MATH 4321 Tutorial 1 Solutions

1 Review

1. Essential elements of a game: players, actions, outcomes, preferences.

2. Representing a game: Game matrix and game tree.Game tree: 1. The game usually contains several procedures. 2. The whole process of thegame is clear.Game matrix: 1. Straightforward. 2. Easy to find saddle points and equilibria.

3. Rational preference relation: a preference relation that is complete and transitive.

4. Payoff function: A payoff function u : X → R is used to represent a preference relation.u(x) ≥ u(y) iff x � y for any outcome x and y.

5. Expected payoff:

• Discrete case: Let u(x) : X → R be the player’s payoff function over outcomes X ={x1, x2, ..., xn} and let p = {p1, p2, ..., pn} be the probability such that pk = P[x = xk].The player’s expected payoff over probability p is defined by

E[u(x)|p] =n∑

k=1

pku(xk).

• Continuous case: Let u(x) : X → R be the player’s payoff function over outcomesX = [x, x] and let the probability be given by a cumulative distribution F (x) withdensity f(x). Then the player’s expected payoff is defined by

E[u(x)] =

∫ x

x

u(x)f(x)dx.

A rational player chooses an action that maximizes the expected payoff.

6. Pure strategy vs mixed strategy: A deterministic plan of action. The player takes theaction with probability 0 or 1. (in contrast to mixed strategy)

7. Normal-form game: A normal-form game includes three components as follows:

• A finite set of players, N = {1, 2, · · · , n}.• A collection of sets of pure strategies, {S1, S2, · · · , Sn}.• A set of payoff functions, {v1, v2, · · · , vn}, each assigning a payoff value to each combination

of chosen strategies, that is, a set of functions vi : S1 × S2 × · · · × Sn → R for eachi ∈ N .

8. Saddle point in pure strategies: A particular row i∗ and column j∗ is a saddle pointunder pure strategies iff

aij∗ ≤ ai∗j∗ ≤ ai∗j (smallest in a row and largest in a column)

for all rows i = 1, ..., n and columns j = 1, ...,m.Note: A saddle point in pure strategies may not exist and may not be unique.

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9. Value of a zero-sum game under pure strategies: Given a game matrix A = (aij), thelower value of the game is

v− = maxi

minjaij. (the largest minimum)

The upper value of the game is

v+ = minj

maxiaij. (the smallest maximum)

Note: It is known that v− ≤ v+.

A zero-sum game has a saddle point and has a value under pure strategies iff v+ = v−. Thevalue of the zero-sum game is the payoff at the saddle point. Write

v = v(A) = v− = v+.

10. Best responses: Player i’s best response s∗i to the strategies s−i chosen by the other playersis the strategies that yield him the greatest payoff. i.e.

πi(s∗i , s−i) ≥ πi(si, s−i), ∀ si 6= s∗i .

11. Lemma for multiple saddle points: In a two-person zero-sum game, suppose (σ1, σ2)and (τ1, τ2) are two saddle strategies, then (σ1, τ2) and (τ1, σ2) are also saddle strategies.Also, their payoffs are the same. (If two corners of the rectangle are saddle points, so arethe other two)

12. Mixed strategies for zero-sum games: Define the set of n-component probabilityvectors:

Sn = {(z1, z2, ..., zn) : 0 ≤ zk ≤ 1, k = 1, ..., n andn∑

k=1

zk = 1}.

A mixed strategy is defined by probability vectors X = (x1, ..., xn) ∈ Sn for player Iand Y = (y1, ..., ym) ∈ Sm for player II, where xi = P[Player I chooses row i] and yj =P[Player II chooses column j].Note: The strategies of the players are dependent on the payoff aij in game matrix A =(aij)n×m but are independent of the opponent’s strategies.

13. Stochastic steady state: Each player chooses his actions probabilistically according tothe same unchanging distribution.

14. Expected payoff under mixed strategies: The expected payoff to Player I of the matrixgame is

E(X, Y ) =n∑

i=1

m∑j=1

aijP[I uses row i and II uses column j]

=n∑

i=1

m∑j=1

aijP[I uses row i]P[II uses column j] (due to independence)

=n∑

i=1

m∑j=1

aijxiyj = (x1 ... xn)A

y1...ym

= XAY T .

