Part 7 · Chapter 29

Game Theory and Decision Analysis

Chapter 28 optimized against a random world — one that is uncertain, but indifferent. This chapter confronts two harder situations. In the first, the future is uncertain and you have no probabilities at all, only a table of what each choice pays in each possible state; decision analysis supplies the criteria for choosing anyway. In the second, the "state of the world" is chosen by an opponent — a rival who is also optimizing, and against you; game theory asks what it means to play well when the other side is trying to make you play badly. Nature does not adapt to your plan, but an adversary does, and that single difference reshapes the entire notion of an optimal decision.

Optimization Techniques Prof. Mithun Mondal Reading time ≈ 60 min
i What you'll learn
  • The payoff matrix, and the difference between deciding against passive nature and a strategic opponent.
  • Decision criteria under uncertainty — maximin, maximax, Hurwicz, Laplace and minimax regret.
  • Decisions under risk: expected monetary value, opportunity loss, and the expected value of perfect information.
  • Two-person zero-sum games, pure strategies, and the saddle point where maximin meets minimax.
  • Dominance reduction and mixed strategies for games with no saddle point.
  • The minimax theorem, its link to linear programming, and Nash equilibrium in non-zero-sum games.
Section 29-1

Nature vs an Opponent

Both halves of this chapter start from the same object: a payoff matrix. Its rows are the decision maker's actions, its columns are the possible states, and each entry is what that action pays in that state. Everything turns on one question about the columns — who chooses which one occurs?

Decision analysisGame theory
Columns chosen byNature — indifferent, non-strategicAn opponent — strategic, adversarial
Opponent's aimNone; states just happenTo minimise your payoff
ProbabilitiesUnknown (uncertainty) or known (risk)Not assumed; the opponent is rational
Solution conceptA criterion (maximin, EMV, …)An equilibrium (saddle point, Nash)

The distinction is not cosmetic. Against nature you may reasonably be optimistic, or assign probabilities, or gamble on a good column appearing — nature will not punish you for it. Against an opponent, optimism is fatal: whatever you hope they will do, they will do the opposite if it hurts you. This is why game theory gravitates to pessimistic, security-first reasoning, and why the two fields, despite sharing a matrix, arrive at such different notions of a good decision.

The same maximin, two different meanings. In decision analysis, maximin is one criterion among several — the cautious option, defensible but not obligatory. In game theory against a rational adversary, maximin is not a mood but a necessity: it is the most you can guarantee when the other side is actively working against you. The mathematics is identical; the justification is completely different.
Section 29-2

Decision Criteria Under Uncertainty

When the states are governed by nature and no probabilities are available, there is no single "correct" choice — only criteria that encode different attitudes to risk. Five are standard, and a well-posed decision problem should be examined through several, because they can disagree.

CriterionRule (for a maximisation of profit)Attitude
Maximin (Wald)Maximise the row minimum — best worst-casePessimistic / cautious
MaximaxMaximise the row maximum — best best-caseOptimistic / aggressive
HurwiczMaximise \(\alpha(\text{row max}) + (1-\alpha)(\text{row min})\)A blend, tuned by optimism \(\alpha\)
LaplaceAssume equal probabilities; maximise the row averageIndifference (insufficient reason)
Minimax regret (Savage)Minimise the maximum regret (opportunity loss)Fear of hindsight

The first four read directly off the payoff rows. The fifth needs a transformation worth understanding, because it recurs under risk as opportunity loss. The regret of an action in a state is how much better the best action for that state would have done — what you would kick yourself for, in hindsight, having missed:

Regret (opportunity loss)
\[ r_{ij} = \Big(\max_{k} \, v_{kj}\Big) - v_{ij} \]

Build the regret matrix column by column — subtract each entry from the best entry in its column — then choose the action whose largest regret is smallest. The Hurwicz criterion deserves a second look too: its optimism coefficient \(\alpha \in [0,1]\) is a dial, with \(\alpha = 0\) collapsing to pure maximin and \(\alpha = 1\) to pure maximax. As Example 1 shows, sliding \(\alpha\) can flip the recommended action — a reminder that the "answer" here is a function of temperament as much as of the numbers.

Section 29-3

Decisions Under Risk & EVPI

When probabilities can be assigned to the states — from data, models or judgement — the problem moves from uncertainty to risk, and a single dominant criterion emerges: choose the action with the best expected monetary value, the probability-weighted average payoff.

