# | claim | justification |
1 | Suppose row i is a dominant row and column j is a dominant column. | assumption for proving conditional |
2 | Row i is a dominant row. | 1 |
3 | v_{ij} is at least as large as any other value in column j. | 2 |
4 | Column j is a dominant column. | 1 |
5 |
v_{ij} is at least as small as any other value in row i. |
4 |
6 |
(R_{i}, C_{j}) is an equilibrium pair. |
3 and 5, minimax equilibrium test |
7 |
If row i is a dominant row and column j is a dominant column, then (R_{i}, C_{j}) is an equilibrium pair. |
1–6 |
# | claim | justification |
1 | Suppose (R_{i}, C_{j}) is a strategy pair that passes the equilibrium test. | assumption for proving conditional |
2 | The payoff determined by (R_{i}, C_{j}) equals the minimal value of its row and the maximal value of its column. | 1, definition of equilibrium test |
3 | The payoff determined by (R_{i}, C_{j}) equals the minimal value of its row. | 2 |
4 | The column player cannot do better by unilaterally changing his strategy. | 3 |
5 |
The payoff determined by (R_{i}, C_{j}) equals the maximal value of its column. |
2 |
6 |
The row player cannot do better by unilaterally changing his strategy. |
5 |
7 |
Neither player can do better by unilaterally changing his strategy. |
4 and 6 |
8 |
(R_{i}, C_{j}) is in equilibrium in the originally defined sense. |
7, original definition of equilibrium |
9 |
If (R_{i}, C_{j}) is a strategy pair that passes the equilibrium test, then it is in equilibrium in the originally defined sense. |
1–8 |
10 |
Suppose (R_{i}, C_{j}) is in equilibrium in the originally defined sense. |
assumption for proving conditional |
11 |
Neither player can do better by unilaterally changing his strategy. |
10, original definition of equilibrium |
12 |
The row player cannot do better by unilaterally changing his strategy. |
11 |
13 |
The payoff determined by (R_{i}, C_{j}) equals the maximal value of its column. |
12 |
14 |
The column player cannot do better by unilaterally changing his strategy. |
11 |
15 |
The payoff determined by (R_{i}, C_{j}) equals the minimal value of its row. |
14 |
16 |
The payoff determined by (R_{i}, C_{j}) equals the minimal value of its row and the maximal value of its column. |
13 and 15 |
17 |
(R_{i}, C_{j}) passes the equilibrium test. |
16, definition of equilibrium test |
18 |
If (R_{i}, C_{j}) is in equilibrium in the originally defined sense, then it passes the equilibrium test. |
10–17 |
19 |
If (R_{i}, C_{j}) is a strategy pair that passes the equilibrium test, then it is in equilibrium in the originally defined sense, and conversely. |
9 and 18 |
C_{1} | C_{2} | |
R_{1} | 1 | 2 |
R_{2} | 1 | 3 |
(R_{1}, C_{1}) is an equilibrium pair, and yet row 1 is dominated.
top of p. 135—tables 5-16 and 5-17:
(Instead of verifying that the mentioned strategy pair is the equilibrium strategy pair, we’ll just verify that the mentioned strategy pair is an equilibrium strategy pair.)
table 5-16:
To verify that [(½ R_{1}, ½ R_{2}), (½ C_{1}, ½ C_{2})] is an equilibrium pair, we need to verify that neither the row player nor the column player can do better by unilaterally changing his strategy.
First consider the game from the row player’s point of view. Consider, in particular, the row player’s pure strategies, R_{1} and R_{2}. If the column player is playing strategy (½ C_{1}, ½ C_{2}), then the expected utility of R_{1} is (½)(1) + (½)(–1), or ½ – ½, or 0. And—still assuming that the column player is playing strategy (½ C_{1}, ½ C_{2})—the expected utility of R_{2} is (½)(–1) + (½)(1), or –½ + ½, or 0. Because the expected utilities of R_{1} and R_{2} are both 0, the expected utilities of all their mixtures is 0, too, and so any mixture of R_{1} and R_{2} is as good (for the row player) as any other. Thus, the strategy (½ R_{1}, ½ R_{2}) is as good (for the row player) as any other. So, the row player cannot do better by unilaterally changing his strategy.