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15. Saddle point in mixed strategies for zero-sum games: A saddle point in mixedstrategies is a pair (X∗, Y ∗) of probability vectors X∗ ∈ Sn, Y ∗ ∈ Sm, that satisfies

E(X, Y ∗) ≤ E(X∗, Y ∗) ≤ E(X∗, Y ), ∀X ∈ Sn and Y ∈ Sm.

16. Equality of payoff theorem: If any optimal mixed strategy for a player has a strictlypositive probability of using a row or column, then that row or column played against anyoptimal opponent strategy will yield the value. i.e.

(i) x∗i > 0 =⇒ E(i, Y ∗) = E(X∗, Y ∗) = v(A);

(ii) y∗j > 0 =⇒ E(X∗, j) = E(X∗, Y ∗) = v(A). Also,

(iii) E(X∗, j) > v(A) =⇒ yj = 0;

(iv) E(i, Y ∗) < v(A) =⇒ xi = 0.

17. Theorem: Let A = (aij) be an n ×m game with value v(A). Let w be a real number. Ifwe can find X∗ and Y ∗ such that E(i, Y ∗) ≤ w ≤ E(X∗, j), i = 1, 2, · · · , n, j = 1, 2, · · · ,m,then w = v(A) and (X∗, Y ∗) is a saddle point for the game.

18. Lemma: The value of a zero-sum game is given by

v(A) = minY ∈Sm

max1≤i≤n

E(i, Y ) = maxX∈Sn

min1≤j≤m

E(X, j),

or v(A) = max1≤i≤n

E(i, Y ∗) = min1≤j≤m

E(X∗, j).

19. Notation: For an n×m matrix A = (aij) we denote the jth column vector of A by Aj andthe ith row vector of A by iA. So

Aj =

a1j...anj

and iA = (ai1 ... aim).

We may also write

E(i, Y ) = iAYT =

m∑j=1

aijyj, E(X, j) = XAj =n∑

i=1

xiaij, E(i, j) = aij.

Note too that

E(X, Y ) =n∑

i=1

xiE(i, Y ) =m∑j=1

yjE(X, j).

2 Problems

1. (sequential decisions) A patient is very sick and was found that would die in 6 monthsif he goes untreated. The only available treatment is risky surgery. The patient is expectedto live for 12 months if the surgery is successful, but the probability that the surgery willfail and the patient will die immediately is 0.3. The patient now has lived for 3 months andwants to make his decisions.

(a) Draw a decision tree for this decision problem.

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(b) Let v(x) be the patient’s payoff function, where x is the number of months until death.Assuming that v(12) = 1 and v(0) = 0, what is the largest payoff the patient can havefor living 3 months so that having surgery is a best response?For the rest of the problem, assume that v(3) = 0.8.

(c) A test is available that will provide some information that predicts whether or not surgerywill be successful. A positive test implies an increased likelihood that the patient willsurvive the surgery as follows:True-positive rate: The probability that the results of this test will be positive if surgeryis to be successful is 0.90.False-positive rate: The probability that the results of this test will be positive if thepatient will not survive the operation is 0.10.What is the probability of a successful surgery if the test is positive?

(d) Assuming that the patient has the test done, at no cost, and the result is positive, shouldsurgery be performed?

(e) It turns out that the test may have some fatal complications; that is, the patient maydie during the test. The probability of death during the test is 0.005. Draw a decisiontree for this revised problem. Should the patient opt to have the test prior to decidingon the operation?

Solution:

(a) The decision tree is shown as following:

(b) In this case the patient has only 3 months left. Then the payoff if he chooses not tohave the surgery is v(3). Then the decision tree is shown as the following:

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Expected payoff from having the surgery is

0.7× v(12) + 0.3× v(0) = 0.7.

Then having surgery is the best response for the patient when v(3) ≤ 0.7. Then thelargest value of v(3) is 0.7.

(c) We calculate the probability by Bayes’ rule. Denote “test is positive” by P , “test isnegative” by N , “successful surgery” by S and “surgery fails” by F . Then

Pr(successful surgery| test is positive) = Pr(S|P )

=Pr(P |S)Pr(S)

Pr(P |S)Pr(S) + Pr(P |F )Pr(F )

=0.90× 0.7

0.90× 0.7 + 0.10× 0.3=

21

22≈ 0.95.

Then if the test is positive, the surgery will be successful with probability 0.95.

(d) The probability for a successful surgery turns to 0.95. The new game tree is

The expected payoff of having surgery is given by

0.95× v(12) + 0.05× v(0) = 0.95 > 0.8 = v(3).

Therefore, the patient will still choose to have the surgery.

