CC BY
11MATHEMATICAL SCIENCES
SOLUTION OF MIXED STRATEGIC GAMES
Otesh M.
Master student of Korkyt Ata Kyzylorda University. Kyzylorda Kazakhstan
Seitmuratov A.
Doktor of Physical and Matematical Sciences, Professoz, Korkyt Ata Kyzylorda University. Kyzylorda. Players can use several strategies to find such solutions. A random selection of your own strategies is called a
mixed strategy Kazakhstan Zhakulova K.T.
Teacher, Karaganda Engineering College. Karaganda. Kazakhstan
Janyssova D.
Senior Lecturer, The Korkyt Ata Kyzylorda University, Kyzylorda. Kazakhstan
Abstract
The article analyzes the combined strategies used in game theory, and describes the combined strategy of randomly choosing one's own strategies using an example. Observance of the minimal strategy gives each player
a gain of no more than X and a loss of no less than / . If the value of the game is V = X = / , then the game is called fully definite or point games. If there is a point in the game matrix, then the game solution is known. Each player follows an effective strategy. If one player follows his own effective strategy, the other player cannot abandon his strategy.
Keywords: mixed strategy, conflict resolution, game theory, minimax strategy, mathematical expectation.
The question arises of how to find a game solution that does not have a fixed point in the matrices. If there is no sense in the game, it will be impossible to find an effective solution to the game using pure strategies. In
these games X < /. It will be important for each player to increase their winnings or decrease the winnings of their opponents.
That is, the use of pure strategies by player A
A15 A2 ' Am with probability ui, U2 5 • •• ,um is called his mixed strategy. Often the mixed strategy of the first player is defined by vector
U = {u1 , u 2,•••,um ), and the second player's strategy is defined by vector Z = {zr , z2 ,•••,zm ). U > 0, i = 1, m z, > 0, j = 1n
J 7 7
n n
Zu = 1 ILzj = 1
i=1 j=1
Pure strategies can be considered as a special case of mixed strategies and can be defined by a vector corresponding to 1 essence of a pure strategy.
T T* r-7*
The game solution is a pair of U , Z mixed strategies that have the following properties: if one player follows his own effective strategy, the other player cannot abandon his strategy.
According to the effective solution, the payoff is called the V cost of games and satisfies the following inequality:
X< V </
Each zero-sum game has a mixed strategic deci-
sion.
t t* ( * * * \
U =\u u * J and
Z =\zz*,. .,z* J are a couple of effective
strategies. If a pure strategy is included in a mixed strategy with a nonzero probability, it is called active.
If one side of the game adheres to its mixed effective strategy, and the other player does not deviate from his active strategies, the winnings in the game will not change and will be equal to the cost of the game.
These conclusions are of great practical importance, since they are a real model for finding effective strategies for nonstationary problems.
For a game without a fixed point, there is a game solution according to Neumann's theorem, and it is determined by a pair of mixed strategies
U* = U,u2) and Z* = (z*,Z*).
Let's find a theorem on active strategies to find
them. If player
A uses his U effective strategies,
then in any active strategy out of B his average payoff is equal to the cost of the game. In a game
without a
2 X 2 waiting point, any pure enemy strategy is active.
Win A (loss B) is a random variable whose mathematical expectation is the value of the game.
Consequently, the average gain A in an effective strategy falls on the first and second strategies of the opponent.
Let the game be represented by the following payout matrix:
A =
a,
V a 21
a
a
22 /
If player A uses his effective mixed strategy U* = (u*, u*), then player B can choose one of the pure strategies, so that the value of A's payoff remains unchanged.That is, player A has an effective combination of U* = (u*, u*), and player B has
Bj pure (according to the first column of the payout
matrix) strategies, the average payoff of A is equal to
the cost of the game v , which is:
* *
au * + a21u * = v In the strategy B2 of player B , the average
payoff of A is equal to the same game value, that is,
* *
al2ul + a22u2 = v. And if we take into account
* * -*
UI + u2 = I, then we come to the following system of equations:
au * + au 2 = v,
* *
auu * + a^u * = v,
22M2
**
u * + u * = 1
* *
ai1Z1 + ai2z* = v.
* *
a21Z1 + a22Z* = V. i Z1 + = 1
* *
anUi + a2Xu * = v,
* *
^ ^22^2 — v, u* + u * = 1.
u1 (a11 " a12 ) = u2 (a22 " a21) u1 = 1 — (a22 —a21)/(a11 —a12)= 1 —u2
u
u2 (a22 a21) a11 a12 u2 (a11 a12 )>
u2 =(a11 a12 )/(a22 a21 + a11 a
12J
u — ( a^ 1
)/(a22 a21 ^ a11 a12 )
* *
- u * + u * = v,
* *
i u* -u* = v,
* * 1 u * + u * = 1.
(1)
By solving (1), one can find U effective strategies and game values.
To determine an effective strategy for B, we write a system of similar equations:
(2)
From the third equation: u* = I — m* . Let's put the
rest:
f* 1 *
— u * +1 — u * = v,
**
u * — I + u * = v, We transform the equation:
|2m1* + v = I, [2m* — v = I,
Add equations: 4m* = 2, from him
* I A * I
u, = —, v = 0, u* = —. 1 2 2
We obtain the system of equations (2) for B :
— z* + z * = 0,
* * /~\
zI* z2* = 0,
**
ZI + Z2 = ^
* * From him, Zj = z2 =
2
That is, mixing pure strategies by choosing cover 1
with a probability of — is an effective strategy for
every player. The guaranteed payoff for each player is zero. The optimal probability of player A, equal to xi, i = 1,2,..m, can be determined by solving the following maximin problem.
maxi min I £ ax, £ a 2 x,... £ amxl
i=1
i=1
i=1
x + x, +... + x = 1,
12 m ?
