TWO COIN MORRA This game is layed by two layers, R and C. Each layer hides either one or two silver dollars in his/her hand. Simultaneously, each layer guesses how many coins the other layer is holding. If R guesses correctly and C does not, then C ays R an amount of money equal to the total number of dollars concealed by both layers. If C guesses correctly and R does not, then R ays C an amount of money equal to the total number of dollars concealed by both layers. If both layers guess correctly or incorrectly, no money exchanges hands. Clearly, each layer must decide how many coins to hide and what number to guess. We will use the notation (,2) to mean that a layer hides dollar and guesses 2. We can reresent the ossible outcomes of a round of lay by a ayoff matrix, indicating how much R will receive from C given the strategies followed by each layer. A negative number means that C receives money from R. R (,) (,2) (2,) (2,2) C (,) (,2) (2,) (2,2) 2 2 This is an examle of a finite two-erson zero-sum game. Finite refers to the fact that each layer has a finite number of strategies. Zero-sum refers to the fact that what one layer gains in wealth, the other loses. Suose R decides to use only strategy (,2). This is an examle of a ure strategy. Since the minimum entry in that row is 2, R can guarantee losing no more than $2 er round, and this will haen if C consistently uses strategy (,). But what if R decides to use either strategy (,2) or (2,), each half of the time, but randomly mixed so that there is no detectable attern? For examle, he could fli a coin to decide which of the two strategies to use. This is an examle of a mixed strategy. To calculate the exected outcome, we mix the second and third rows by multilying them each by /2 and
TWO COIN MORRA 2 adding them together. The result is [ 2 2 [ + 2 [ 2 2. Since the minimum entry is /2, R can exect to lose no more than half a dollar er round on the average, and he can exect this to haen if C consistently chooses strategy (2,2). Can R do better than this? If R uses each strategy / of the time, then R mixes his rows by multilying each of them by / and adding them together, giving [ 2 + [ 2 [ + [ + [ Since the minimum entry is /, R would exect to lose no more than a quarter dollar er round on the average, and he can exect this to haen if C consistently uses only strategies (,2) and (2,2). Problem. Try to find a mixed strategy for R which is even better. In articular, try to find nonnegative numbers, 2,, that sum to one.
TWO COIN MORRA such that the minimum entry in [ 2 [ + 2 2 [ + [ + is at least zero. Can you find a mixture where the minimum entry is larger than zero? C s strategies can be studied in a similar manner. For examle, suose C decides to use only strategy (2,2). The numbers in the matrix reresent amounts that C ays R, so C is interested in looking at the maximum entry to see how badly he will do. Since the maximum entry in column is, C can exect to lose at most $ er round, and this haens if R consistently uses strategy (,2). If C uses each of his strategies a fourth of the time, we must mix the columns by multilying each of them by / and adding them together. This yields 2 + 2 + + = / / / / Since the maximum entry is /, by using this mixed strategy, C can exect to lose no more than a quarter dollar er round on the average, and he can exect this if R consistently uses only strategies (,2) and (2,2). Of course, by symmetry, we can exect the best strategy for R to be the same as the best strategy for C. Problem.2 Consider the following matching game. Each layer hides either a nickel or a dime. If the two coins match, R gets C s coin. If they don t match, C gets R s coin. What are the otimal strategies for each layer? Can they each exect to break even on the average? (If so the game is called fair.).
2 SADDLEPOINTS 2 SADDLEPOINTS Consider the game with the following ayoff matrix R I II III C I II III 2 5 2 This game has an interesting roerty. Determine the smallest entry for each row. The largest of these is 2, and it occurs in row. Now determine the largest entry for each column. The smallest of these is also equal to 2 and it occurs in column 2. Because these two numbers are equal, this game is said to have a saddleoint. R I II III C I II III 2 5 2 5 2 2 2 This suggests that the otimal strategy for R is to use only strategy I, and the otimal strategy for C is to use only strategy II. For if R uses only I, R can exect to win at least $2 er round, and if C uses only II, C can exect to lose at most $2 er round. In general, we have the following theorem. Theorem 2. If the row minimum m in some row r equals the column maximum M in some column c, then the otimal strategy for R is to use only strategy r, the otimal strategy for C is to use only strategy c, and under otimal lay the amount that R can exect to receive on the average is m = M. We will establish this theorem by considering a sufficiently general examle. Suose we have a matrix in which the maximum entry in row 2 is 5 and the minimum entry in column is also 5.