🔑
Expected value and opportunity loss
\[ \text{EMV}(a_i) = \sum_j p_j\, v_{ij}, \qquad \text{EOL}(a_i) = \sum_j p_j\, r_{ij} \]

Maximising EMV and minimising the expected opportunity loss (EOL) always select the same action — they are two views of one decision. The EOL of the best action is never zero unless one action is best in every state; that residual is exactly the cost imposed by uncertainty.

That residual has a name and a use. The expected value of perfect information is what you could gain if an oracle told you the true state before you chose. Compute the expected payoff with perfect information by taking the best action in each state and weighting by probability; the EVPI is the improvement that buys over acting on EMV alone — and it equals the minimum EOL exactly.

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Expected value of perfect information
\[ \text{EVPI} = \underbrace{\sum_j p_j \max_i v_{ij}}_{\text{expected value with perfect info}} - \underbrace{\max_i \text{EMV}(a_i)}_{\text{best without it}} = \min_i \text{EOL}(a_i) \]

EVPI is a ceiling on the price of any forecast, survey or test: no information source, however good, can be worth more than perfect information. It is the same idea met in Chapter 28, now read off a decision table instead of a scenario tree.

When decisions come in sequence — test, then decide; or decide, then observe, then decide again — the natural tool is the decision tree. Square nodes are decisions the analyst controls; circular nodes are chance events governed by probability; branches carry payoffs. The tree is solved by rollback: starting from the leaves, take expectations at chance nodes and maxima at decision nodes, folding the future back to the present until the first decision's value — and the best opening move — is known.

decide a₁ a₂ p 50 5 55 10 □ max · ○ expectation · rollback
A decision tree solved by rollback
Hurwicz: α·max + (1−α)·min α α=0 maximin α=1 maximax the dial from pessimist to optimist
The Hurwicz optimism dial
Section 29-4

Zero-Sum Games & Saddle Points

Now the column is chosen by an adversary. A two-person zero-sum game has two players with exactly opposed interests: whatever one wins, the other loses, so a single payoff matrix — read as gains to the row player \(A\) and losses to the column player \(B\) — describes the whole game. \(A\) picks a row to maximise the payoff; \(B\) picks a column to minimise it; and each must choose knowing the other is a rational optimizer.

Security-first reasoning is now forced. Player \(A\), assuming the worst, evaluates each row by its minimum and picks the row whose minimum is largest — the maximin, the most \(A\) can guarantee. Player \(B\), symmetrically, evaluates each column by its maximum and picks the column whose maximum is smallest — the minimax, the most \(B\) can be forced to concede. When these two coincide, the game is settled.

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Saddle point & value of the game
\[ \max_i \min_j v_{ij} \;=\; \min_j \max_i v_{ij} \;=\; V \]

When the maximin equals the minimax, that common number \(V\) is the value of the game, and the cell achieving it is a saddle point — simultaneously the smallest entry in its row and the largest in its column. Both players have an optimal pure strategy (a single row, a single column), and neither can profit by deviating unilaterally. It is an equilibrium in the strictest sense.

The name is geometric. A saddle point is a mountain pass: the lowest point along the ridge and the highest point along the valley, at once. \(A\) climbing the valley and \(B\) descending the ridge meet exactly there, and neither can do better by moving. Not every game has one, though — and when maximin \(<\) minimax, there is a gap, no pure-strategy equilibrium exists, and the players must randomise, which is the subject of the next section.

saddle = V B: minimax ↓ A: maximin ↑
The saddle point as a mountain pass
B₁ B₂ B₃ row min A₁ 3 5 4 3 A₂ 2 6 1 1 A₃ 4 7 6 4 col max 4 7 6 maximin = minimax = 4 saddle at (A₃, B₁)
Locating the saddle point in the matrix
Section 29-5

Dominance & Mixed Strategies

Before solving any game, prune it. A row is dominated if some other row is at least as good for \(A\) in every column — \(A\) would never play the worse one, so delete it. A column is dominated if some other column is at least as good for \(B\) (that is, smaller in every row) — delete it too. Dominance often shrinks a large matrix to a \(2\times2\) that can be solved by hand, and never removes an optimal strategy.