Now consider the game from the column player’s point of view. Consider, in particular, the column player’s pure strategies, C_{1} and C_{2}. If the row player is playing strategy (½ R_{1}, ½ R_{2}), then the expected utility of C_{1} is (½)(1) + (½)(–1), or ½ – ½, or 0. And—still assuming that the row player is playing strategy (½ R_{1}, ½ R_{2})—the expected utility of C_{2} is (½)(–1) + (½)(1), or –½ + ½, or 0. Because the expected utilities of C_{1} and C_{2} are both 0, the expected utilities of all their mixtures is 0, too, and so any mixture of C_{1} and C_{2} is as good (for the column player) as any other. Thus, the strategy (½ C_{1}, ½ C_{2}) is as good (for the column player) as any other. So, the column player cannot do better by unilaterally changing her strategy.
Since neither the row player nor the column player can do better by unilaterally changing his or her strategy, [(½ R_{1}, ½ R_{2}), (½ C_{1}, ½ C_{2})] is an equilibrium pair.
To compute the value of the game, we find the expected utility of the playing of [(½ R_{1}, ½ R_{2}), (½ C_{1}, ½ C_{2})], by computing the weighted average of the values of the four possible outcomes of the game: (½)(½)(1) + (½)(½)(–1) + (½)(½)(–1) + (½)(½)(1) = ¼ – ¼ – ¼ + ¼ = 0.
table 5-17:
To verify that [(½ R_{1}, ½ R_{2}), (½ C_{1}, ½ C_{2})] is an equilibrium pair, we need to verify that neither the row player nor the column player can do better by unilaterally changing his strategy.
First consider the game from the row player’s point of view. Consider, in particular, the row player’s pure strategies, R_{1} and R_{2}. If the column player is playing strategy (½ C_{1}, ½ C_{2}), then the expected utility of R_{1} is (½)(22) + (½)(–18), or 11 – 9, or 2. And—still assuming that the column player is playing strategy (½ C_{1}, ½ C_{2})—the expected utility of R_{2} is (½)(–18) + (½)(22), or –9 + 11, or 2. Because the expected utilities of R_{1} and R_{2} are both 2, the expected utilities of all their mixtures is 2, too, and so any mixture of R_{1} and R_{2} is as good (for the row player) as any other. Thus, the strategy (½ R_{1}, ½ R_{2}) is as good (for the row player) as any other. So, the row player cannot do better by unilaterally changing his strategy.
Now consider the game from the column player’s point of view. Consider, in particular, the column player’s pure strategies, C_{1} and C_{2}. If the row player is playing strategy (½ R_{1}, ½ R_{2}), then the expected utility of C_{1} is (½)(22) + (½)(–18), or 11 – 9, or 2. And—still assuming that the row player is playing strategy (½ R_{1}, ½ R_{2})—the expected utility of C_{2} is (½)(–18) + (½)(22), or –9 + 11, or 2. Because the expected utilities of C_{1} and C_{2} are both 2, the expected utilities of all their mixtures is 2, too, and so any mixture of C_{1} and C_{2} is as good (for the column player) as any other. Thus, the strategy (½ C_{1}, ½ C_{2}) is as good (for the column player) as any other. So, the column player cannot do better by unilaterally changing her strategy.
Since neither the row player nor the column player can do better by unilaterally changing his or her strategy, [(½ R_{1}, ½ R_{2}), (½ C_{1}, ½ C_{2})] is an equilibrium pair.
To compute the value of the game, we find the expected utility of the playing of [(½ R_{1}, ½ R_{2}), (½ C_{1}, ½ C_{2})], by computing the weighted average of the values of the four possible outcomes of the game: (½)(½)(22) + (½)(½)(–18) + (½)(½)(–18) + (½)(½)(22) = 22/4 – 18/4 – 18/4 + 22/4 = 8/4 = 2.
problems on p. 137:
To verify that [(½ R_{1}, ½ R_{2}), (½ C_{1}, ½ C_{2})] is an equilibrium pair, we need to verify that neither the row player nor the column player can do better by unilaterally changing his strategy.
