Index

1. Introduction to Game Theory
2. Key terms
3. 3. Maximin Strategy (Player A - Maximizing Player)
4. 4. Minimax Strategy (Player B - Minimizing Player)
5. 5. Saddle Point
6. Understanding Practically
7. Example 2 A Zero Sum Game where there is a saddle point
8. The Graphical Method of solving $2timesn$ and $mtimes2$ games.
9. Example 1 $2timesn$ case.
10. Example 2 $mtimes2$ case
11. Principle of Dominance
12. Partial Dominance
13. 🎯 Game Theory — Reasoning Trace (Complete Solving Guide) -- Final Summary.

Introduction to Game Theory

1. What is Game Theory?

Game theory studies conflict and cooperation between intelligent decision-makers. In OR, we focus on 2-person zero-sum games where:

• Two players compete
• One player's gain = Other player's loss
• Total payoff = 0

Example: Chess, poker, military strategy

Key terms:

2. Payoff Matrix

A matrix showing Player A's payoffs for all strategy combinations

           Player B
           B₁    B₂    B₃
Player A  
   A₁     a₁₁   a₁₂   a₁₃
   A₂     a₂₁   a₂₂   a₂₃
   A₃     a₃₁   a₃₂   a₃₃

• Rows = Player A's strategies
• Columns = Player B's strategies
• Values = A's gain (B's loss)

3. Maximin Strategy (Player A - Maximizing Player)

Player A wants to maximize their minimum gain.

Steps:

1. Find minimum in each row (worst case for A in that row)
2. Select the row with maximum of these minimums
3. This is Player A's maximin strategy

Maximin Value = max(row minimums)

4. Minimax Strategy (Player B - Minimizing Player)

Player B wants to minimize their maximum loss (or minimize A's maximum gain).

Steps:

1. Find maximum in each column (worst case for B in that column)
2. Select the column with minimum of these maximums
3. This is Player B's minimax strategy

Minimax Value = min(column maximums)

5. Saddle Point

A saddle point exists when:

At the saddle point:

• Element is minimum in its row (best for A against B's strategy)

• Element is maximum in its column (best for B against A's strategy)

If saddle point exists:

• Game has a pure strategy solution
• Value of game = value at saddle point
• Both players should use their pure strategies

Understanding Practically

All that hit you all at once? Yeah, same here.

So, let's slow down, and understand with an example.

Step 1: Understanding the setup

We have two players:

We have two players:

• A → the row player → wants to maximize their gain.

• B → the column player → wants to minimize A’s gain (since B’s loss = A’s gain).

That’s why it’s called a zero-sum game:
whatever A gains, B loses equally.

The game is represented as a payoff matrix — payoffs to A.

Example (general form):

Step 2: Formula for $2\times 2$ Game (Mixed Strategy Case)

If no saddle point exists, we use:

where:

• $p=\text{probability that plays}{A}_1$
• $q=\text{probability that B plays}{B}_1$
• $V=\text{value of the game (expected to gain A)}$

Step 3: Example 1 -- Finding Saddle Point

Let's take this example:

A wants to maximize, so first check worst-case for each row:

Before we proceed, let's understand the philosophy here.

Both players are rational and risk-aware:

• each tries to maximize their guaranteed gain and
• simultaneously minimize their possible loss.

So the logic is symmetric — just mirrored:

• A: “Whatever B does, I want to make sure I’m not left with something too small.”

• B: “Whatever A does, I want to make sure I don’t lose too much.”

That’s precisely what “maximin” and “minimax” capture.

Let's assume that B moves first. (Column access)

So assuming that B goes first, and tries to stop A from getting the higher value, but it doesn't know which row A will go for first.

Say, it goes with column $B_1$

Going with the maximin rule that B follows, if B's thinking would be already grab the highest value per column and limit A to a "lower higher value", then B would already try to lock down the maximum, which is 4.

Now, with column $B_2$ , the maximum would be 3.

Now B looks at those worst cases (3 and 4) and picks the smaller one to limit A’s win:

This way, B can ensure A never earns more than 3.

