0.7(1 − x). The solution is x* = .

Similarly, for player II the lines cross where E(1, (y, 1 − y)) = E(2, (y, 1 − y)), or − y + (1 − y) = y + 0.7(1 − y).

The mixed strategy solution is X* = Y* = (, ), v = .

Once column 2 is gone, row 1 may be dropped. Then we apply the graphical method to get X* = (0, , ) and Y* = (, 0, ). The value of the game is v = .

Column 3 immediately dominates every other column. Then it doesn’t matter what row player I chooses because the payoff is always −1.

To see where these come from since they are all 2 × 2 games without pure saddle points, simply find where the two payoff lines cross for each player.

For (a) we must solve 4x − 9(1 − x) = − 3x + 6(1 − x) x = , and 4y − 3(1 − y) = − 9y + 6(1 − y), which gives y = . Then plugging in to either payoff line we get v = − . The other parts are similar.

The reduced matrix is . The graph for this matrix for player II is

The solution of the original game is, therefore, X* = (, , 0), Y* = (, , 0) and v(A) = .

where g = a4 + a5 a1 − a3.

We use the fact that E(i, Y*) = v if xi* > 0.

Then since there is a 0 in every row and all the ak′ s > 0, the maximum minimum must be v− = 0. Since there is an ak > 0 in every column, the maximum in every column is ak. The minimum of those is a1 and so v+ = a1.

and we conclude that v = . We needed the facts that < ∞, and the fact ≠ 0, since ak > 0 for all k = 1, 2, ….

Next xi ≥ > 0, and

which means it must be true, since each term is nonnegative, xi = , i = 1, 2, …. Similarly, X* = Y*, and the components of both optimal strategies are , i = 1, 2, ….

Then

Sending n → ∞ on the left side and using the fact = ∞, we get v = 0. Note that we know ahead of time that v ≥ 0 since

Let X = (x1, x2, …) be any mixed strategy for player I. Then, it is always true that E(X, j) = xjaj ≥ v = 0 for any column j. By Theorem 1.3.8 this says that X is optimal for player I. On the other hand, if Y* is optimal for player II, then E(i, Y*) = ai yi ≤ v = 0, i = 1, 2, … . Since ai > 0, this implies yi = 0 for every i = 1, 2, …. But then Y = (0, 0, …) is not a strategy and we conclude player II does not have an optimal strategy. Since the space of strategies in an infinite sequence space is not closed and bounded, we are not guaranteed that an optimal mixed strategy exists by the minimax theorem.

Now, if (X*, Y*) is a saddle then v = E(X*, Y*) and

But we have seen that the two ends of this long inequality are the same. We conclude that

Conversely, if max1 ≤ i ≤ nE(i, Y*) = min1 ≤ j ≤ m E(X*, j) = a, then we have the inequalities

This immediately implies that (X*, Y*) is a saddle and a = E(X*, Y*) = value of the game.

The figure indicates that the two lines determining the optimal strategies will cross where − 2x + 3(1 − x) = 8x + (− 2)(1 − x). This tells us that x* = and then v = − 2 + 3 (1 − ) = . We have X* = (, ).

Next, since the two lines giving the optimal X* come from columns 1 and 2, we may drop the remaining columns and solve for Y* using the matrix The two lines for player II cross where 3y − 2(1 − y) = − 2 y + 8(1 − y), or y* = . Thus, Y* = (, , 0, 0).

The maximum of this over 0 ≤ x ≤ 1 occurs when x = 0, which means the best response strategy for player I is X* = (0, 1), with expected payoff 4.

The saddle point is X* = (, ) = Y*. This is obtained from solving 3x − 2(1 − x) = − 2 x + (1 − x), and 3y − 2(1 − y) = − 2y + (1 − y). The value of this game is v = − , so this is definitely a game you should not play.

which is maximised at x = 0. This results in an expected payoff to you of zero. is not a winning strategy for the stranger.

and

These are the best responses since the second partials fxx = gyy = − 2 > 0.

Next,

The maximum of f(x, y*) is , achieved at x = , which means it is true that f(x* y*) ≥ f(x, y*) for all x.

If we use the graphical method, the lines cross where x + 10(1 − x) = 8x x = . The rest also checks.

The defense will guard against the pass more.

which gives x = x* = . Then

Since A* = ,

the formulas match.

The rest of the steps are similar to find Y* and v(A).

and v(A) = = 5.943. Finally, to calculate Y*, we have

This results in Y* = (0.6792, 0.30188, 0.01886, 0). Thus player I will attack target 3 about 40% of the time, while player II will defend target 3 with probability of only about 2%.

(b) The inverse of A is

The formulas then give X* = (, 0, ), Y* = (, , ) and v = . There is another optimal Y* = (, 0, ), which we can find by noting that the second column is dominated by a convex combination of the first and third columns (with λ = ). This optimal strategy for player II is not obtained using the formulas. It is in fact optimal since

Since E(2, Y) = 2 < v(A), we know that the second component of an optimal strategy for player I must have x2 = 0.

(c) Since

the formulas give the optimal strategies

(d) The matrix has the last column dominated by column 1. Then player I has row 2 dominated by column 1. The resulting matrix is . This game can be solved with 2 × 2 formulas or graphically. The solution for the original game is then

We also see that X, (A + b), YT = X A YT + b.

If (X* Y*) is a saddle for A, then

and

for any X, Y. Together this says that X*, Y* is also a saddle for A + b. The converse is similar.

Using the formulas we also have v(A) = 13 and Y* = X*. This leads us to suspect that each row and column, in the general case, is played with probability .

Now, let S denote the common sum of the rows and columns in an n × n magic square. For any optimal strategy X for player I and Y for player II, we must have

Since E(X, j) = aijxi ≥ v, adding for j = 1, 2, …, n, we get

Similarly, since E(i, Y) = aijyj ≤ v, adding for i = 1, 2, …, n, we get

Putting them together, we have v(A) = .

Finally, we need to verify that X* = Y* = (, …, ) is optimal. That follows immediately from the fact that

and similarly E(i, Y*) = v(A), i = 1, 2, …, n. We conclude using Theorem 1.3.8.

Then, setting q = we have

If n = 5, we have q = , so

Note that you search with decreasing probabilities as the box number increases.

Then, we have

We may use the formulas to get (A) = , X* = Y* = (, , ), and the cat and rat play each of the rows and columns of the reduced game with probability . This corresponds to dcbi, djke, ekjd for the cat, each with probability , and abcj, aike, hlia each with probability for the rat. Equivalent paths exist for eliminated rows and columns.