When no saddle point survives, no single row is safe: whatever pure strategy \(A\) commits to, \(B\) can exploit it. The escape is to become unpredictable — to play a mixed strategy, choosing rows at random with carefully set probabilities so that \(B\) cannot profit from guessing. The optimal mix makes \(A\)'s expected payoff the same whichever column \(B\) picks; that indifference is exactly what removes \(B\)'s ability to exploit.

🔑
Mixed-strategy solution of a 2×2 game
\[ p = \frac{v_{22}-v_{21}}{(v_{11}+v_{22})-(v_{12}+v_{21})}, \qquad q = \frac{v_{22}-v_{12}}{(v_{11}+v_{22})-(v_{12}+v_{21})}, \qquad V = \frac{v_{11}v_{22}-v_{12}v_{21}}{(v_{11}+v_{22})-(v_{12}+v_{21})} \]

Here \(p\) is \(A\)'s probability of playing row 1 (and \(1-p\) row 2), \(q\) is \(B\)'s probability of playing column 1, and \(V\) is the value of the game. The formulas come straight from setting \(A\)'s expected payoff equal against each of \(B\)'s columns — the "oddments" method is just a bookkeeping shortcut for the same equations.

The graphical view makes the logic vivid. Plot \(A\)'s expected payoff against \(p\) as \(p\) runs from 0 to 1: against each of \(B\)'s columns it is a straight line, and the two lines cross. To the left of the crossing one column is worse for \(A\); to the right, the other. \(B\), minimising, always pushes \(A\) onto the lower of the two lines — so \(A\)'s guaranteed payoff is the lower envelope, and its highest point is exactly the crossing. The optimal \(p\) is where the lines meet, and its height is the value \(V\).

payoff p (prob. of A₁) 0 0 1 vs B₁: 2+2p vs B₂: 3−2p p = 1/4 V=2.5 B forces A onto the lower envelope (teal); its highest point is the value
Graphical solution of a 2×2 mixed-strategy game
Section 29-6

Minimax Theorem & Nash Equilibrium

That every finite two-person zero-sum game has a solution is not obvious — it is a theorem, and a landmark one. Von Neumann's minimax theorem (1928) guarantees that once mixed strategies are allowed, the maximin and the minimax always meet: there is a value \(V\) that \(A\) can guarantee to get and \(B\) can guarantee to hold \(A\) to, simultaneously.

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The minimax theorem
\[ \max_{\mathbf{x}} \min_{\mathbf{y}} \ \mathbf{x}^{\mathsf T} V \mathbf{y} \;=\; \min_{\mathbf{y}} \max_{\mathbf{x}} \ \mathbf{x}^{\mathsf T} V \mathbf{y} \]

over all mixed strategies \(\mathbf{x}\) (for \(A\)) and \(\mathbf{y}\) (for \(B\)). Allowing randomisation closes the maximin–minimax gap that pure strategies could not. Crucially, finding the optimal mix is a linear program — every zero-sum game reduces to an LP, and the strong duality of Chapter 10 is the minimax theorem in disguise, the two players' problems being primal and dual.

Zero-sum is a special world, though — pure opposition, so one number \(V\) describes it. Most real conflicts are non-zero-sum: the players' interests overlap as well as clash, outcomes can be jointly good or jointly bad, and there is no single "value." The generalising idea is John Nash's, and it is the cornerstone of modern game theory.

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Nash equilibrium
A choice of strategies, one per player, such that no player can improve their own payoff by unilaterally changing strategy.

At equilibrium every player is doing the best they can given what the others are doing. A saddle point is exactly the Nash equilibrium of a zero-sum game; Nash's theorem (1950) extends the guarantee — every finite game has at least one equilibrium, possibly in mixed strategies — to any number of players and any payoffs.

The famous cautionary tale is the prisoner's dilemma. Two parties each choose to cooperate or defect; for each, defecting pays more whatever the other does, so "defect" is a dominant strategy and mutual defection is the unique Nash equilibrium — yet both would have been better off cooperating. The equilibrium is stable but collectively poor, which is the whole point: individual rationality and group welfare can diverge, and no player can fix it alone. Example 6 works the numbers; the phenomenon underlies price wars, arms races, and the tragedy of the commons alike.