Now in the next turn, A moves.

A assumes B will always pick the column that hurts A most (smallest in the row).

So for $A_1$ it will try to minimize it's loss for however gain it can get from that row. If A knows that it will get a lesser value, so it will try to get a "higher lesser value".

So A already becomes pessimistic per row and gets the minimum value per row.

If that's the case then the maximin rule fits here perfectly since for $A_1$ A will go for 2.

Similarly, for row $A_2$, $A$ will go for 1.

Now:

This way, A can guarantee itself at least 2.

Checking for saddle point.

As we know:

A saddle point exists when:

We have:

• $\text{Maximin}=2$
• $\text{Minimax}=3$

They don’t coincide → no saddle point → the game is unstable under pure strategies.
So, both must mix their choices probabilistically to reach equilibrium.

Step 4: Solve using formulae

From the payoff matrix:

$a=2$, $b=4$, $c=3$, $d=1$.

Now, we apply the formulae:

• $p=\text{probability that plays}{A}_1$
• $q=\text{probability that B plays}{B}_1$
• $V=\text{value of the game (expected to gain A)}$

So, the probability of $A$ playing $A_1$ is 50%
The probability of $B$ playing $B_1$ is 75%
And $A$'s expected gain and B's loss is 2.5

Step 5: Interpretation

• Since there was no saddle point, both players must randomize their choices.

• On average, A will win 2.5 units per game if both play optimally.

• If one player deviates, the other can exploit it to their advantage.

Example 2: A Zero Sum Game where there is a saddle point

Same as before:

Step 1: Find A's row minima

A knows B will always pick the worst column (smallest value) in whatever row A chooses.

• Row $A_1$ → $min(4,1,2)$ = 1
• Row $A_2$ → $min(3,2,0)$ = 2

Now, taking a maximum of those minima:

So A can guarantee at least 2, no matter what B does.

Step 2: Find B's column maxima

B knows A will always pick the best row (largest value) in whatever column B chooses.

• Column $B_1\to max(4,3)$ = 4
• Column $B_2\to max(1,2)$ = 2

Now, taking a minimum of those maxima:

So B can limit A’s gain to at most 2.

Step 3: Compare to check if saddle point exists or not

✅ They coincide!

That means we’ve found a saddle point.

Step 4: Identify Saddle Point cell.

It’s the cell where the values of both the maximin and the minimax (2) appear in the matrix:

So the saddle point is at the intersection of $A_2$ and $B_2$

Step 5: Interpretation.

• A’s optimal strategy → play A₂ every time.
• B’s optimal strategy → play B₂ every time.
• Value of the game $V=2$.

No mixing or probability needed — this pair ($A_2$, $B_2$) is stable.
If either player deviates, they’ll get a worse outcome.

The Stability Intuition

At the saddle point:

• A can’t improve by switching rows (because B’s choice B₂ already gives them their best guarantee).

• B can’t improve by switching columns (because A₂ already limits A’s maximum).

It’s literally like a saddle shape if you imagine the payoff as a surface:

• It’s max along A’s direction (rows)

• It’s min along B’s direction (columns)

That’s why the term saddle point is used.

The Graphical Method of solving $2\times n$ and $m\times 2$ games.

Why do we use the graphical method?

When:

• There is no saddle point, and
• The game is 2 x n (A has 2 strategies) or m x 2 (B has 2 strategies)

because in these cases, one variable (probability) can be plotted easily on a 2D graph.

Example 1: $2\times n$ case.

Let's understand this better with the help of an example.

Given game:

Step 1: What we're trying to find

• A is the row player (wants to maximize their payoff).
• B is the column player (wants to minimize A’s payoff).

Since there's no saddle point, $A$ will mix between $A_1$ and $A_2$.

Let:

Step 2: Expected payoff for A

For each of B’s choices, we find A’s expected payoff (depending on p).

That will be as follows:

for each column $j$

where each $A_1{B}_j$ represents the most minimum payoffs that B allows, and each $A_2{B}_j$ represents the most payoffs that A gains from B.