For example, if Burr decides to fire at 10 paces, and Hamilton fires at 10 as well, then the expected payoff to Burr is zero since the duelists have the same accuracies and payoffs (it’s a draw if they fire at the same time). If Burr decides to fire at 10 paces and Hamilton decides to not fire, then the outcome depends only on whether or not Burr kills Hamilton at 10. If Burr misses, he dies. The expected payoff to Burr is

Similarly, Burr fires at 2 paces and Hamilton fires at 6 paces, then Hamilton’s survival depends on whether or not he missed at 6. The expected payoff to Burr is

Calculating the lower value we see that v− = 0 and the upper value is v+ = 0, both of which are achieved at row 3, column 3. We have a pure saddle point and both players should wait until 2 paces to shoot. Makes perfect sense that a duelist would not risk missing.

Now we use the method of symmetry to solve the game with matrix B. Obviously, v(B) = 0, and if we let P = (p1, …, p6) denote the optimal strategy for player I, we have

This system of inequalities has one and only one solution given by

As far as the solution of the game with matrix B goes, we have v(B) = 0, and P is the optimal strategy for both players I and II.

Now to find the solution of the original game, we have (now with a new meaning for the symbols pi) n = 2, m = 3 and

Then b = p1 + p2 = , b = q1 + q2 + q3 = = . Now define the strategy for the original game with A,

and

Finally, v(A) = = = .

The problem is making the statement that this is indeed the solution of our original game. We verify that by calculating using the original matrix A,

By Theorem 1.3.8 we conclude that X*, Y* is optimal and v(A) = .

For method 2, the LP problems are as follows:

Both methods result in the solution

This results in X* = (0, and v(A) = − . It looks like John should not bother with State 1, and spend most of his time in State 2. For Dick, the problem is

The result is Y* = (, 0, 0, ). Dick should skip States 2 and 3 and put everything into States 1 and 4. John should think about raising more money to be competitive in State 4, otherwise, John suffers a net loss of − .

Referring to the Figure 1.1 we have the strategies for rat: 1 = hgf, 2 = hlk, 3 = abc, and for cat, A = djk, B = ekj.

The solution is the cat should take small loops and the rat should stay along the edges of the maze. X* = (, ), Y* = (, 0, ) and v = .

The numbers in the table are obtained from the rules of the game. For example, if LUC plays AB and UIC plays XY, then the expected payoff to LUC on each match is zero and hence on both matches it is zero.

Solving this game with Maple/Mathematica/Gambit, we get the saddle point

so LUC should play AA and BB with equal probability, and UIC should play XX, YY with equal probability. Z is going to sit out. The value of the game is v = −1, and it looks like LUC loses the tournament.

To use Maple to get these matrices:

Accuracy functions:
 > p1: = x- > piecewise(x = 0, .2, x = .2, .4, x = .4, .4, x = .6, .4, x = .8, 1);
 > p2: = x- > piecewise(x = 0, 0.6, x = .2, .8, x = .4, .8, x = .6, .8, x = .8, 1);
Payoff function
 > u1: = (x, y)- > 
      piecewise(x < y, 1*p1(x) + 
              (-1)*(1-p1(x))*p2(y) + 
              (0)*(1-p1(x))*(1-p2(y)), 
      x > y, (-1)*p2(y) + (1)*(1-p2(y))*p2(x) + 0*(1-p2(y))*(1-p1(x)), 
      x = y, 0*p1(x)*p2(x) + (1)*p1(x)*(1-p2(x)) + (-1)*(1-p1(x))*p2(x)
      + 0*(1-p1(x))*(1-p2(x)));
 > with(LinearAlgebra):
 > A: = Matrix([[u1(0, 0), u1(0, .4), u1(0, .8)], 
          [u1(.4, 0), u1(.4, .4), u1(.4, .8)], 
          [u1(.8, 0), u1(.8, .4), u1(.8, .8)]]);

The solution of this game is

Then, X* =, (0.43, 0.086.0.152, 0.326, 0, 0, 0), Y* = (0.043, 0.217, 0.413, 0.326, 0, 0, 0), v = 2.02174. This says 43% of the time farmer I should concede nothing, while farmer II should concede 2 about 41% of the time. Since the value to farmer I is about 2, farmer II will get about 4 yards. The rules definitely favor farmer II.

In the second case, in which the judge awards 3 yards to each farmer if they make equal concessions, each farmer should concede 2 yards and the value of the game is 3. Observe that the only difference between this game's matrix and the previous game is that the matrix has 3 on the main diagonal. This game, however, has a pure saddle point.

For example, if we take a = , and truncate it at column 5, we get the matrix

and the graph is shown.

The solution of this game is

Next, we calculate E(X*, Y*) = X*A, Y*T = (3λ −1) for any λ. Thus, we must have v(A) = (3 λ −1) and v(A) = − , if λ = 0.

Now, we have to check that (X*, Y*) is a saddle point and v(A) is really the value. For that, we check

The last inequalities are not true (unless λ = 0), since

Finally, we check

The last inequalities are not all true because − ≤ − + λ = v, but

and that contradicts 0 ≤ λ ≤ 1. Thus, everything falls apart.

The solution of the game is X* = (0, 0, 0, 1), Y* = (, , 0), v(A) = 2.

This reflects the rules that if Wiley waits at the right exit, Speedy will be caught. If Wiley waits between B and C, Speedy is caught with probability p if he exits from B or C, but he gets away if he comes out of A. If Wiley overlooks all three exits he has equal probability q of catching Speedy no matter what Speedy chooses.

The solution of the game by linear programming is X* = (, 0, 0, , 0), Y* = (, , ), and v(A) = . Since B and C are symmetric, another possible Y* = (, , ). A little less than half the time Wiley should wait outside A, and the rest of the time he should overlook all three exits. Speedy should exit from A with probability and either B or C with probability .

This game is easily solved using 2x + 3(1 − x) = 4x + (1 − x) x = , and 2y + 4(1 − y) = 3y + (1 − y), which gives y = . The optimal strategies are

Sonny should head for Queens with probability , and Barzini should head for Brooklyn with probability . On average, Sonny rescues between 2 and 3 capos.

for i = 1, 2, …, n, j = 1, 2, …, m.

for all strategies X, Y. Calculating the payoffs, we get

for any X = (x, 1 − x), 0 ≤ x ≤ 1. Similarly,

Since EI(X*, Y*) = EII (X*, Y*) = −, we have shown (X*, Y*) is a Nash equilibrium.