B: Cooperate B: Defect A: Coop A: Def (10, 10) jointly best (2, 12) (12, 2) (5, 5) Nash eq. defect dominates for both → (Def, Def) stable, yet worse for both than (Coop, Coop)
The prisoner's dilemma
the matrix vs nature vs opponent uncertainty maximin…regret risk EMV · EVPI zero-sum saddle · mixed non-zero Nash one matrix, two worlds, four solution concepts
The decision-and-game landscape
Section 29-7

Synthesis & Comparison

The chapter's whole territory organises around the two questions of Section 29-1 — who controls the columns, and what you know about them.

SituationColumns fromWhat you knowSolve with
Decision under uncertaintyNaturePayoffs only, no probabilitiesMaximin / maximax / Hurwicz / Laplace / regret
Decision under riskNaturePayoffs and probabilitiesMax EMV; EVPI bounds the value of information
Zero-sum game (saddle)OpponentFully opposed payoffsPure strategies at the saddle point
Zero-sum game (no saddle)OpponentFully opposed payoffsDominance, then mixed strategies (an LP)
Non-zero-sum gameOpponent(s)Overlapping interestsNash equilibrium

Two threads run through it all. The first is that maximin is the safe default whenever the environment might be against you — a mere option against passive nature, an obligation against a rational adversary. The second closes a loop opened much earlier in the course: solving a game is solving a linear program, so the duality theory of Part 2 turns out to be the same mathematics as von Neumann's minimax. The optimization techniques of this book are not a list of separate tricks; under uncertainty and conflict they keep revealing themselves to be one another in different clothing — which is the note Part 7 has been sounding all along, and the one Chapter 30 will close on with real engineering case studies.

Section 29-8

Worked Examples

Examples 1–2 use one profit matrix, three actions against three states of nature: \(A_1 = (50,40,5)\), \(A_2 = (10,45,55)\), \(A_3 = (30,35,40)\) across states \(S_1,S_2,S_3\).

1 Five criteria, three answers

Problem. With no probabilities, apply maximin, maximax, Hurwicz (\(\alpha=0.5\)), Laplace, and minimax regret.

Solution. Row minima are \(A_1{=}5,\ A_2{=}10,\ A_3{=}30\), so maximin picks \(A_3\) (30). Row maxima are \(A_1{=}50,\ A_2{=}55,\ A_3{=}40\), so maximax picks \(A_2\) (55). Hurwicz at \(\alpha=0.5\): \(A_1{=}0.5(50){+}0.5(5){=}27.5\), \(A_2{=}32.5\), \(A_3{=}35\) — picks \(A_3\); note that raising \(\alpha\) to \(0.7\) gives \(A_2{=}41.5 > A_3{=}37\), flipping the choice, which shows how much rides on temperament. Laplace averages: \(A_1{=}31.7,\ A_2{=}36.7,\ A_3{=}35\) — picks \(A_2\). For minimax regret, the column maxima are \((50,45,55)\); the regret rows are \(A_1{=}(0,5,50)\), \(A_2{=}(40,0,0)\), \(A_3{=}(20,10,15)\), with maximum regrets \(50,40,20\) — so minimax regret picks \(A_3\) (20). The tally: three criteria say \(A_3\), two say \(A_2\). There is no single right answer under pure uncertainty — only a right answer for a stated attitude to risk.

2 Under risk: EMV and the price of a forecast

Problem. Now assign \(P(S_1){=}0.5\), \(P(S_2){=}0.3\), \(P(S_3){=}0.2\). Find the best action by EMV, then the EVPI.

Solution. \(\text{EMV}(A_1)=0.5(50)+0.3(40)+0.2(5)=25+12+1=38\); \(\text{EMV}(A_2)=5+13.5+11=29.5\); \(\text{EMV}(A_3)=15+10.5+8=33.5\). The best is \(A_1\) with EMV \(=38\) — under these probabilities the once-cautious-looking \(A_1\) wins, because the most likely state \(S_1\) is where it excels. For EVPI, the expected value with perfect information takes the best action per state — \(A_1{=}50\) in \(S_1\), \(A_2{=}45\) in \(S_2\), \(A_2{=}55\) in \(S_3\) — giving \(0.5(50)+0.3(45)+0.2(55)=25+13.5+11=49.5\). Then \(\text{EVPI}=49.5-38=11.5\). Cross-check via opportunity loss: the regret rows weighted by probability give \(\text{EOL}(A_1)=0.5(0)+0.3(5)+0.2(50)=11.5\), \(\text{EOL}(A_2)=20\), \(\text{EOL}(A_3)=16\); the minimum EOL is \(11.5\), exactly the EVPI. No forecast for these states should ever cost more than \(11.5\).