Each of these is a linear equation in $p$.
They represent the payoff A will receive if B sticks to that strategy.

Now, from the payoff matrix:

So, now we have three linear equations:

• $E_{B_1}=5-3p$
• $E_{B_2}=3+p$
• $E_{B_3}=2+4p$

Plotting the equations.

• X-axis: $p$ (ranging from 0 to 1)
• Y-axis: Expected payoff $E$ (based on the values of $p$)

So the table would be as follows:

Now that we have two points per line, we can draw them and find out the intersection points.

Using the concepts of Module 1 -- Basic Linear Programming Problems (LPP) and Applications#Topic 4 Graphical Method of Solving Linear Programming Problems we can plot a diagram and find the intersection points in the diagram between the lines as follows:

This is the graph.

Now, we need to find the intersection points of all the lines, pairwise.

Solving for:

• $E_{B_1}=E_{B_2}$

• $E_{B_2}=E_{B_3}$

• $E_{B_1}=E_{B_3}$

So, now that we have the probability values, the expected payoffs for each value will be:

• At $p=\frac{1}{3}$:

$E_{B_1}=4$, $E_{B_2}=E_{B_3}=3.33$

• At $p=\frac{3}{7}$

$E_{B_1}=E_{B_3}=3.7143$, $E_{B_2}=3.4286$

• At $p=\frac{1}{2}$

$E_{B_1}=E_{B_2}=3.5$, $E_{B_3}=4$

Getting the envelope values per expected payoffs

Since we are working of A, who follows the maximin rule, first we will take a minimum for all 3 columns of all the expected payoffs.

• $min(4,3.33,3.33)=3.33$
• $min(3.7143,3.4286,3.4286)=3.4286$
• $min(3.5,3.5,4)=3.5$

Now the maximum of these envelope values is 3.5, which is at $p=\frac{1}{2}=0.5$

That's the final answer, the optimal point, $p=0.5$, i.e. the probability that $A$ plays $A_1$ is 50% and the expected payoff of $A$ is $3.5$

Example 2: $m\times 2$ case

Now B (the column player) has 2 strategies,
and A (the row player) has m strategies.

Let’s make it concrete with a 3×2 game (just enough to visualize).

These are the payoffs to A.

Step 1 — What we need to find

We want:

• B’s probability $q$: the chance B plays B₁
• (so $B_2$ is $1-q$)
• The value of the game $V$ — A’s expected payoff when both play optimally.

Step 2 — Formula for expected payoff per row

For each row (A’s strategy),
the expected payoff to A, given B plays with probability q, is:

That’s the same pattern as earlier, just flipped.

Step 3 — Plug values for each row

What these mean

Each equation tells you how A’s expected payoff changes as B varies its strategy between B₁ and B₂.

• The first (6 - 4q) slopes down as q increases.
• The second (3 + q) slopes up slowly.
• The third (1 + 4q) slopes up fast.

Plotting the equations

If you were to plot this:

• X-axis → $q$ (0 to 1)
• Y-axis → $E$ (expected payoff)

You’d get three lines:

Visually:

• A₁ starts highest (6) → goes down to 2

• A₂ starts mid (3) → goes up to 4

• A₃ starts low (1) → goes up to 5

• A wants the payoff as high as possible.

• B wants the payoff as low as possible.

So, for any given q:

• A will choose the row that gives the maximum E (the topmost line).

• B’s “effective” payoff to A is that maximum value (since A can always switch to the highest).

Find the intersection points

For:

(a)

(b)

(c)

Expected payoff values per equation:

For:

• $q=0.6$

$E_{A_1}=6-4(0.6)=3.6$
$E_{A_2}=3+0.6=3.6$
$E_{A_3}=1+4(0.6)=5.6$

• $q=0.67$

$E_{A_1}=6-4(0.67)=3.32$
$E_{A_2}=3+0.67=3.67$
$E_{A_3}=1+4(0.67)=3.68$

• $q=0.625$

$E_{A_1}=6-4(0.625)=3.5$
$E_{A_2}=3+0.625=3.625$
$E_{A_3}=1+4(0.625)=3.5$

Finding the upper envelope values

Since we are working for B here, who follows the minimax rule, we will first find the maximum out of all the payoff values to find the 3 upper envelope values:

Now taking the minimum of this would get us $3.625$, which is $V$ and thus the maximum value that $B$ will allow $A$ to get.