Then we have two pure Nash equilibria at (10, 1) and (5, 1).

The intersection of the two sets constitutes all the Nash equilibria. Mathematica characterizes the intersection as

You can see from the figure that there are lots of Nash equilibria. For example x* = , y* = , is one.

and

Graphing these two functions shows that they intersect at (0, 0), (, ), (1, 1):

The Nash equilibria are X1 = (, ), Y1 = (, ), X2 = (0, 1) = Y2, X3 = (1, 0) = Y3.

and

Here’s the figure:

The only Nash is X = (, ), Y = (), with payoffs − , . The rational reaction sets intersect at only this Nash.

Then X* = (). In order for this to work, X* must be a legitimate strategy with positive components.

This requires that A-C + D-B ≠ 0 and D ≠ B. Then Y* = () and the components must be positive.

Each of these numbers is ≤ . Next,

Each of these are actually equal to . We conclude that X*, Y* is indeed a Nash equilibrium.

There are three pure Nash equilibria: (1) row 1, column 3, (2) row 2, column 2, and (3) row 3, column 1.

To use the equality of payoffs theorem to find these you must consider all possibilities in which at least two rows (or columns) are played with positive probability. For example, suppose that columns 2 and 3 are used by player II with positive probability. Equality of payoffs then tells us

Hence, X = (x1, 0.5 x1, 1-1.5 x1) is a solution as long as 0 > x1 < .

If x1 = 0, we get X = (0, 0, 1), which is part of a pure Nash but is not obtained from equality of payoffs. If x1 = then X = (, , 0). By equality of payoffs we then obtain

Then, Y = (y1, − y1, ) is the solution if 0 ≤ y1 < . Observe that y1 = is not consistent with our original assumption that y2 > 0, y3 > 0, but y1 = 0 is. In that case, Y = (0, , ) and we have the Nash equilibrium X = (, , 0), Y = (0, , ). Any of the remaining Nash equilibria in the tables are obtained similarly.

If we assume Y = (y1, y2, y3) is part of a mixed Nash with yj > 0, j = 1, 2, 3, then equality of payoffs gives the equations

The first and fourth equations give vI = 0.25. Then the third equation gives x1 = , then x2 = , and finally x3 = . Since X = (, , ), we then use equality of payoffs to find Y. The equations are as follows:

This has the same solution as before Y = X, vI = vII = 0.25, and this is consistent with our original assumption that all components of Y are positive.

Similarly,

The Nash equilibrium is X = (, ), Y = (, ) with payoffs EI = − , EII = . The government should aid half the paupers, but 80% of paupers should be bums (or the government predicts that 80% of welfare recipients will not look for work). The rational reaction sets intersect only at the mixed Nash point.

and for any strategy X Sn,

Similarly,

and for any Y Sn,

and

These are both legitimate strategies. Finally, vI = , vII = .

and

(X*, Y*) is a Nash equilibrium if they are strategies, and then

Of course, if A or B does not have an inverse, then det (A) = 0 or det (B) = 0.

For the game in (1), A* = and B* = . We get

and

There are also two pure Nash equilibria, X1* = (0, 1), Y1* = (0, 1) and X2* = (1, 0), Y2* = (1, 0).

For the game in (2), A* = and B* = . We get

and

In this part of the problem, A does not have an inverse and det (A) = 0. There are also two pure Nash equilibria, X1* = (0, 1), Y1* = (0, 1) and X2* = (1, 0), Y2* = (1, 0).

If a > 0, b > 0, there are three Nash equilibria: X1 = Y1 = (1, 0), X2 = Y2 = (0, 1) and, the mixed Nash, X3 = Y3 = (). The mixed Nash is obtained from equality of payoffs:

If a < 0, b < 0, there are three Nash equilibria: X1 = (1, 0), Y1 = (0, 1), X2 = (0, 1), Y2 = (1, 0) and, the mixed Nash, X3 = Y3 = (). The rational reaction sets are piecewise linear lines as in previous figures which intersect at the Nash equilibria.

and

Similarly,

and the maximum of g is achieved at (y1, y2), where

The best response set, BR2(x), for player II is the set of all pairs (y1(x), y2(x)).

This is obtained by hand or using the simple Mathematica commands:

Then, for player II

The Nash equilibrium is X* = (, ), Y* = (, , 0). Then y1 = , y2 = , x = , and by definition BR1(, ). Also, (, ) BR2().

The Mathematica commands to solve this and all the 2 × 2 games are the following:

The last command produces the graph:

Adding the first three equations and using the fourth gives us v = 30. Then solving the first three equations gives Y* = (, , ). Since the game is symmetric, X* = Y*. The expected payoff to each player is 30.

The best response for player I will be to choose xi = 1 corresponding to the largest coefficient in EI. The only way to get a mixed strategy as a best response for player I is if the coefficients are all equal, and then they must all be 30, as we have seen. Thus, at least one coefficient must be greater than 30. Assume 50y2 + 40y3 > 30. Then, x1 = 1, x2 = x3 = 0 and player I’s expected payoff will be 50y2 + 40y3 > 30. Thus, no matter what, if player II does not use the mixed Nash, then player I’s expected payoff using the best response will be more than 30. This also applies for player II. Thus, if any player deviates from the Nash equilibrium, the best response gives a larger expected payoff to the other player.

where Jk = (1 1 1 · · · 1) is the 1 × k row vector consisting of all 1’s. In addition, p* = EI(X*, Y*) = X*AY*T, and q* = EII(X*, Y*) = − X*AY*T = − p*. You can see that this problem is the same as method 2 of linear programming for solving a game. Thus, Lemke–Howson reduces to method 2.

The Nash equilibrium found using Lemke–Howson is X* = (, , 0), Y* = (0, , ), and the value of the game is v(A) = .

The corresponding respective payoffs are as follows:

Using P we calculate the expected payoff for each player:

The payoff to player II is similar—just change the matrix.

Now P1 and P2 correspond to the two pure Nash equilibria. Under P1 and P2, the social welfare is 6. Also P3 corresponds to the mixed Nash equilibrium with social welfare 5. Since P = X*TY* = (xiyj), if (X*, Y*) is a Nash equilibrium, then Pi, i = 1, 2, 3, is a correlated equilibrium.

Next consider P4. We have

Thus P4 is a correlated equilibrium. The social welfare under P4 is 6.

