Strategic Cost Management · Game Theory

Graphical Method for 2xn and mx2 Games

Updated 11 October 2026 · Fact-checked

The graphical method solves a zero-sum game where one player has only two strategies. Plot the expected payoff against each opponent strategy as a straight line over probability 0 to 1. For a 2xn game, find the highest point of the lower envelope. For an mx2 game, find the lowest point of the upper envelope. Then solve the resulting 2x2 game.

Understand Graphical Method for 2xn and mx2 Games

A two-person zero-sum game has a payoff table. Whatever one player gains, the other loses. If the table has no saddle point, neither player can rely on one pure strategy. Each must mix strategies using probabilities. This is a mixed strategy.

The graphical method works when one player has only two strategies. That means a 2xn game (A has 2 rows, B has n columns) or an mx2 game (A has m rows, B has 2 columns). With two strategies, one probability p decides the whole mix. A plays the first strategy with probability p and the second with 1 − p. So you can draw everything on one axis from 0 to 1.

For a 2xn game, A's expected payoff against each column of B is a straight line in p. B will always answer with the column that is worst for A. So A's guaranteed payoff is the lower envelope, the minimum of all lines at each p. A wants the highest point of that envelope. That point gives the value of the game and A's best p.

For an mx2 game, you reverse the view. B mixes its two columns with probability q and 1 − q. Each row of A gives a line in q. A will pick the row that is worst for B, so B faces the upper envelope, the maximum of the lines. B wants the lowest point of that envelope.

The optimal point is where two lines cross. Only those two lines matter. Their strategies form a 2x2 game, and you solve it with the usual mixed-strategy formulas. All other strategies get probability zero.

Key rules to remember

Expected payoff line, 2xn game: E(Bj) = a1j × p + a2j × (1 − p) = a2j + (a1j − a2j) × p; A plays row 1 with probability p and row 2 with 1 − p. Payoffs are to A. Draw one line for each column j.
Expected payoff line, mx2 game: F(Ai) = ai2 + (ai1 − ai2) × q; B plays column 1 with probability q and column 2 with 1 − q. Draw one line for each row i.
Optimal point rule: 2xn: maximise the lower envelope. mx2: minimise the upper envelope.; Use the envelope itself, not just any intersection of two lines.
Solving the 2x2 sub-game (no saddle point): For rows (a b / c d): p(row 1) = (d − c) ÷ (a − b − c + d); V = (ad − bc) ÷ (a − b − c + d); Valid when the 2x2 game has no saddle point. Use the same idea for B's probabilities.
Probability check: p + (1 − p) = 1 and every probability ≥ 0; Strategies not in the final 2x2 sub-game get probability 0.

How to solve Graphical Method for 2xn and mx2 Games questions

Follow the same order for any 2xn or mx2 question. Call A the maximising player and B the minimising player.

1Check for a saddle point first. Find the row minima and column maxima. If maximin equals minimax, the pure strategy answer is the solution and no graph is needed.
2Apply the dominance rule to remove dominated rows or columns. This can cut the game down before you plot.
3Decide the case. If A has two rows, write A's expected payoff against each column as a function of p. If B has two columns, write B's expected loss against each row as a function of q.
4Find each line's value at 0 and 1, or its equation. Plot or compare the lines over 0 to 1.
5For a 2xn game, trace the lower envelope and locate its highest point. For an mx2 game, trace the upper envelope and locate its lowest point. Confirm with a check that no other line cuts below (or above) that point.
6Take the two lines that meet at that point. Set their equations equal and solve for p (or q). Substitute back to get the value of the game.
7Form the 2x2 sub-game from those two strategies. Solve it for the other player's probabilities. Check that the value comes out the same.
8State the answer in full: both players' strategies with probabilities (zero for unused ones) and the value of the game.

Quickest way: Compare lines numerically instead of drawing

When to use it: Use it when the table is small and you need speed. Always draw a rough sketch if the examiner asks for a graph.