MATRIX NOTATION 5 R 2 C 2 a b c d e f 5 Since 5 is the minimum entry in row 2, we know that d 5. But since 5 is the maximum entry in column, we know also that d 5. So d = 5. Suose R uses only strategy 2. Since 5 is the minimum row entry, R can exect to win at least $5 er round. Now suose R tries to do better by mixing his strategies according to, 2,, where, 2,, and + 2 + =. Multily the three rows by these numbers and add them together. Look at the third entry. It is a + 2 d + f. Since 5 is the column maximum, a 5 and f 5. Remember that d = 5. Since, 2,, we conclude a 5, 2 d = 5 2, and f 5. So a + 2 d + f 5 + 5 2 + 5 = 5( + 2 + ) = 5() = 5. Thus the third entry of the row mixture is no greater than 5, and hence the minimum entry of the row mixture is also no greater than 5. Therefore R cannot exect to win more than $5 er round on the average even if he mixes his strategies. A similar argument works for C. Suose C uses only strategy. Since 5 is the maximum column entry, C can exect to lose at most $5 er round. Now suose C tries to do better by mixing his strategies according to q, q 2, q, q, where q, q 2, q, q and q + q 2 + q + q =. Multily the four columns by these numbers and add them together. Look at the second entry. It is q b + q 2 c + q d + q e. Since 5 is the row minimum, b, c, e 5. Remember that d = 5. Since q, q 2, q, q, we conclude q b 5q, q 2 c 5q 2, q d = 5q, and q e 5q. So q b + q 2 c + q d + q e 5q + 5q 2 + 5q + 5q = 5(q + q 2 + q + q ) = 5() = 5. Thus the second entry of the column mixture is at least 5, and hence the maximum entry of the column mixture is also at least 5. Therefore C cannot exect to lose less than $5 er round on the average even if he mixes his strategies. MATRIX NOTATION Given a matrix game with ayoff matrix A = (a ij ). Suose R uses mixed strategy (,..., m ) where,..., m, + + m =, and C uses 5
MATRIX NOTATION 6 mixed strategy (q,..., q n ) where q,..., q n, q + + q n =. Then R can exect to receive m i= nj= i a ij q j er round on the average. This can be seen in two ways. If R mixes the rows according to his strategy, the resulting row is [ a + + m a m a n + + m a mn = [ m i= i a i m i= i a in. But C mixes his strategies according to q,..., q n, so the exected amount R wins er round on the average will be ( m i= i a i )q + + ( m i= i a in )q n = m i= nj= i a ij q j. The second way to see this is, if C mixes the columns according to his strategy, the resulting column is q a + + q n a nj= n a j q j. =. q a m + + q n a nj= mn a mj q j But R mixes his strategies according to,..., m, so the exected amount that R wins er round on the average will be ( n j= a ij q j ) + + m ( n j= a mj q j ) = m i= nj= i a ij q j. This is more easily exressed using matrix multilication. mixed strategy as a m matrix Write R s = [ m and C s mixed strategy as an n matrix q = q. q n
OPTIMAL STRATEGIES 7 Then if R mixes rows according to his strategy, the resulting row is A. Now if C mixes columns according to his strategy, the result is an exected ayoff of (A)q = Aq to R er round on the average. Similarly, if C mixes columns according to his strategy, the resulting column is Aq. Now if R mixes rows according to his strategy, the result is an exected ayoff of (Aq) = Aq to R er round on the average. OPTIMAL STRATEGIES If R uses mixed strategy, then R can exect to win at least the minimum of the row A er round on the average. Suose this minimum is the cth entry and equals w. Then w = Aq where. q =. with a in entry c and s everywhere else. Now consider any other q = where q,..., q n and q + + q n =. Then A [w w so q. q n (A)q [w wq = wq + + wq n = w(q + + q n ) = w. We have shown that (A)q w for all strategies q. So w = (A)q. So R can exect to win at least Aq. Obviously R wants to choose to make this minimum as large as ossible, so R s goal is to find max Aq.