Finally consider P5. We have

Thus, P5 is also a correlated equilibrium. The social welfare under P5 is .

P5 is the correlated equilibrium that maximizes the social welfare. This comes from the solution of

They are all Pareto-optimal because it is impossible for either player to improve their payoff without simultaneously decreasing the other player’s payoff, as you can see from the figure:

None of the Nash equilibria are payoff-dominant. The mixed Nash (X3, Y3) risk dominates the other two.

(X3, Y3) is payoff-dominant and Pareto-optimal.

Clearly, X3, Y3 is payoff-dominant and Pareto-optimal. Neither (X1, Y1) nor (X2, Y2) are Pareto-optimal relative to the other Nash equilibria, but they each risk dominate (X3, Y3).

The game matrix is

There is a pure Nash at (4, 1).

The probability a player takes a branch is indicated below the branch.

Aggie has three pure strategies:

(a1) Fire at 20 paces;

(a2) Pass at 20 paces; if Baggie passes at 20, fire at 10 paces;

(a3) Pass at 20 paces; if Baggie passes at 20, pass at 10 paces; if Baggie passes at 10, fire at 0 paces.

Baggie has three pure strategies:

(b1) If Aggie passes, fire at 20 paces;

(b2) If Aggie passes, pass at 20 paces; if Aggie passes at 10, fire at 10 paces;

(b3) If Aggie passes, pass at 20 paces; if Aggie passes at 10, pass at 10 paces.

The game matrix is then

To see where the entries come from, consider (a1) versus (b1). Aggie is going to throw her pie and if Baggie doesn’t get hit, then Baggie will throw her pie. Aggie’s expected payoff will be

because Aggie hits Baggie with probability and gets + 1, but if she misses, then she is going to get a pie in the face.

If Aggie plays (a2) and Baggie plays (b3), then Aggie will pass at 20 paces; Baggie will also pass at 20; then Aggie will fire at 10 paces. Aggies’ expected payoff is then

Here is a slightly modified version of this game in which both players can miss their shot with a resulting score of 0. The value of the game is still to Aggie. It is still optimal for Aggie to pass at 20, and for Baggie to take her shot at 20. If Baggie misses, she is certain to get a pie in the face.

Since V > 2CF, there are two pure Nash equilibria: (1) at (CT, -CF) and (2) at (− CT, CF). If we consider the tree, we see that woman 2 will always Agree if woman 1 claims Mine. This eliminates the branch Object by woman 2. Next, since CT > -CT, woman 1 will always call Mine, and hence woman 1 will end up with the child. The Nash equilibrium (CT, − CF) is the equilibrium that will be played by backward induction. Note that we can’t obtain that conclusion just from the matrix.

Similarly, assume that woman 2 is the true mother. Then the tree becomes

Since CT − V > − CT, woman 2 always Objects if woman 1 calls Mine. Eliminating the Agree branch for woman 2, we now compare (− CF, CT) and (− CF − W, CT − V). Since − CF > − CF − W, we conclude that woman 1 will always call Hers. Backward induction gives us that woman 1 always calls Hers (and if woman 1 calls Mine, then woman 2 Objects). Either way, woman 2, the true mother, ends up with the child.

Player I has perfect information but player II only knows if player I has raised $1 or $2, or has folded. This is a zero sum game with value to player I and − to player II. The Nash equilibrium is the following.

If player I gets an Ace, she should always Raise $2, while if player I gets a Two, she should Raise $2 50% of the time and Fold 50% of the time. Player II’s Nash equilibrium is if player I Raises $1, player II should Call with probability and Fold with probability ; if player I Raises $2, then player II should Call 50% of the time and Fold 50% of the time. A variant equilibrium for player II is that if player I Raises $1 then II should Call 100% of the time.

By dominance, the matrix reduces to the simple game

This game has a Nash equilibrium at (0, 6). We will see it is subgame perfect.

The subgame perfect equilibrium is found by backward induction. At node 1:2, BAT will play Out with payoff (− 2, 4) for each player. At 1:3, BAT plays Passive with payoffs (2, 3). At 2:1, PM has the choice of either payoff 4 or payoff 3 and hence plays Tough at 2:1. Now player BAT at 1:1 compares (0, 6) with (− 2, 4) and chooses Out. The subgame perfect equilibrium is thus BAT plays Out, and PM has no choices to make. The Nash equilibrium at (0, 6) is subgame perfect.

Let’s first consider the equivalent strategic form of this game. The game matrix is

First, note that the first column for player II (Moe) is dominated strictly and hence can be dropped. But that’s it for domination.

There are four Nash equilibria for this game. In the first Nash, Curly always chooses to order quiche, while Moe chooses to cave. This is the dull game giving an expected payoff of 2.1 to Curly and 0.9 to Moe. An alternative Nash is Curly always chooses to order a beer and Moe chooses to cave, giving a payoff of 2.9 to Curly and 0.9 to Moe. It seems that Moe is a wimp.

There are three pure strategies for Jack: (1) Real, (2) Fake, then Real and (3) Fake, then Fake. Jill also has three pure strategies: (1) Swat, (2) Pass, then Swat and (3) Pass, then Pass.

We can use Gambit to solve the problem, but we will use the invertible matrix theorem instead. We get

For example, (1) versus (1) means Jack will put down the real fly and Jill will try to swat the fly. Jack’s expected payoff is

The game tree from the first part is easily modified noting that the probability of hitting or missing the fly only applies to the real fly. Here is the figure with Gambit’s solution below the branches.

The solution of the game is

 > restart:A: = Matrix([[1, 1, 1, 1], [0, 3, 3, 3], [0, 2, 5, 5], [0, 2, 4, 7]]);
 > B: = Matrix([[0, 0, 0, 0], [3, 2, 2, 2], [3, 5, 4, 4], [3, 5, 7, 6]]);
 > P: = Matrix(4, 4, symbol = p);
 > Z: = add(add((A[i, j] + B[i, j])*p[i, j], i = 1..4), j = 1..4):
 > with(simplex):
 > cnsts: = {add(add(p[i, j], i = 1..4), j = 1..4) = 1}:
 > for i from 1 to 4 do
               for k from 1 to 4 do
                cnsts: = cnsts union {add(p[i, j]*A[i, j], j = 1..4) > = 
                          add(p[i, j]*A[k, j], j = 1..4)}:
                end do:
end do:
 > for j from 1 to 4 do
                 for k from 1 to 4 do
               cnsts: = cnsts union {add(p[i, j]*B[i, j], i = 1..4) > = 
                        add(p[i, j]*B[i, k], i = 1..4)}:
               end do:
end do:
 > maximize(Z, cnsts, NONNEGATIVE);
 > with(Optimization):
 > LPSolve(Z, cnsts, assume = nonnegative, maximize);

The result is p1, 1 = 1 and pi, j = 0 otherwise. The correlated equilibrium also gives the action that each player Stops.