(e) To simplify the decision tree, we first examine decisions of the patient under someconditions:

• If the patient does not take the test, then the probability of a successful surgeryis 0.7. The expectation of having surgery is 0.7 (calculated in (b)) < 0.8 = v(3).Then the patient will choose not to have surgery if he does not take the test.

• If the patient takes the test and the test turns out to be negative, then we can easilycalculate (similar to (c)) that the probability of a successful surgery is 7

34≈ 0.21.

The expected payoff of having surgery is less than 0.8. Then the patient will alsonot have surgery if the test result is negative.

Then the decision tree is shown as the following:

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We need to get the probability values Pr(P ) and Pr(N) first. From (c) we know that

Pr(S|P ) = Pr(P⋂

S)Pr(P )

= 0.95 and Pr(P |S) = Pr(P⋂

S)Pr(S)

= 0.9. Then we get Pr(P⋂S) =

0.9× Pr(S) = 0.63 and thus Pr(P ) = Pr(P⋂

S)0.95

= 6395≈ 0.66, Pr(N) = 1− 0.66 = 0.34.

Then the expected payoff of taking the test should be

0.995× 0.66× 0.95× v(12) + 0.995× 0.66× 0.05× v(0) + 0.995× 0.34× v(3)

+0.005× v(0) ≈ 0.62 + 0 + 0.27 + 0 = 0.89.

Since expected payoff 0.89 > 0.8 = v(3), we conclude that the test should be takenalthough it is risky.

2. (pure-strategy zero-sum game) In a football game, the offense has two strategies: runor pass. The defense also has two strategies: defend against the run, or defend againstthe pass. Because of the pace of the game, the decisions of the two players must be madesimultaneously. If the defense makes the right choice, the offense cannot gain any yard ifshe chooses to pass; and can only gain 3 yards if she chooses to run. On the other hand, ifthe defense fails to make the right choice, the offense will successfully gain 6 yards and xyards if she chooses to run and to pass, respectively. The payoff of each outcome can be thenumber of yards gained.

(a) Model this as a zero-sum game.

(b) Assuming that x > 0, find the value of x so that this game has a saddle point in purestrategies.

Solution:

(a) This game can be modelled as a zero-sum game. The reason is whenever one player gainscertain number of yards, the other will loss the same number of yards correspondingly.So their payoffs under the same outcome sum up to zero. We let the offense be the row

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player and the defense be the column player and get the following game matrix:

defend run defend passrun 3 6pass x 0

(b) Our purpose is to find x guaranteeing that v− = v+.

defend run defend passrun 3 6 min

ja1j = 3

pass x 0 minja2j = 0

maxiai1 max

iai2 = 6 v− = 3

v+ = min(maxiai1, 6)

We find from above analysis that in order to get v− = v+, the value of maxiai1 should

be 3, which implies that 0 < x ≤ 3.

3. (optimal mixed strategies) Consider the game with matrix

A =

−2 3 5 −23 −4 1 −6−5 3 2 −1−1 −3 2 2

.Someone claims that the strategies X∗ = (1

9, 0, 8

9, 0) and Y ∗ = (0, 7

9, 29, 0) are optimal.

(a) Is that correct? Why or why not?

(b) If X∗ = (1333, 533, 0, 15

33) is optimal and v(A) = −26

33, find Y ∗.

Solution:

(a) According equality of payoff theorem, if X∗ and Y ∗ are a pair of optimal strategies withx1 > 0 and x3 > 0, we must have E(1, Y ∗) = E(3, Y ∗) = v. However, E(1, Y ∗) = 31

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and E(3, Y ∗) = 259

. E(1, Y ∗) 6= E(3, Y ∗). Then we can see X∗ and Y ∗ are not optimalstrategies.

(b) Again, we use equality of payoff theorem. We have x1 > 0, x2 > 0, x3 = 0, x4 > 0 andvalue v(A) = −26

33, which implies that

E(1, Y ∗) = −2y1 + 3y2 + 5y3 − 2y4 = −26

33,

E(2, Y ∗) = 3y1 − 4y2 + y3 − 6y4 = −26

33,

E(4, Y ∗) = −y1 − 3y2 + 2y3 + 2y4 = −26

33,

y1 + y2 + y3 + y4 = 1.

Then we get y1 = 5299, y2 = 8

33, y3 = 0, y4 = 23

99, so Y ∗ = (52

99, 833, 0, 23

99).

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