The probability of $B$ playing $A_3$ is 62.5%.

Principle of Dominance

The principle of dominance is a shortcut to simplify large game matrices.

It says: if one strategy is always worse than another, it will never be chosen — so we can eliminate it from the matrix.

It’s like pruning unnecessary branches in decision making.

The idea, in simple words

For A (the maximizing player)

• Each row belongs to A — it’s one of A’s possible strategies.

• To check if a row is dominated, look at its cell values one by one.

• If every cell in this row has a value less than or equal to the corresponding cell of another row (and at least one is strictly less),
  → the row is dominated and can be removed.

In short:

“If one row’s numbers are all smaller than another row’s numbers, the smaller row is useless — A will never play it.”

For B (the minimizing player)

• Each column belongs to B — it’s one of B’s strategies.

• To check if a column is dominated, compare its cell values vertically.

• If every cell in this column is ==greater than or equal to ==the corresponding cell of another column (and at least one strictly greater),

  → that column is dominated and can be removed.

In short:

“If one column’s numbers are all bigger than another column’s numbers, the bigger column is useless — B will never play it.”

⚙️ Example 1 — Row dominance (A’s case)

Compare cell by cell between A₁ and A₂:
A₂’s cells (3, 5, 4) are all greater than A₁’s (2, 4, 3).
→ A₂ dominates A₁ → remove A₁.

⚙️ Example 2 — Column dominance (B’s case)

Compare cell by cell between B₂ and B₃:
B₃’s cells (5, 4) are both smaller than B₂’s (6, 5).
→ B₃ dominates B₂ → remove B₂.

Partial Dominance

Sometimes, no single row (or column) is completely better than another — but a combination of two (or more) strategies can produce values that are as good or better than another row or column in every cell.

That’s called partial dominance.

Case 1 — For A (maximizing player)

• A’s strategies = rows.

• Each row has several cell values.

• If no single row dominates another, check whether a weighted mix of two rows (say, 50%-50% or any ratio) produces cell values that are all greater than or equal to another row’s cell values.

• If yes, that “dominated” row can be removed.

In short:

“If mixing two of A’s rows gives cell values that are never smaller than another row’s values, that other row becomes useless.”

Example : A's partial dominance

No single row dominates A₃ directly.

But if A mixes A₁ and A₂ equally (50%–50%):

• B₁ → (4 + 2) / 2 = 3
• B₂ → (3 + 5) / 2 = 4

or:

So this combination gives the pair (3, 4),
which is equal to or better than A₃’s (3, 4).
→ A₃ is partially dominated by the combination of A₁ and A₂ → remove A₃.

✅ Result: A₃ is redundant — A can achieve everything it does using a mix of A₁ and A₂.

Example: B's partial dominance

No single column dominates another.
Now test if combining B₁ and B₂ (say, 50% each) gives smaller cell values than B₃.

Combination (B₁, B₂ → average):

• A₁ → (3 + 6) / 2 = 4.5
• A₂ → (4 + 2) / 2 = 3

Compare with B₃ → (4, 3)

→ For A₁: 4.5 > 4 ❌ (slightly worse)
→ For A₂: 3 = 3 ✅

So this mix doesn’t dominate B₃ yet — but if we tweak the ratio (maybe 40%–60%), we might get smaller values for both rows.

If such a ratio exists, then B₃ would be partially dominated by (B₁, B₂).

The idea, simplified

Practical Tip

You rarely need to calculate the exact mixing ratio in exams.
Usually, the question either gives you an obvious equal-weight example (like 50%-50%) or you just need to show that such a combination exists logically — not solve for it exactly.

🎯 Game Theory — Reasoning Trace (Complete Solving Guide) -- Final Summary.