According to the instructions in the problem we take the payoffs for each player to be 2, if they are sole survivor, if there are two survivors, 0 if all three stooges survive and −1 for the particular stooge who is killed. The payoffs at the end of the first round are −1 to any killed stooge in round 1, and the expected payoffs if there are two survivors, determined by analyzing the two person duels first. This is very similar to using backward induction and subgame perfect equilibria.

Begin by analyzing the two person duels Larry vs. Moe, Moe vs. Curly, and Curly vs. Larry. We assume that round one is over and only the two participants in the duel survive. We need that information in order to determine the payoffs in the two person duels. Thus, if Larry kills Curly, Larry is sole survivor and he gets 2, while Curly gets −1. If Larry misses, he’s killed by Curly and Curly is sole survivor.

1. Curly versus Larry:

In this simple game, we easily calculate the expected payoff to Larry is − and to Curly is .

2. Moe versus Curly:

The expected payoff to Moe is , and to Curly is − .

3. Larry versus Moe:

In this case, if Larry shoots Moe, Larry gets 2 and Moe gets −1, but if Larry misses Moe, then Moe gets his second shot. If Moe misses Larry, then there are two survivors of the truel, and each gets . The expected payoff to Moe is , and to Curly is − .

Next consider the tree below for the first round incorporating the expected payoffs for the second round.

The expected payoff to Larry is 0.68, Moe is 0.512, and Curly is 0.22. The Nash equilibrium says that Larry should deliberately miss with his first shot. Then Moe should fire at Curly; if Curly lives, then Curly should fire at Moe–and Moe dies. In the second round, Larry has no choice but to fire at Curly, and if Curly lives, then Curly will kill Larry.

Does it make sense that Larry will deliberately miss in the first round? Think about it. If Larry manages to kill Moe in the first round, then surely Curly kills Larry when he gets his turn. If Larry manages to kill Curly, then with 80% probability, Moe kills Larry. His best chance of survival is to give Moe and Curly a chance to shoot at each other. Since Larry should deliberately miss, when Moe gets his shot he should try to kill Curly because when it’s Curly’s turn, Moe is a dead man.

To find the probability of survival for each player assuming the Nash equilibrium is being played we calculate:

1. For Larry:

2. For Moe:

3. For Curly:

The expected payoffs to each player are for I and for II. The Nash equilibrium is the following.

If both players are rational, I should Pass, then II should Pass with probability , then I should Pass with probability , then player II should Pass with probability 0.

If I is rational and II is an altruist, then I should Pass with probability 1, then II Passes, then I Passes with probability , then II Passes.

If I is an altruist and II is rational, then I Passes, II Passes with probability , I Passes, and II Passes with probability 0.

If they are both altruists, they each Pass at each step and go directly to the end payoff of 12.80 for I and 3.20 for II.

Thus,

for every Thus, automatically satisfies the definition of a Nash equilibrium. A maximum of both payoffs is a much stronger requirement than a Nash point.

The function

There is a unique Nash equilibrium at since

gives and these points provide a maximum of each function with the other variable fixed because the second partial derivatives are −2 < 0. On the other hand,

Thus, maximizes neither of the payoff functions. Furthermore, there does not exist a point that maximizes both u1, u2 at the same point (q1, q2).

The graphs of the rational reaction sets are as follows:

The function

This equation becomes −0.41 γ3 + γ2 + 0.41 γ − 0.5 = 0 which implies γ = 0.585. The Nash equilibrium position for each candidate is (γ*, γ*) = (0.585, 0.585). With that position, each candidate will get 50% of the vote.

and solving for x and y results in x = 4, y = 4 as the Nash equilibrium. The payoffs are u1(4, 4) = u2(4, 4) = 16.

First, calculate maxt1≥0 u1(t1, t2) and maxt2≥0 u2(t1, t2). For example,

and the maximum is achieved at the set valued function

This is the best response of country 1 to t2. Next, calculate the best response to t1 for country 2,

If you now graph these sets on the same set of axes, the Nash equilibria are points of intersection of the sets. The result is

For example, this says that either (1) country 1 should concede immediately and country 2 should wait until first time or (2) country 2 should concede immediately and country 1 should wait until first time

and

and hence, concavity in the appropriate variable for each payoff function. Solving the first derivatives set to zero gives

Then

We need to show that

Recall that 1−i denotes that all the players except player i are playing 1. Note that since when all players choose 1, and What is a better xi for player i to switch to? In order to be better it must satisfy

Note that Hence, we must have

That is impossible. Hence, (1, 1, …, 1) is a Nash equilibrium.

Each farmer has the payoff function

Assume that Q < 1, 950, 000. Take a partial derivative and set to zero to get qi = 562, 500 bushels each. Then Q = 1, 687, 500 < 1, 950, 000. So an interior pure Nash equilibrium consists of each farmer sending 562, 500 bushels each to market and using 437, 500 bushels for feed. The price per bushel will be which is greater than the government-guaranteed price.

If each producer ships 562, 500 bushels, the profit for each producer is

Now suppose producer 1 ships his entire crop of q1 = 106 bushels. Assuming the other producers are not aware that producer 1 is shipping 106 bushels. If the other producers still ship 562, 500 bushels the total shipped will be 2, 125, 000 > 1, 950, 000 and the price per bushel would be subsidized at 2. The revenue to each producer is then

This means all the producers make less money but 2 and 3 make significantly less.

It’s very unlikely that 2 and 3 would be happy with this solution. We need to find 2 and 3’s best responses to 1’s shipment of 106. Assume first that 106 + q2 + q3 ≤ 1, 950, 000. We calculate

The problem for q3 is similar. Taking a derivative, setting to zero, and solving for q2, q3 results in q2 = q3 = (1.25 × 106)/3 = 416, 667. That is the best response of producers 2 and 3 to producer 1 shipping his entire crop assuming Q ≤ 1, 950, 000. The price per bushel at total output 1, 833, 334 bushels will be $2.78 and the profit for each producer will be u1 = 2, 780, 000, u2 = u3 = 1, 158, 334.