Write each line as a + b × p, using the payoff of the second row as the intercept and (first row − second row) as the slope.
Work out each line's value at p = 0 and p = 1. These are just the two columns of payoffs.
Spot the likely pair: for a 2xn game, a rising line and a falling line that form the top of the lower envelope.
Solve that pair. Then substitute the p value into every other line to check none is lower (for 2xn) or higher (for mx2).
If the check fails, try the next pair along the envelope. Then solve the 2x2 sub-game and confirm that both players get the same value.

Common mistakes in Graphical Method for 2xn and mx2 Games

Picking the highest intersection of any two lines instead of the highest point of the lower envelope.
Several lines cross and the first crossing found looks right.
Fix: Put the p value into every line. If any line is below the crossing, that crossing is not on the envelope. Use the pair that actually forms the envelope peak.
Using the lower envelope for an mx2 game, or the upper envelope for a 2xn game.
Students memorise one picture and forget whose point of view the lines show.
Fix: In 2xn, lines show A's gain, and B pushes it down, so A maximises the lower envelope. In mx2, lines show B's loss, and A pushes it up, so B minimises the upper envelope.
Skipping the saddle point check.
The question says 'solve graphically', so students jump to plotting.
Fix: Always find maximin and minimax first. If a saddle point exists, the mixed-strategy method gives a wrong or pointless answer.
Writing a probability for the strategy of the player who is not on the axis, or giving non-zero probability to an unused strategy.
Students forget that only two lines meet at the optimum.
Fix: Give probability zero to every strategy outside the final 2x2 sub-game. Make sure each player's probabilities add to 1.
Errors in the slope and intercept, for example mixing up which row has probability p.
The row labelled A1 gets p in one line and 1 − p in another.
Fix: Fix once: A1 gets p, A2 gets 1 − p (or B1 gets q, B2 gets 1 − q). Test each line at p = 0 and p = 1. It should return the payoffs of row 2 and row 1.
Not stating the full solution.
Students stop once p is found.
Fix: Give A's mix, B's mix and the value of the game. Mention which strategies are not used.

Worked examples

Example 1

Solve the following 2x3 game graphically. Payoffs are to A (the maximising player), who has rows A1 and A2. B has columns B1, B2 and B3.

B1 B2 B3
A1 1 3 11
A2 8 5 2

Find the optimal strategies of both players and the value of the game.

Show the solution

Saddle point check. Row minima are 1 (A1) and 2 (A2), so maximin = 2. Column maxima are 8, 5 and 11, so minimax = 5. They differ, so there is no saddle point.
Let A play A1 with probability p and A2 with 1 − p.
Expected payoff against B1 = 1p + 8(1 − p) = 8 − 7p. Against B2 = 3p + 5(1 − p) = 5 − 2p. Against B3 = 11p + 2(1 − p) = 2 + 9p.
At p = 0 the lines have values 8, 5, 2, so the lower envelope starts at 2 (B3). At p = 1 the values are 1, 3, 11, so it ends at 1 (B1).
The B1 and B3 lines cross at 8 − 7p = 2 + 9p, which gives p = 3/8. At p = 3/8, the B2 line = 5 − 0.75 = 4.25. B1 and B3 give 5.375, which is above 4.25. So this point is not on the lower envelope.
The B3 and B2 lines cross at 2 + 9p = 5 − 2p, which gives 11p = 3, so p = 3/11. At this point B1 gives 8 − 21/11 = 67/11, which is above 49/11. So B2 and B3 form the envelope peak.
Value = 2 + 9 × 3/11 = 2 + 27/11 = 49/11. So A plays A1 with 3/11 and A2 with 8/11. B1 is not used.
Now solve the 2x2 sub-game for B using columns B2 and B3. Let B play B2 with probability q and B3 with 1 − q. Against A1: 3q + 11(1 − q) = 11 − 8q. Against A2: 5q + 2(1 − q) = 2 + 3q.
Set equal: 11 − 8q = 2 + 3q, so 9 = 11q, giving q = 9/11. Then B3 = 2/11. Value = 2 + 3 × 9/11 = 49/11, which matches.