First consider the game from the row player’s point of view. Consider, in particular, the row player’s pure strategies, R_{1} and R_{2}. If the column player is playing strategy (½ C_{1}, ½ C_{2}), then the expected utility of R_{1} is (½)(a) + (½)(–b), or ½(a – b). And—still assuming that the column player is playing strategy (½ C_{1}, ½ C_{2})—the expected utility of R_{2} is (½)(–b) + (½)(a), or ½(–b + a), or ½(a – b). Because the expected utilities of R_{1} and R_{2} are both ½(a – b), the expected utilities of all their mixtures is ½(a – b), too, and so any mixture of R_{1} and R_{2} is as good (for the row player) as any other. Thus, the strategy (½ R_{1}, ½ R_{2}) is as good (for the row player) as any other. So, the row player cannot do better by unilaterally changing his strategy.
Now consider the game from the column player’s point of view. Consider, in particular, the column player’s pure strategies, C_{1} and C_{2}. If the row player is playing strategy (½ R_{1}, ½ R_{2}), then the expected utility of C_{1} is (½)(a) + (½)(–b), or ½(a – b). And—still assuming that the row player is playing strategy (½ R_{1}, ½ R_{2})—the expected utility of C_{2} is (½)(–b) + (½)(a), or ½(–b + a), or ½(a – b). Because the expected utilities of C_{1} and C_{2} are both ½(a – b), the expected utilities of all their mixtures is ½(a – b), too, and so any mixture of C_{1} and C_{2} is as good (for the column player) as any other. Thus, the strategy (½ C_{1}, ½ C_{2}) is as good (for the column player) as any other. So, the column player cannot do better by unilaterally changing her strategy.
Since neither the row player nor the column player can do better by unilaterally changing his or her strategy, [(½ R_{1}, ½ R_{2}), (½ C_{1}, ½ C_{2})] is an equilibrium pair.
C_{1} | C_{2} | |
R_{1} | 2 | 1 |
R_{2} | 1 | 2 |
section 5-3c:
table 5-25:
EU(C_{1}) = EU(C_{2})
p(3) + (1 – p)(–7) = p(1) + (1 – p)(4)
3p – 7 + 7p = p + 4 – 4p
10p – 7 = –3p + 4
13p = 11
p = 11/13EU(R_{1}) = EU(R_{2})
q(3) + (1 – q)(1) = q(–7) + (1 – q)(4)
3q + 1 – q = –7q + 4 – 4q
2q + 1 = –11q + 4
13q = 3
q = 3/13EU(game)
= (11/13)(3/13)(3) + (11/13)(1 – 3/13)(1) + (1 – 11/13)(3/13)(–7) + (1 – 11/13)(1 – 3/13)(4)
= (11/13)(3/13)(3) + (11/13)(10/13)(1) + (2/13)(3/13)(–7) + (2/13)(10/13)(4)
= (33/169)(3) + (110/169)(1) + (6/169)(–7) + (20/169)(4)
= 99/169 + 110/169 – 42/169 + 80/169
= 247/169
= 19/13table 5-26:
EU(C_{1}) = EU(C_{2})
p(4) + (1 – p)(5) = p(20) + (1 – p)(–3)
4p + 5 – 5p = 20p – 3 + 3p
–p + 5 = 23p – 3
–24p = –8
p = 8/24
p = 1/3EU(R_{1}) = EU(R_{2})
q(4) + (1 – q)(20) = q(5) + (1 – q)(–3)
4q + 20 – 20q = 5q – 3 + 3q
–16q + 20 = 8q – 3
–24q = –23
q = 23/24EU(game)
= (1/3)(23/24)(4) + (1/3)(1 – 23/24)(20) + (1 – 1/3)(23/24)(5) + (1 – 1/3)(1 – 23/24)(–3)
= (1/3)(23/24)(4) + (1/3)(1/24)(20) + (2/3)(23/24)(5) + (2/3)(1/24)(–3)
= (23/72)(4) + (1/72)(20) + (46/72)(5) + (2/72)(–3)
= 92/72 + 20/72 + 230/72 – 6/72
= 336/72
= 14/3
p. 141:
To verify that [(2/3 mountain, 1/3 plain), (4/9 mountain, 5/9 plain)] is an equilibrium pair, we need to verify that neither the row player nor the column player can do better by unilaterally changing his strategy.