🧩 1. Identify what type of game you’re solving

• Check if the problem says two-person zero-sum game → use these steps.

• Check the size of the payoff matrix:

  • 2×2 → use saddle point or mixed strategy formulas.
  • 2×n or m×2 → use graphical method.
  • larger → apply principle of dominance first to reduce.

⚙️ 2. Check for Saddle Point (Pure Strategy)

1. For each row, find its minimum cell value (A assumes B minimizes).

2. Find the maximum of these minima → A’s Maximin.

3. For each column, find its maximum cell value (B assumes A maximizes).

4. Find the minimum of these maxima → B’s Minimax.

5. If Maximin = Minimax, saddle point exists:

   • That value = Value of the Game (V).
   • The corresponding cell = equilibrium point.
   • ✅ Stop here — no mixing needed.

🧮 3. If No Saddle Point → Apply the Principle of Dominance

• For A (row player):
  Compare all cell values per row.
  If every cell in one row is less than or equal to the corresponding cell in another row,
  → remove that row (it’s dominated).

• For B (column player):
  Compare all cell values per column.
  If every cell in one column is greater than or equal to the corresponding cell in another column,
  → remove that column (it’s dominated).

• Partial dominance:
  If a combination of two rows (or columns) produces cell values that are never worse than another,
  remove the dominated one.

• Repeat until no more rows or columns can be removed.

• If reduced to 2×2 → go to Step 4.

⚙️ 4. 2×2 Game — Mixed Strategy Formula

Let the matrix be:

Then use:

where:

• $p$ = probability that A plays $A_1$
• $q$ = probability that B plays $B_1$
• $V$ = value of the game

Check that $0\le p,q\le 1$.
If not, a pure strategy dominates and that row/column can be removed.

📈 5. 2×n Game — Graphical Method (A mixes)

• Let A play $A_1$ with probability $p$ and $A_2$ with $(1-p)$.

• For each of B’s strategies (each column):

  $$E_{B_j}=A_{1j}p+A_{2j}(1-p)$$

• Plot $E_{B_j}$ vs $p$ (p on x-axis, E on y-axis).

• The lowest line at each p shows what B can enforce (the lower envelope).

• The highest point on that lower envelope = optimal point.

At that point:

• $p^{\ast }$ = optimal probability for A.
• The y-value at that intersection = $V$, the value of the game.

📉 6. m×2 Game — Graphical Method (B mixes)

• Let B play $B_1$ with probability $q$ and $B_2$ with $(1-q)$.

• For each of A’s strategies (each row):

  $$E_{A_i}=A_{i1}q+A_{i2}(1-q)$$

• Plot $E_{A_i}$ vs $q$ (q on x-axis, E on y-axis).

• The highest line at each q shows what A can get (the upper envelope).

• The lowest point on that upper envelope = optimal point.

At that point:

• $q^{\ast }$ = optimal probability for B.
• The y-value at that intersection = $V$, the value of the game.

🧠 7. Interpretation

• If there’s a saddle point → that strategy pair is the equilibrium.
• If mixed → use the probabilities $p$ and $q$ found from formulas or graphs.
• The value of the game $V$ represents:
  • $V>0$ → A’s expected gain (B’s loss).
  • $V<0$ → A’s expected loss (B’s gain).

🪶 8. Sanity Check Before Final Answer

• Are $p$ and $q$ between 0 and 1?
• Does $V$ logically fit the data (no negative values if not expected)?
• If dominance was applied, confirm that removed rows/columns truly made no difference.
• For graphs, make sure the intersection used lies on the active envelope (lower or upper as applicable).

✅ Short Mental Checklist

1️⃣ Write payoff matrix (for A).
2️⃣ Find Maximin and Minimax.
3️⃣ If equal → Saddle → Done.
4️⃣ Else → Apply Dominance.
5️⃣ If 2×2 → Use Formula.
6️⃣ If 2×n or m×2 → Use Graphical Method.
7️⃣ Read $p$, $q$, and $V$, interpret results.