Practise › Questions › Game theory: play safe and stable solutions
Game theory: play safe and stable solutions questions
A two-player zero-sum game is set out as a single matrix of pay-offs. Ask what each player can guarantee themselves, and the two answers either meet, settling the game, or they do not.
6 original questions · 24 marks · the game theory: play safe and stable solutions notes · Decision Mathematics 2
Every question here is written for this library rather than taken from a past paper. Write your answer out before opening the worked one: the answers award marks point by point, and the marks are easier to see when you have something of your own to compare against.
A pay-off matrix has the entry 5 where the first row meets the second column. State what this means for each player.
Worked answer
The matrix is written from the row player's point of view, so if those choices are made the row player gains 5 and the column player loses 5. The game is zero-sum, so the column player is looking for small entries, not large ones. B1 for the row player's gain of 5, B1 for the column player's loss.A game has row minima 4 and 2, and column maxima 7, 4 and 9. State whether it is stable and give the value if it is.
Worked answer
Maximin = max(4, 2) = 4 and minimax = min(7, 4, 9) = 4. They agree, so the game is stable with value 4, occurring where the first row meets the second column. M1 for comparing maximin with minimax, A1 for the value.In a zero-sum game the row player's pay-offs are R1 (4, 2, 6), R2 (1, 5, 3) and R3 (3, 3, 2). Find both play-safe strategies and say whether the game is stable.
Worked answer
Row minima are 2, 1 and 2, so the row player's maximin is 2, playing R1 or R3: both guarantee the same worst case.
Column maxima are 4, 5 and 6, so the column player's minimax is 4, playing C1.
Write the row minima beside the matrix and the column maxima beneath it. Those two lists are where the method marks are; a bare pair of strategies scores one mark at most.
Maximin 2 does not equal minimax 4, so the game is not stable and a mixed strategy is needed. All that has been established is that the value lies between 2 and 4. M1 A1 for the row minima and maximin, M1 A1 for the column maxima and minimax, A1 for the play-safe strategies, A1 for the game not being stable.In a zero-sum game the row player's pay-offs are R1 (5, 3, 7), R2 (6, 4, 8) and R3 (2, 1, 3). Show that the game has a stable solution and give the value and the strategies.
Worked answer
Row minima are 3, 4 and 1, so the maximin is 4, playing R2.
Column maxima are 6, 4 and 8, so the minimax is 4, playing C2.
They agree, so the game is stable with value 4, at R2 with C2. Neither player can improve by moving alone. Switching row loses the row player 1 or 3, and switching column costs the column player 1 or 4. M1 A1 for the maximin, M1 A1 for the minimax and the stable value.In a zero-sum game the row player's pay-offs are R1 (5, 3, 7), R2 (6, 4, 8) and R3 (2, 1, 3). Reduce the matrix as far as possible using dominance, and comment on the result.
Worked answer
R2 beats R1 in every column, since 6 > 5, 4 > 3 and 8 > 7, so R2 dominates R1. R2 also beats R3 everywhere, so R2 dominates R3: delete both rows.
In the single remaining row, the column player wants the smallest of 6, 4 and 8, so C2 dominates C1 and C3 and both go.
The matrix reduces to the single entry 4, so the game is stable with value 4 at R2 with C2. M1 A1 for the row dominance, M1 A1 for the column dominance and the value. Dominance reaches the answer without any comparison of maximin with minimax. Quote the inequalities that justify each deletion; 'looks worse' is not a justification.In a zero-sum game the row player's pay-offs are R1 (6, 0, 7), R2 (4, k, 5) and R3 (3, 1, 8), where k is an integer with 1 ≤ k ≤ 9. Find all the values of k for which the game has a stable solution, and state the value of the game in each case.
Worked answer
Work in terms of k. The row minima are 0, min(4, k) and 1, so the maximin is k when k ≤ 4 and 4 when k ≥ 4.
The column maxima are 6, max(1, k) and 8, so the minimax is min(6, k) for k ≥ 1, which is k when k ≤ 6 and 6 when k ≥ 6.
Setting the two equal, they agree only while both equal k, that is for k = 1, 2, 3, 4, and then the value of the game is k, at R2 with C2. M1 A1 for the maximin in terms of k, M1 A1 for the minimax in terms of k, A1 for the four values of k, A1 for the value of the game. For k = 5 the maximin is 4 against a minimax of 5, and for k ≥ 6 it is 4 against 6, so those games are not stable.
Note that maximin never exceeds minimax. The entry where the play-safe row meets the play-safe column is at least that row's minimum and at most that column's maximum, so the two agree only when that entry is smallest in its row and largest in its column, which is exactly a saddle point. Answers that produce a maximin above the minimax have an arithmetic error somewhere.
The same practice on paper: the printable workbook for this topic, questions and a worked answer book.
Practise game theory: play safe and stable solutions one question at a time
The player marks nothing for you. It shows one question, waits, then shows the worked answer so you can mark yourself, and brings a question back sooner when it went badly.