First consider the game from the row player’s point of view. Consider, in particular, the row player’s pure strategies, mountain and plain. If the column player is playing strategy (4/9 mountain, 5/9 plain), then the expected utility of mountain (for the row player) is (4/9)(–50) + (5/9)(100), or –200/9 + 500/9, or 300/9, or 100/3. And—still assuming that the column player is playing strategy (4/9 mountain, 5/9 plain)—the expected utility of plain (for the row player) is (4/9)(200) + (5/9)(–100), or 800/9 – 500/9, or 300/9, or 100/3. Because the expected utilities of mountain and plain are both 100/3, the expected utilities of all their mixtures is 100/3, too, and so any mixture of mountain and plain is as good (for the row player) as any other. Thus, the strategy (2/3 mountain, 1/3 plain) is as good (for the row player) as any other. So, the row player cannot do better by unilaterally changing his strategy.
Now consider the game from the column player’s point of view. Consider, in particular, the column player’s pure strategies, mountain and plain. If the row player is playing strategy (2/3 mountain, 1/3 plain), then the expected utility of mountain (for the column player) is (2/3)(–50) + (1/3)(200), or –100/3 + 200/3, or 100/3. And—still assuming that the row player is playing strategy (2/3 mountain, 1/3 plain)—the expected utility of plain (for the column player) is (2/3)(100) + (1/3)(–100), or 200/3 – 100/3, or 100/3. Because the expected utilities of mountain and plain are both 100/3, the expected utilities of all their mixtures is 100/3, too, and so any mixture of mountain and plain is as good (for the column player) as any other. Thus, the strategy (4/9 mountain, 5/9 plain) is as good (for the column player) as any other. So, the column player cannot do better by unilaterally changing her strategy.
Since neither the row player nor the column player can do better by unilaterally changing his or her strategy, [(2/3 mountain, 1/3 plain), (4/9 mountain, 5/9 plain)] is an equilibrium pair.
We also have to find the value of this game. We do that as follows:
EU(game)
= (2/3)(4/9)(–50) + (2/3)(1 – 4/9)(100) + (1 – 2/3)(4/9)(200) + (1 – 2/3)(1 – 4/9)(–100)
= (2/3)(4/9)(–50) + (2/3)(5/9)(100) + (1/3)(4/9)(200) + (1/3)(5/9)(–100)
= (8/27)(–50) + (10/27)(100) + (4/27)(200) + (5/9)(–100)
= –400/27 + 1000/27 + 800/27 – 500/27
= 900/27
= 100/3
problems on pp. 143–144
# | claim | justification |
1 | Suppose Able plays (1/2 R_{1}, 1/2 R_{2}). | assumption for proving conditional |
2 | The expected utility, for Baker, of playing C_{2} is (1/2)(0) + (1/2)(2) | 1, table 5-29 |
3 | The expected utility, for Baker, of playing C_{2} is 1. | 2, simplified |
4 | 1 is greater than 1/2. | math |
5 |
The expected utility, for Baker, of playing C_{2} is greater than 1/2. |
3 and 4 |
6 |
If Able plays (1/2 R_{1}, 1/2 R_{2}), then Baker can do better than 1/2 by playing C_{2}. |
1–5 |
do not call | call | |
do not call | –1, –1 | 2, 1 |
call | 1, 2 | 0, 0 |
confess | do not | |
confess | –15, –10 | –5, –25 |
do not | –16, –1 | –6, –2 |
# | claim | justification |
1 | If the row player confesses, then confessing is better for the column player. | table, row 1 |
2 | If the row player does not confess, then confessing is better for the column player. | table, row 2 |
3 | Confessing is the dominant strategy for the column player. | 1 and 2 |
4 | If the column player confesses, then confessing is better for the row player. | table, column 1 |
5 |
If the column player does not confess, then confessing is better for the row player. |
table, column 2 |
6 |
Confessing is the dominant strategy for the row player. |
4 and 5 |
7 |
(confess, confess) is an equilibrium pair. |
3 and 6 |
8 |
The payoff, for each player, of (confess, confess) is worse than the payoff of (do not, do not). |
table, cells for (confess, confess) and (do not, do not) |
9 |
The game is a prisoner’s dilemma. |
7 and 8 |
The situation can be set up as follows:
$2.00 $1.90 $1.80 $1.70 $1.60 $2.00 $25, $25 $0, $40 $0, $30 $0, $20 $0, $10 $1.90 $40, $0 $20, $20 $0, $30 $0, $20 $0, 10 $1.80 $30, $0 $30, $0 $15, $15 $0, $20 $0, $10 $1.70 $20, $0 $20, $0 $20, $0 $10, $10 $0, $10 $1.60 $10, $0 $10, $0 $10, $0 $10, $0 $5, $5 Consider the following dominance considerations:
- For each player, $2.00 is dominated (by $1.90, and also by $1.80, for that matter). So neither player will play $2.00
- For each player, if the row and column for $2.00 is ruled out, $1.90 is dominated (by $1.80, and indeed $1.70). So neither player will play $1.90.