OPTIMAL STRATEGIES 8 Starting all over again: If C uses mixed strategy q, then C can exect to lose at most the maximum of the column Aq. Suose this maximum is the rth entry and equals w. Then w = Aq where = [ with a in entry r and s everywhere else. Now consider any other = [ m where,..., m and + + m =. Then so Aq ẉ. w ẉ (Aq). w = w + + m w = ( + + m )w = w. We have shown that (Aq) w for all strategies. So w = max (Aq). So C can exect to lose at most max Aq. Obviously C want to choose q to make this maximum as small as ossible, so C s goal is to find max Aq. Theorem. max Aq max Aq, where,..., m, + + m =, q,..., q n, q + + q n =. Proof: Consider any and q. Suose Aq occurs when q = q. Suose max Aq occurs when =. Then Aq = Aq Aq Aq = max Aq. So no matter what is chosen and what q is chosen, the left-hand side is always no more than the right-hand side. So even the largest left-hand side is no more than the smallest right-hand side. Hence max Aq max Aq. Theorem.2 If and q can be found so that Aq = max Aq, then is otimal for R and q is otimal for C (where and q are each nonnegative vectors whose comonents sum to ).
5 THE GRAPHICAL SOLUTION OF 2 N GAMES 9 So Proof: If there were a better, then max Aq > Aq Aq = max Aq. Aq > max which contradicts the revious theorem. Similarly, if there were a better q, then So max Aq = max Aq Aq Aq > max Aq. Aq > max Aq max Aq max Aq which contradicts the revious theorem. The above theorem hels to verify that roosed mixed strategies are otimal, although it does not indicate how to find them in the first lace. 5 THE GRAPHICAL SOLUTION OF 2 n GAMES If R has just two ure strategies, there is an easy way to determine R s otimal mix grahically. For examle, consider the game R 2 C 2 2 2 2 2 Draw two arallel vertical axes. For each column, lot the first entry on the first axis and the second entry on the second axis. Connect the two oints by a line segment and label the segment by the number of the column. Now darken the line segments which bound the figure from below; then find and mark with a dot the highest oint on this darkened boundary.
5 THE GRAPHICAL SOLUTION OF 2 N GAMES 2 The height of the dot as measured against the vertical axes is the value of the game. In the above examle, the value of the game is /7. The osition of the dot between the two axes indicates which mixture of his two strategies R should use. For examle, if the dot is /7 of the way from the first axis to the second, as it is above, R should use his second strategy /7 of the time and his first strategy /7 of the time. The lines which intersect at the dot identify the strategies C should use in his mixture. In the above examle, C should use only strategies 2 and. In the above examle, the dot is at the intersection of lines 2 and. This means that R should choose and 2 so that in the row mixture, the 2nd and th entries turn out to be equal. Hence 2 = 2 + 2 = 2 / 2 = / = /7 2 = /7. The value of the game is 2 = 2 + 2 = /7. C should use only strategies 2 and and should choose q 2 and q so that in the column mixture,
5 THE GRAPHICAL SOLUTION OF 2 N GAMES the two column entries turn out to be equal. Hence q 2 2q = q 2 + q 2q 2 = 5q q 2 /q = 5/2 q 2 = 5/7 q = 2/7. Again, the value of the game is q 2q 2 = q + q = /7. So by the last theorem of the revious section, we have found the otimal strategies for both layers. The above method can be modified to solve m 2 games as well, but then we have to interchange the role of the columns and the rows, darken the line segments which bound the grah from above, and locate the lowest oint on this boundary.