Now assume that Q ≥ 1, 950, 000. In this case, p = 2 and the profit for each producer is u1 = 2, 000, 000, u2 = 2 × q2, u3 = 2 × q3. The maximum for producers 2 and 3 is achieved with q2 = q3 = 106 and all 3 producers ship their entire crop to market. This is clearly the best response of producers 2 and 3 to q1 = 106.

The second partials give us

both of which are < 0. That means that we can find the Nash equilibrium where the first partials are zero. Solving, we get the general solution

and that is the Nash equilibrium as long as 0 ≤ x* < I, 0 ≤ y* < T.

Now since we conclude that g(x) = 1, 0 ≤ x ≤ 1. This is the density for a uniformly distributed random variable on [0, 1]. Thus, the Nash equilibrium is that each investor chooses an investment level at random.

Finally, we will see that vI = vII = 0. But that is immediate from

The payoffs to each firm is then which is greater than the profit under competition. Furthermore, the cartel price will be the same as the monopoly price since the total quantity produced is the same.

Thus firm 1 should produce more than firm 2 as a best response. Similarly, firm 2 has an incentive to produce more than firm 1. The cartel collapses.

Since any critical point will provide a maximum. Set αi = Γ − ci. We have to solve the system of equations In matrix form, we get the system

It is easy to check that

and then

After a little algebra, we see that the optimal quantity that each firm should produce is

If ci = c, i = 1, 2, …, N, then The profits then also approach zero.

Taking a derivative and setting to zero gives

and it is easy to check that provides the maximum. Also

We can take advantage of the fact that there is symmetry between the two firms since they have the same price function and costs. That means, for firm 1,

The maximum is achieved at s Thus, the Nash equilibrium is and

Observe that q1 is the expected value of the best response production quantities when the costs for firm 2 are ci, i = 1, 2, 3 with probability pi, i = 1, 2, 3. The system of best response equations has following solution:

Solving these two equation results in

Next, firm 1 will choose q1 to maximize

Taking a derivative and setting to zero, we get Finally, plugging this q1 into the best response functions q2, q3, we get the optimal Stackelberg production quantities:

The profits for each firm are as follows:

We will show that u2(c, c) ≥ u2(c, p2), for any p2 ≠ c. Now, u2(c, c) = 0 and

In every case u2(c, c) = 0 ≥ u2(c, p2). Similarly u1(c, c) ≥ u1(p1, c) for every p1 ≠ c. This says (c, c) is a Nash equilibrium and both firms make zero profit.

and

To explain the profit function for firm 1, if p1 < p2, firm 1 will be able to sell the quantity of gadgets q = min{Γ − p1, K}, the smaller of the demand quantity at price p1 and the capacity of production for firm 1.

If p1 = p2, so that both firms are charging the same price, they split the market and since there is enough capacity to fill the demand.

If p1 > p2 in the standard Bertrand model, firm 1 loses all the business since they charge a higher price for gadgets. In the limited capacity model, firm 1 loses all the business only if, in addition, p2 ≥ Γ − K; that is, if K ≥ Γ − p2, which is the amount that firm 2 will be able to sell at price p2, and this quantity is less than the production capacity.

Finally, if p1 > p2, and K < Γ − p2, firm 1 will be able to sell the amount of gadgets that exceed the capacity of firm 2. That is, if K < Γ − p2, then the quantity demanded from firm 2 at price p2 is greater than the production capacity of firm 2 so the residual amount of gadgets can be sold to consumers by firm 1 at price p1. But note that in this case, the number of gadgets that are demanded at price p1 is Γ − p1, so firm 1 can sell at most Γ − p1− K < Γ − p2 − K.

What about Nash equilibria? Even in the case c1 = c2 = 0 there is no pure Nash equilibrium even at This is known as the Edgeworth paradox.

Solve to get the best response function:

Next, solve to get

and then, the optimal price for firm 2 is

The profit functions become

and

Note that with the Stackelberg formulation and the formulation in the preceding problem the Nash equilibrium has positive prices and profits.

The Maple commands to do these calculations are as follows:

> u1:=(p1, p2)->(G-p1+b*p2)*(p1-c1);
> u2:=(p1, p2)->(G-p2+b*p1)*(p2-c2);
> eq1:=diff(u2(p1, p2), p2)=0;
> solve(eq1, p2);
> assign(p2, %);
> f:=p1->u1(p1, p2);
> eq2:=diff(f(p1), p1)=0;
> solve(eq2, p1);
> simplify(%);
> assign(p1, %);
> p2;simplify(%);
> factor(%);
> assign(a2, %);
> u1(p1, a2);
> simplify(%);
> u2(p1, a2);
> simplify(%);

For the data of the problem, we have

and recall that v1 ≥ v2 ··· ≥ vN. We have to show that ui(v1, …, vN) gives a larger payoff to player i if player i makes any other bid bi ≠ vi. We assume that v1 = v2 so the two highest valuations are the same. Now for any player i if she bids less than v1 = v2 she does not win the object and her payoff is zero. If she bids bi > v1 she wins the object with payoff vi − bi < v1 − bi < 0. If she bids bi = v1, her payoff is In all cases, she is worse off if she deviates from the bid bi = vi as long as the other players stick with their valuation bids.

Differentiate, set to zero, and solve to get

Since all bidders will actually pay their own bids and each bid is the expected payment from each bidder is

Since there are N bidders, the total expected payment to the seller will be

The unnormalized allocations satisfy

Since 1 − nα > 0, xi − α ≥ 0 and In terms of the unnormalized allocations,

Since b < nc and n > |S| both terms in the max are negative so If S = N,

since b < nc. By definition, we conclude

The normalized characteristic function is

The normalized least core is The unnormalized least core is

The normalized least core comes from the inequalities

which decides is the first ε. Then Then implies and hence

and x1 + x2 + x3 = 4. These imply that 3 − ε ≤ x1 + x2 ≤ 5 + ε and then −2 ≤ 2ε. Consequently, we derive that ε1 = −1 and the least core is

That’s a contradiction, and hence C(0) = .

and so x1 ≤ 0. But since −x1 = v(1) − x1 ≤ 0, we have x1 = 0.

For example, for S = 1, players 2 and 3 get their capacity first which is 135; the 15 left over go to player 1.