- For each player, if the rows and columns for $2.00 and $1.90 are ruled out, $1.80 is dominated (by $1.70). So neither player will play $1.80.
- For each player, if the rows ands columns for $2.00, $1.90, and $1.80 are ruled out, $1.70 is dominated (by $1.60). So neither player will play $1.70.
- For each player, if the rows and columns for $2.00, $1.90, $1.80, and $1.70 are ruled out, then all that is left is $1.60. That leads them to split the lowest total profit.
First, we have this descriptive matrix:
hold fast veer hold fast dead hero, dead hero live hero, chicken veer chicken, live hero no change, no change Second, we have the following utility assignments:
- live hero: 1
- no change: 0
- dead hero: –1
- chicken: –2
Combining these two pieces of information, we have this matrix:
hold fast veer hold fast –1, –1 1, –2 veer –2, 1 0, 0 For each player, hold fast dominates, and so (hold fast, hold fast) is the equilibrium pair, and yet the payoff of (hold fast, hold fast) is worse, for each player, than the payoff of (veer, veer). So this game is a variant of the prisoner’s dilemma.
First, we have this descriptive matrix:
move sack stay home move sack mule is fed by both of us, mule is fed by both of us mule is fed by me, mule is fed by the other guy stay home mule is fed by the other guy, mule is fed by me mule is not fed, mule is not fed Second, we have the following utility assignments:
- mule is fed by the other guy: 2
- mule is fed by both of us: 1
- mule is fed by me: 0
- mule is not fed: –1
Combining these two pieces of information, we have this matrix:
move sack stay home move sack 1, 1 0, 2 stay home 2, 0 –1, –1 There are two equilibrium strategy pairs—(move sack, stay home) and (stay home, move sack)—but one is better for one player, and the other is better for the other players. So we have a clash of wills.
Again, we have this descriptive matrix:
move sack stay home move sack mule is fed by both of us, mule is fed by both of us mule is fed by me, mule is fed by the other guy stay home mule is fed by the other guy, mule is fed by me mule is not fed, mule is not fed Second, we have the following utility assignments:
- mule is fed by the other guy: 2
- mule is fed by both of us: 1
- mule is not fed: 0
- mule is fed by me: –1
Combining these two pieces of information, we have this matrix:
move sack stay home move sack 1, 1 –1, 2 stay home 2, –1 0, 0 For each player, there is a dominant strategy—stay home—but the outcome of (stay home, stay home) is worse, for each, than the outcome of (move sack, move sack). So we have a prisoner’s dilemma.
EU(remaining a cheater)
= (1 – p)u + p[q(1) + (1 – q)u]
= u – pu + p[q + u – qu]
= u – pu + pq + pu – pqu
= u + pq – pqu
= u + pq(1 – u)EU(becoming a cooperator)
= (1 – p)[(1 – q)u + q(0)] + p[r(u’) + (1 – r)u]
= (1 – p)[u – qu] + p[ru’ + u – ru]
= u – qu – pu + pqu + pru’ + pu – pru
= u – qu + pqu + pru’ – pru
= u – qu(1 – p) + pr(u’ – u)
= u + pr(u’ – u) – qu(1 – p)
p = 1/2
q = 3/4
r = 1/4
u = 1/4
u’ = 3/4
then the expected gain from remaining a cheater is
(1/2)(3/4)(1 – 1/4)
= (3/8)(3/4) = 9/32
and the expected gain from becoming a cooperator is
(1/2)(1/4)(3/4 – 1/4) – (3/4)(1/4)(1 – 1/2)
= (1/8)(2/4) – (3/16)(1/2)
= 2/16 – 3/32
= 4/32 – 3/32
= 1/32
Since the expected gain from remaining a cheater is more than the expected gain from becoming a cooperator, it is more rational to remain a cheater. (This is not surprising, considering that there is a fairly high value for q (the possibility that a cheater can exploit a cooperator) and a low value for r (the probability of being able to cooperate with a fellow cooperator).