X >0,i = 1,2,...,m. We will mark this problem to bring it to linear
programming.
1 m m m
v = min I £ aax,, £ a2x , . . ,£
ainXi
From him,
Let's go back to the «Search» report. Game without a fixed point, payout matrix:
A =
'1 -1A v1 -1/
a = — 1,
ß = 1
We are looking for a game solution in a mixed strategy. For player A, construct a system of equations (1):
£^a,Jx, >v. j = 1,2,...,n.
i=1
That is, the game for A is written as follows:
m
v~ £ ax ^ 0j = 1,2,.a
i=1
X + X, + ... + X = 1,
1 2 m
x >0,i = 1,2,...,m. The cost of the game can be negative or positive.
1
m
m
m
<
1
1
1
/
min s max
x
The y, ^2,..., yn effective strategies of player B are determined by solving the following problem:
^ n n n
E 1 , E a21
V j=1 j=1 j=1
yi + y 2 +... + yn = 1 y, >0,j = 1,2,...,n.
The task for B has the following form:
n
v-Ea^y, >0,i = 1,2,...,m,
,=1
Task.
y + y2 + •••+ y« = 1
F = 25x + 34x2 ^ max
X + x2 < 42 X + 2x2 < 48 X + 4x < 72
x > 0, X ^ 0
F - 25x - 34x2 = 0
X1 ^h X2 ^^ X3 — 42 xx + 2x2 + x4 = 48 xx + 4 x2 + x5 = 72 Table 1
<
<
Basis Empty members X| X2 X3 X4 X5
x3 42 □ 1 1 0 0
X4 48 1 2 0 1 0
X5 72 1 4 0 0 1
F 0 -25 -34 0 0 0
Let us find out if there are negative numbers in Fs, except for empty terms, if there are no such numbers, the problem is solved. We have - 25 < 0 (choose the smallest modulus between -25 and -34), look for positive numbers above that number. If there are no such numbers, then there is no solution to the problem. We have three numbers above -25: 1; 1 and 1.
We calculate:
. 142 48 721 42 „„ min \ —, —, — [> = — = 42
I 1 1 1 I 1
The element at the intersection of row (^3 ) and column (x ) is called a guide. We have it equal to 1.
(If the guiding element is equal to m ^ 1, then we divide all the elements of the line by this m). The unknown X1 is entered into the base, and the unknown
x3 is subtracted from it.
Complete the second simplex table. Row (x3 ) in
the first table now has row (x ). We then convert rows
(x ), (x5), and (F) of the first table so that their elements in column (xx) are converted to 0. For this,
1. From the elements of row (x ) we take the
elements of the corresponding row (x4 ) and write the
results of the difference into row (x4 ) of the second table;
2. From the elements of row (x ), we obtain the
elements of the corresponding row (x5) and write the
results of the difference into row (x5) of the second table;
3. Multiply the elements of row (x ) by 25 and add them with the corresponding elements of row (F) and write the result in row (F) of the second table.
We write the results into the following simplex table:
Table 2
Basis Empty members X| X2 X3 X4 X5
X 42 1 1 1 0 0
X4 6 0 -1 1 0
X5 30 0 3 -1 0 1
F 1050 0 -9 25 0 0
(F) has a negative number, that is, -9. Therefore, we will continue to look for an effective solution. There are three positive numbers above -9: 1; 1 and 3.
We calculate:
. f42 6 30] 6 , min [ — —, — J = — = 6 I I I 3 J I
The element located at the intersection of row (x4 ) and column (x2 ) is a guideline and is equal to 1. Unknown (x2 ) is entered into the database, and unknown (x4 ) is removed from it.
Complete the third simplex table. Row (x4) of the second table now contains row (x2). We then
transform the elements of rows (xx), (x5), and (F)
in column (x2 ) of the second table to 0. That is,
1. Subtract the elements of the corresponding row ) from the elements of row (x2 ) and write the result to row (xx) of the third table;
2. Multiply the elements of line (x2 ) by 3 and subtract from the elements of line (x5), write the results to line (x5) of the third table;
3. Multiply the elements of line (x2 ) by 9 and add them with the corresponding elements of line (F), write the results in line (F) of the third table.
We write the results into the following simplex table:
Table 3
Basis Empty members x^ x2 x3 x4 x5
x^ 36 1 0 2 -1 0
x2 6 0 1 -1 1 0
x5 12 0 0 2 -3 1
F 1104 0 0 16 9 0
Line (F) has no negative numbers. An effective solution was found: Fnax = 1104 Xi = 36,
X2 — 6 , X3 — X4 — 0 , Xj —12 .
References
1. Blackwell D., Girshik M. Theory of games and statistical decisions. - M.: IL, 1958.
2. Danilov V.I. Lectures on game theory. - M., 2001.
3. Dyubin G.N., Syuzdal V.G. Introduction to Applied Game Theory. - M.: Nauka, 1981.
4. Kovalenko A.A. Collection of problems in game theory. - Lviv, 1974.
5. Seitmuratov A., Taimuratova L. Conditions of extreme stress state// News of the National Academy of Sciences of the Republic of Kazakhstan. Series of geology and technology sciences.Volume 5, Number 437 (2019), 202 - 206 https://doi.org/10.32014/2019.2518-170X.143
6. Krushevsky A.V. Game theory. - Kiev: Vishcha school, 1977.
7. Moulin E. Game theory. - M., 1985.
8. Petrosyan L.A., Zenkevich N.A., Semina E.A. Game theory. - M.: Higher. Shk., 1998.