3 A game with a saddle point

Problem. Solve the zero-sum game (gains to \(A\)): rows \(A_1{=}(3,5,4)\), \(A_2{=}(2,6,1)\), \(A_3{=}(4,7,6)\) against \(B\)'s columns \(B_1,B_2,B_3\).

Solution. Player \(A\) takes each row's minimum: \(A_1{\to}3\), \(A_2{\to}1\), \(A_3{\to}4\); the maximin is \(4\) (row \(A_3\)). Player \(B\) takes each column's maximum: \(B_1{\to}4\), \(B_2{\to}7\), \(B_3{\to}6\); the minimax is \(4\) (column \(B_1\)). Since maximin \(=\) minimax \(=4\), the game has a saddle point at cell \((A_3,B_1)=4\) — check that \(4\) is the smallest in its row and the largest in its column, and it is. The value of the game is \(V=4\): \(A\) should always play \(A_3\), \(B\) should always play \(B_1\), and neither gains by deviating. Both optimal strategies are pure — no randomisation needed.

4 No saddle point: a mixed strategy

Problem. Solve \(\begin{smallmatrix} 4 & 1 \\ 2 & 3 \end{smallmatrix}\) (gains to \(A\), rows \(A_1,A_2\), columns \(B_1,B_2\)).

Solution. Row minima are \(1\) and \(2\) (maximin \(=2\)); column maxima are \(4\) and \(3\) (minimax \(=3\)). Since \(2 \ne 3\), there is no saddle point, so both players randomise. The denominator is \((v_{11}+v_{22})-(v_{12}+v_{21}) = (4+3)-(1+2) = 4\). Then \(A\)'s probability of row 1 is \(p=(v_{22}-v_{21})/4=(3-2)/4=\tfrac14\), so \(A\) plays \((\tfrac14, \tfrac34)\). \(B\)'s probability of column 1 is \(q=(v_{22}-v_{12})/4=(3-1)/4=\tfrac12\), so \(B\) plays \((\tfrac12,\tfrac12)\). The value is \(V=(v_{11}v_{22}-v_{12}v_{21})/4=(12-2)/4=2.5\). Verify the indifference that defines the mix: if \(A\) plays \((\tfrac14,\tfrac34)\), then against \(B_1\) the payoff is \(\tfrac14(4)+\tfrac34(2)=2.5\), and against \(B_2\) it is \(\tfrac14(1)+\tfrac34(3)=2.5\) — identical, so \(B\) cannot exploit \(A\). The value \(2.5\) sits neatly between the maximin \(2\) and minimax \(3\), as it must.

5 Dominance down to a solvable game

Problem. Solve the \(3\times3\) game rows \(A_1{=}(6,3,8)\), \(A_2{=}(2,5,4)\), \(A_3{=}(1,2,3)\) by reducing it first.

Solution. Look for dominated rows (bad for \(A\)): row \(A_3=(1,2,3)\) is beaten in every column by \(A_1=(6,3,8)\) (\(6{\ge}1,\ 3{\ge}2,\ 8{\ge}3\)), so \(A\) never plays \(A_3\) — delete it. Now look for dominated columns (bad for \(B\), i.e. larger is worse since \(B\) minimises): in the remaining rows, column \(B_3=(8,4)\) is beaten by \(B_1=(6,2)\) (\(6{\le}8,\ 2{\le}4\)), so \(B\) never plays \(B_3\) — delete it. What remains is the \(2\times2\) game \(\begin{smallmatrix} 6 & 3 \\ 2 & 5 \end{smallmatrix}\). Its denominator is \((6+5)-(3+2)=6\); no saddle (maximin \(3\), minimax \(5\)); so \(p=(5-2)/6=\tfrac12\), \(q=(5-3)/6=\tfrac13\), and \(V=(30-6)/6=4\). \(A\) plays \((\tfrac12,\tfrac12,0)\), \(B\) plays \((\tfrac13,\tfrac23,0)\), value \(4\). Dominance turned an awkward \(3\times3\) into a one-line solve without discarding anything optimal.