Hence, x2 ≤ 60 + ε, x1 ≤ 45 + ε ⇒ 75 − ε ≤ x1 + x2 ≤ 105 + 2ε that gives ε1 = −10. Then x1 + x2 = 85, x3 = 65, and since x1 ≤ 35, x2 ≤ 50, it must be true that x1 = 35, x2 = 50. The least core is C(−10) = {(35, 50, 65)} the exact same allocation as before but with a nonempty core. Every player is ok with this allocation.

Add these up and use the fact that to get the requirement

With this contradiction, we see that n > 2 ⇒ C(0) = .

Another way to see this is to use a result from an earlier Problem 6.17 that gives us a criterion to use to determine when the core is empty. The criterion says

In the garbage game, we have

implies that n − 1 > 1 or n > 2. We conclude that when n > 2, from Problem 6.17, the core of the garbage game is empty.

Next,

and that gives us the first ε1 = −50. This implies

Checking all possible combinations of inequalities results in the first ε from the inequalities

and Then, the first least core is

Next, we calculate the excesses for

Thus, we may eliminate coalitions 3 and 12 since their excesses cannot be reduced further. The next least core is

We get

The first ε making Then we calculate the next least core is This contains only one point and is the nucleolus.

To get the least core, we have

Using the inequalities 90 − ε ≤ x1 ≤ 105.5 − ε, we get the first Then x1 = 97.75 and x2 + x3 = 72.75.

We need the next least core. We calculate the excesses for to get

The next least core will work with coalitions {2, 3, 12, 13} since only those coalitions have excesses which may be lowered. We have

The two inequalities 16 − x3 ≤ ε, 37.25 − x2 ≤ ε lead to 53.25 − 2ε ≤ x2 + x3 = 72.75 and ε2 = −9.75 as the first ε making X2 ≠ . Then 37.25 + 9.75 = 47 ≤ x2, and 16 + 9.75 = 25.75 ≤ x3 imply that x2 = 47, x3 = 25.75.

The nucleolus is

It is interesting to compare this with the common assumption that the fair allocation should be that each player will get the amount of 170, 500 proportional to the amount they invest. That would lead to the allocation 170, 500 = 108063.38, and 170, 500 = 43225.35, and 170, 500 = 19211.27. But, this does not take into account that player 3 is a very important investor. It is her money that pushes the grand coalition into the 5.5% rate of return. Without player 3 the most they could get is 5%. Consequently, player 3 has to be compensated for this power. The proportional allocation doesn’t do that.

The characteristic function is the number of hours saved by a coalition. Single coalitions save nothing v(i) = 0, i = 1, 2, 3, 4, and

For instance v(124) = 7 since Shemp and Curly overlap 4 hours and Shemp and Moe overlap 3 hours.

The first least core is determined from

A decided allocation is Next, to get X2 first calculate the excesses for and we see that we may eliminate coalitions S = 4, S = 123.

We then obtain and Again we calculate the excesses and we determine we may eliminate coalitions S = 1, S = 14, S = 1234.

Finally, we calculate using only the coalitions S = 2, 3, 124, 134, and we get That is the nucleolus.

The least core is

This reduces to so this is the nucleolus.

In the original terms of costs, we have

Then

Of course, this can also be done tediously by hand. The result is

and that’s one fair way to divide the total of 302.

The Shapley allocation is the same as the nucleolus. Looks like waxing is lucrative.

Then the nucleolus is C(−1) = {(1, 1, 2)}. Thus, Moe and Curly both save 1 hour while Larry saves 2. Since Curly arrives first, he can now work 9-12. Then Larry works 12-3, and Moe works 3-5.

The Shapley allocation is the same as the least core. For example,

Here is the table giving the Shapley allocation:

The Shapley value is and this point is not in C(0) since and

The nucleolus of this game is {(115, 0, 15)} which is quite different from the Shapley value. Under the nucleolus, of the 130 available if the players cooperate, the seller gets 115 for her object, and player 3 gets to keep 15 of her 130. She pays the seller 115. Under Shapley, the seller gets 81.67, player 3 pays a total of 130-31.67 = 98.33 of which 81.67 goes to the seller to buy the object and 16.67 goes to player 2 to go away. Do you think this is what would actually happen?

Adding the three-player coalitions, we see that

Then x1 + x2 + x3 + x4 = 2 ⇒ x1 = 2, x2 = x3 = x4 = 0. Thus, C(0) = {(2, 0, 0, 0)}.

Since the core contains only one point, it is the nucleolus. Player 1 is allocated everything while players 2, 3, and 4 get nothing. Doesn’t seem fair.

Then The Shapley value is then

which has more than one point. That means we have to calculate the least core:

Thus, ε ≥ −5 ⇒ ε1 = −5 ⇒ x1 = x2 = 15 and C(−5) = {(15, 15)}, an even split.

The Shapley allocation is

which is the same as the nucleolus.

A direct calculation shows that

Plugging into Shapley’s formula gives

we compute

This characteristic function is superadditive, and they all have an incentive to join the carpool.

Eliminating x3 we get x2 ≤ 12 + ε, x1 ≤ 15 + ε ⇒ 13 − ε ≤ x1 + x2 ≤ 27 + 2ε. This leads to and then

Then Since 31 + 22 = 53, we conclude Thus,

The Shapley allocation is, therefore, We get Larry should pay Moe and Curly should pay Moe

The corresponding cost savings are as follows:

By going through similar calculations, we get

The Shapley value is

Larry, Curly, and Shemp should pay Moe the difference in the amount it costs them to drive on their own and the amount of the savings attributable to that player.

1. Larry should pay Moe under Shapley and under nucleolus.

2. Curly should pay Moe under Shapley and under nucleolus.

3. Shemp should pay Moe under Shapley and under nucleolus.

Next, count up the number of times each player is critical for a coalition. For player 1 it is 7 times; player 2, 3, and 4, exactly 1 time each. The total number of winning coalitions for each critical player is 10. Hence,

Player 1’s power index is 7 times that of the other players even though she has only twice as many votes (as player 2).

By calculus, the solution is The bargained solution is that each contestant should Split. That seems fair and natural.

the same solution as before. This means that in order to achieve the threat bargained solution, they should agree to always play row 1, column 1, or if they play many times, half the time they should play row 2, column 1, and half the time row 1, column 2. The expected payoffs are the same.

In fact take a derivative of (u − 2)(6 − u − 2) and set to zero to see that u = 3.

Security point (2, 2)

Next, since the Pareto-optimal boundary is u + v = 6, we have mp = −1, b = 6. We calculate value(−mp A − B) and the saddle strategies for that game to get value(A − B) = 2, Xt = (1, 0), Yt = (1, 0).