Note, by the way, that the foregoing approach just compares the utility gains, relative to the status quo, of entering in the specified scenario as either a cheater or a cooperator. You can also compare the total utilities (not just utility gains) to be expected from either course of action; if you did that, then you would just add the value of u to each of the two values. Since u = 1/4, the total utility of remaining a cheater is 9/32 + 1/4, or 9/32 + 8/32, or 17/32; and the total utility of becoming a cooperator is 1/32 + 1/4, or 1/32 + 8/32, or 9/32.
# | claim | justification |
1 | Suppose some method—call it M—is dictatorial. | assumption for proving conditional |
2 | There is some citizen—call her C—whose ordering M always yields as the social ranking (regardless of the other aspects of the profile containing C's ordering). | 1, definition of dictatorship |
3 | Suppose that M does not satisfy the Pareto condition. | assumption for deriving a contradiction |
4 | There exists some profile—call it P—in which every citizen ranks one alternative—call it x—above another—call it y—and for which M does not yield a social ordering that ranks x above y. | 3, definition of Pareto condition |
5 | In P, C ranks x above y. | 4 |
6 |
For P, M yields a social ordering that ranks x above y. |
5 and 2 |
7 |
We have a contradiction. |
4 and 6 |
8 |
We can reject the assumption that M does not satisfy the Pareto condition. |
3 and 7 |
9 | M satisfies the Pareto condition. | 8 |
10 | If some method is dictatorial, then it satisfies the Pareto condition. | 1–9 |
(We will assume that the last phrase of this means the same as ‘places the members of S in the same relative positions in the social orderings it yields for both P_{1} and P_{2}’.)
There are a couple of ways to prove this. Here is a direct way:
# | claim | justification |
1 | Suppose, for some method M, that if P_{1} and P_{2} are two profiles and S is any subset of the set of alternatives, then if the citizens’ relative rankings of the members of S are the same for P_{1} and P_{2}, then M places the members of S in the same relative positions in its social orderings for both P_{1} and P_{2}. | assumption for proving conditional |
2 | If P_{1} and P_{2} are two profiles and S is any two-alternative subset of the set of alternatives, then if the citizens’ relative rankings of the members of S are the same for P_{1} and P_{2}, then M places the members of S in the same relative positions in its social orderings for both P_{1} and P_{2}. | 1 |
3 | M satisfies condition I. | 2 |
4 | The condition stated in line 1 implies condition I. | 1–3 |
And here is an indirect way (based on proof by contradiction):
# | claim | justification |
1 | Suppose, for some method M, that if P_{1} and P_{2} are two profiles and S is any subset of the set of alternatives, then if the citizens’ relative rankings of the members of S are the same for P_{1} and P_{2}, then M places the members of S in the same relative positions in its social orderings for both P_{1} and P_{2}. | assumption for proving conditional |
2 | Suppose M does not satisfy condition I. | assumption for deriving a contradiction |
3 | There exists some alternatives x and y, and some profiles F_{1} and F_{2}, such that each citizen ranks x and y in the same order in F_{1} and F_{2}, but M yields social orderings for F_{1} and F_{2} in which x and y are not in the same order. | 2, definition of condition I |
4 | There exists some subset of the set of alternatives such that the citizens’ relative rankings of the members of that subset are the same for F_{1} and F_{2} but M does not place the members of that subset in the same relative positions in its social orderings for both F_{1} and F_{2}. | 3 |
5 | We have a contradiction. | 1 and 4 |
6 |
We can reject the assumption that M does not satisfy condition I. |
2 |
7 |
M satisfies condition I. |
6 |
8 |
The condition stated in line 1 implies condition I. |
1–7 |