6 The prisoner's dilemma

Problem. Two firms each set a High or Low price. Profits (firm A, firm B) are: both High \((10,10)\); A High, B Low \((2,12)\); A Low, B High \((12,2)\); both Low \((5,5)\). Find the Nash equilibrium and compare it to the cooperative outcome.

Solution. Treat "Low" as defect. For firm A, compare its options against each of B's choices: if B plays High, A earns \(12\) by going Low versus \(10\) by High — Low is better; if B plays Low, A earns \(5\) Low versus \(2\) High — Low is better again. So Low dominates High for A regardless of B, and by symmetry the same holds for B. Both play Low, giving the Nash equilibrium \((5,5)\): neither can improve by unilaterally raising its price, since deviating to High while the rival stays Low drops that firm from \(5\) to \(2\). Yet both-High would have paid \((10,10)\) — strictly better for both. The equilibrium is stable but jointly poor: individually rational choices produce a collectively bad outcome, and no firm can escape it alone. This is exactly the logic of a price war, and why cartels need enforcement to hold cooperation together.

Review

Chapter Summary

Nature vs opponent

Both start from a payoff matrix. The columns are chosen by indifferent nature (decision analysis) or a strategic adversary (game theory) — and that changes everything.

Criteria under uncertainty

Maximin (cautious), maximax (optimistic), Hurwicz (dial \(\alpha\)), Laplace (equal odds), minimax regret. They can disagree; the choice encodes an attitude.

Risk & EVPI

Maximise \(\text{EMV}=\sum p_j v_{ij}\); it agrees with minimum EOL. \(\text{EVPI}=\min\text{EOL}\) caps the value of any forecast.

Saddle point

If maximin \(=\) minimax, that value \(V\) is a saddle point with optimal pure strategies — an equilibrium no player can beat by deviating.

Dominance & mixing

Delete dominated rows/columns first. With no saddle, randomise: the \(2\times2\) formulas give \(p,q,V\) by equalising \(A\)'s payoff across \(B\)'s columns.

Minimax & Nash

Von Neumann: every zero-sum game has a value (an LP, dual to itself). Nash extends equilibrium to any game — the prisoner's dilemma shows it can be jointly poor.

Practice

Problems

For decision problems, build the regret matrix carefully — subtract from the column maximum. For games, always check for a saddle point before reaching for mixed strategies, and prune by dominance first. Difficulty rises down the list.

  1. Explain how deciding against nature differs from deciding against an opponent, and why maximin carries different weight in each.
  2. State the maximin, maximax, Hurwicz, Laplace and minimax-regret criteria, and say what attitude to risk each expresses.
  3. An investor faces payoffs \(A_1=(40,20,10)\), \(A_2=(15,30,35)\), \(A_3=(25,25,25)\) over three states. Apply all five criteria (use \(\alpha=0.6\) for Hurwicz) and report each recommendation.
  4. For Problem 3 with \(P=(0.4,0.4,0.2)\), find the best action by EMV, then compute the EVPI and confirm it equals the minimum expected opportunity loss.
  5. Explain what a decision tree's square and circular nodes represent, and describe the rollback procedure for solving one.
  6. Define a saddle point and the value of a game. Solve the zero-sum game rows \((6,2,4)\), \((3,5,3)\), \((7,1,2)\) if a saddle exists.
  7. Solve the \(2\times2\) game \(\begin{smallmatrix} 1 & 4 \\ 5 & 2 \end{smallmatrix}\): check for a saddle, then find the optimal mixed strategies and the value, and verify the indifference condition.
  8. Reduce and solve the game rows \(A_1=(8,5,7)\), \(A_2=(3,6,2)\), \(A_3=(9,4,8)\) using dominance where possible, then any mixed strategy needed.
  9. State the minimax theorem and explain its connection to linear programming duality from Part 2.
  10. Define a Nash equilibrium, and explain why a saddle point is the Nash equilibrium of a zero-sum game.
  11. Build a prisoner's-dilemma payoff table of your own (profits for two competitors choosing High or Low price), identify the dominant strategy and the Nash equilibrium, and show it is worse for both than mutual cooperation.
  12. Explain why every mixed-strategy zero-sum game can be solved as a linear program, and what the primal and dual correspond to for the two players.