Answer: A: A1 with probability 3/11, A2 with 8/11. B: B1 with 0, B2 with 9/11, B3 with 2/11. Value of the game = 49/11 (about 4.45) in A's favour.

Example 2

Solve the following 3x2 game graphically. Payoffs are to A (the maximising player), who has rows A1, A2 and A3. B has columns B1 and B2.

B1 B2
A1 1 8
A2 6 3
A3 3 5

Find the optimal strategies and the value of the game.

Show the solution

Saddle point check. Row minima are 1, 3 and 3, so maximin = 3. Column maxima are 6 and 8, so minimax = 6. They differ, so there is no saddle point. No row is dominated by another.
A has three rows, so use B's view. Let B play B1 with probability q and B2 with 1 − q. B's expected loss against each row is a line in q.
Against A1: 1q + 8(1 − q) = 8 − 7q. Against A2: 6q + 3(1 − q) = 3 + 3q. Against A3: 3q + 5(1 − q) = 5 − 2q.
B wants the lowest point of the upper envelope. At q = 0 the values are 8, 3, 5, so the envelope starts at 8 (A1). At q = 1 the values are 1, 6, 3, so it ends at 6 (A2).
The A1 and A2 lines cross at 8 − 7q = 3 + 3q, so 5 = 10q and q = 1/2. At q = 1/2, A1 = 4.5 and A2 = 4.5. A3 = 5 − 1 = 4, which is below 4.5. So A3 does not lie on the upper envelope at this point, and the lowest point of the upper envelope is at q = 1/2 with value 4.5.
So B plays B1 and B2 each with probability 1/2. Value = 4.5.
Now solve for A using rows A1 and A2. Let A play A1 with probability p and A2 with 1 − p. Against B1: 1p + 6(1 − p) = 6 − 5p. Against B2: 8p + 3(1 − p) = 3 + 5p.
Set equal: 6 − 5p = 3 + 5p, so p = 3/10. Value = 6 − 1.5 = 4.5, which matches. A3 is not used.

Answer: A: A1 with probability 3/10, A2 with 7/10, A3 with 0. B: B1 with 1/2, B2 with 1/2. Value of the game = 4.5 in A's favour.

Exam tips

Do the saddle point check and dominance check first and show them in your answer. This gets marks even if later arithmetic slips.
Label the graph clearly: axis from 0 to 1, each line named by the opponent's strategy, and the optimal point marked. Say whether you used the lower or upper envelope.
Always test the intersection against every other line. This is the step that separates a correct answer from a common wrong one.
In MCQs on this topic, expect to be asked for the value of the game or a probability. Solve the 2x2 sub-game quickly with the formula and double-check by computing the value from both players' sides.
Finish with a full statement of both players' strategies, probabilities that add to 1, and the value of the game.

Practice questions from Game Theory

Graphical Method for 2xn and mx2 Games: frequently asked questions

When can I use the graphical method in game theory?

You can use it when one player has exactly two strategies, after any dominated strategies are removed. That means the game is 2xn or mx2. If both players still have three or more strategies after reduction, you need other methods such as linear programming.

How do I know which two lines to use for the optimal point?

Use the two lines that meet at the peak of the lower envelope (2xn) or the trough of the upper envelope (mx2). Check by substituting the p or q value into all other lines. If another line is lower (2xn) or higher (mx2), the pair is wrong.

Do I need to check for a saddle point before plotting?

Yes. If maximin equals minimax, the game has a pure strategy solution and the value is that common number. Plotting is needed only when there is no saddle point.

What is the difference between solving a 2xn and an mx2 game?

In a 2xn game you plot A's expected gain against each of B's columns and maximise the lower envelope. In an mx2 game you plot B's expected loss against each of A's rows and minimise the upper envelope. After that, both cases reduce to a 2x2 game.