The threat security point is, therefore, (ut, vt) = (4, 2) with threat strategies Xt = (1, 0) = Yt.

Next, we use the formulas

with b = 6. The threat solution is with both players threatening to play row 1, column 1.

The KS line for (u*, v*) = (2, 2) is v − 2 = k(u − 2), or simplified to v = u. This intersects u + v = 6, the Pareto-optimal boundary, at (3, 3). Therefore, (3, 3) is the KS solution.

There is no KS line for (u*, v*) = (4, 2) because it is on the edge of the Pareto line.

If we consider the cooperative game, the characteristic function is v(1) = v(2) = 2, v(12) = 6, which gives nucleolus and Shapley value (3, 3) and matches with the solution for the (2, 2) security point.

We have That is our safety point. The Pareto-optimal boundary has three line segments:

The Nash problem is

The part of the Pareto-optimal boundary for this problem is the line segment Using calculus, we find

Next, we consider the threat solution. We have to find the threat strategies for all three line segments:

1. Then and

and

Since 5.87 [0, 1], this is not the threat solution.

2. Then b = 6, and

Then

This is in the range. Let's check the final segment.

3. In this case and

The safety point is then

Since this too is not the threat solution.

We conclude that the threat solution is and player 1 threatens to always play the second row; player 2 threatens to use the third column unless they both come to their senses.

Finally we determine the KS solution for the safety point

We have Then

is the KS line, and this intersects the Pareto optimal boundary through the segment at the point and that is the KS solution.

with (p, w) S, where

Solving the first equation for p gives

Substitute this p into the equation hw = 0 to get

which implies f′(w) = p0.

2.

3.

In order for X3 to be an ESS, we need

which becomes for 0 < p < px. This is clearly impossible, so X3 is not an ESS.

We calculate for

and

The question is whether (for small enough 0 < p < 1) we have

Since 98x2 − 28x + 2 = 2(7x − 1)2 > 0 for any we conclude that is indeed an ESS. Consequently, in a series of naval battles between the British and French, strong navys will always attack, while weak navys will attack with probability

For X* = (1, 0, 0), you can see this is an ESS because it is strict. Consider next Since the set of best response strategies is Y = (y, 1 − y, 0). Then u(Y, Y) = 4y2 − 4y + 2 and Since it is not true that u(Y, Y) < u(X*, Y), for all best responses Y ≠ X*, X* is not an ESS.

and for every strategy

so X is an ESS.

The figure shows trajectories starting from three different initial points. In the three-dimensional figure, you can see that the trajectories remain in the plane p1 + p2 = 1. The Maple commands used to find the stationary solutions, check their stability, and produce the graph are as follows:

> restart:with(DEtools):with(plots):with(LinearAlgebra):
> A:=Matrix([[3, 2], [4, 1]]); X:=< x1, x2>;
> Transpose(X).A.X;
> s:=expand(\%);
> L:=A.X; f:=(x1, x2)->L[1]-s;g:=(x1, x2)->L[2]-s;
> solve({f(x1, x2)=0, g(x1, x2)=0}, [x1, x2]);
> q:=diff(f(x1, x2), x1)+diff(g(x1, x2), x2);
> with(VectorCalculus):Jacobian(<f(x1, x2), g(x1, x2)>, [x1, x2]);
> j:=simplify(\%);
> a1:=subs(x1=1/2, x2=1/2, j);b1:=subs(x1=1/2, x2=1/2, q);
> Determinant(a1);
> DEplot3d({D(p1)(t)=p1(t)*(3*p1(t)+2*p2(t)
                    -3*p1(t)^2-6*p1(t)*p2(t)-p2(t)^2), 
            D(p2)(t)=p2(t)*(4*p1(t)+p2(t)
                    -3*p1(t)^2-6*p1(t)*p2(t)-p2(t)^2) }, 
            {p1(t), p2(t)}, t=0..10, 
            [[p1(0)=0.1, p2(0)=0.9], [p1(0)=0.6, p2(0)=.4], 
            [p1(0)=1/4, p2(0)=3/4]], scene=[t, p1(t), p2(t)], 
            stepsize=.1, linecolor=t);

For the problem, Determinant(a1)=5>0, and b1=-6<0, so the stability Theorem 7.2.3 allows us to conclude that is asymptotically stable.

and get Observe that if we want to use the formula for X* in Problem 3.25, we must use B = AT.

for any X S3. Now we check if u(X*, X) > u(X, X) for all X S3. We have

The easiest way to check if this is > 0 is to graph the function, or to minimize the function. We’ll do both. First, we have

and the problem becomes

We can do this by calculus or with Mathematica. The minimum is 0 attained at Thus, f(x1, x2) > 0 at any point except at the Nash equilibrium X*. This tells us that X* is indeed an ESS. Here is the graph of f showing it is positive except for the minimum point.

Using p3 = 1 − p1 − p2, and a lot of algebra we get the replicator equations for p1, p2 are as follows:

Then is a stationary solution. To check stability, we use Theorem 7.2.3. These calculations may be done by hand but Mathematica is way easier. Here are the commands to do this:

The third line calculates fp1 + g p2, and the fifth line calculates the determinant of the Jacobian matrix given in the fourth line. By the Theorem 7.2.3, we conclude that is asymptotically stable.

The remaining stationary solutions are (p1 = 0, p2 = 1), (p1 = 0, p2 = 0), and (p1 = 1, p2 = 0). The convergence to is shown in the figure.

Hence, we are in the condition u(X*, X*) = u(Y, X*), and we have to check if u(X*, Y) > u(Y, Y) for every Y ≠ X*. However,

is Thus, it is not true that X* is an ESS.

Then

and According to Theorem 7.2.3, X* is unstable. This is illustrated in the following figure in the case when a = 0.

The case when is not ESS.

The case when a = 2 is shown in the following figure.

The case a = 2, is not ESS

You can see that unless you start exactly at X*, the trajectories converge to one of the pure Nash equilibria.

It still is true that is the unique symmetric Nash equilibrium. We see this by calculating

Now we calculate for any Y = (y1, y2, 1 − y1 − y2)

Again the condition u(X*, X*) = u(Y, X*) holds and we have to check if u(X*, Y) > u(Y, Y) for every Y ≠ X*. We consider,

over all strategies Y and calculate that this minimum is zero, attained only when Y = X*. Thus, in this case X* is an ESS. Here is the figure for the replicator dynamics.

When diagonal < 0, is ESS