MathsDecision Mathematics 2 › Mixed strategies

Mixed strategies

When no single choice is safe, the answer is to be unpredictable in exactly the right proportion. Two straight lines and their crossing point settle a two by two game.

Year FMEDEXCEL 9FM0 D2

Builds on Game theory: play safe and stable solutions and The Simplex algorithm.

IN THIS TOPIC

  • Set up the expected pay-off against each opposing choice as a function of p.
  • Solve a 2 by n or n by 2 game graphically and state the value.
  • Convert a larger game into a linear program for Simplex.

WHAT YOU PROBABLY THINK

In a mixed strategy the player should favour the choice with the larger pay-offs, in proportion to them.

Two lines and a crossing

Suppose the row player plays R1 with probability p and R2 with probability 1 − p. Against each column the expected pay-off is a linear function of p, so plotting them over 0 ≤ p ≤ 1 gives one straight line per column.

The row player is guaranteed at least the lowest of those lines at any given p, so the best choice of p is where that lower boundary is highest, which is at a crossing point. Equating the two relevant expressions gives p, and substituting gives the value of the game. The proportions are set by the geometry, not by the sizes of the pay-offs, which is where the opening claim goes wrong: an unpredictable mix is what makes a strategy safe.

Dominance: the column player would never choose C1, since C3 is better against both rowsC1C2C3R1352R26142 < 3 and 4 < 6, and the column player wants smalldelete C1, leaving a 2 by 2 game
FIG. 1First reduce: the column player would never choose C1, since C3 is smaller against both rows, so C1 goes and a two by two game is left.
Two expected pay-off lines crossing at p = 0.5: the row player's best mix, giving a game value of 3p = 0.53against C2against C3take the lower line, then its highest point
FIG. 2Two expected pay-off lines crossing at p = 0.5, giving the best mix and a game value of 3.

WORKED EXAMPLE

Solving a two by two game

After deleting a dominated column, a game has rows (5, 2) and (1, 4). Find the row player's optimal mix and the value.

Against the first column: 5p + 1(1 − p) = 1 + 4p. Against the second: 2p + 4(1 − p) = 4 − 2p.

Equating: 1 + 4p = 4 − 2p, so 6p = 3 and p = 0.5.

The value is 1 + 4(0.5) = 3, and the other expression gives 4 − 1 = 3 as a check. The row player plays each row half the time.

The column player, and larger games

The column player's mix is found the same way, plotting expected losses against each row and taking the lowest point of the upper boundary. The value comes out the same, which is a useful check: if the two calculations disagree, one of them is wrong.

For a 2 by n or n by 2 game, plot all n lines and read off the highest point of the lower boundary; only two lines meet there, and those are the ones to equate. Larger games are converted into a linear program. Add a constant to every entry if necessary to make them all positive, then maximise the value subject to one constraint per opposing choice, and solve with Simplex. Subtracting the constant again at the end recovers the true value.

YOUR TURN

The column player's mix

For the same two by two game, find the column player's optimal mix and confirm the value.

Show the working

Let the column player play C1 with probability q. Against R1 the expected loss is 5q + 2(1 − q) = 2 + 3q.

Against R2 it is 1q + 4(1 − q) = 4 − 3q.

Equating: 2 + 3q = 4 − 3q, so 6q = 2 and q = 1/3.

The value is 2 + 1 = 3, agreeing with the row player's calculation, so both mixes are correct.

THE EXAM BIT

  • Define p clearly as the probability of one named row, and say which.
  • Write the expected pay-off against every opposing choice as a function of p.
  • Equate only the two expressions that meet at the optimum, and justify which those are from the graph.
  • Check the value by working out the other player's mix as well.

CHECK YOURSELF

Against two columns the expected pay-offs are 2 + 5p and 7 − 3p. Find the optimal p and the value.

Show a hint

Equate the two.

Show the answer

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Plot the expected pay-off against each opposing choice as a line in p; the optimum is the highest point of the lower boundary.

Equate the two lines that meet there for p and the value, and check by finding the other player's mix; larger games go to Simplex.

WORKBOOK

Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.

CHECK YOUR PROGRESS

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  • Set up the expected pay-off against each opposing choice as a function of p.
  • Solve a 2 by n or n by 2 game graphically and state the value.
  • Convert a larger game into a linear program for Simplex.

Open the full revision checklist to see every objective in the course in one place.

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