If in the game each of the opponents uses only one and the same strategy, then the game itself is said to take place in this case. in pure strategies , and used by the player A and player IN a couple of strategies are called pure strategies .
Definition. In an antagonistic game, a pair of strategies ( A i , IN j) is called equilibrium or stable if it is not profitable for any of the players to deviate from their strategy.
It makes sense to use pure strategies when the players A And IN have information about each other's actions and the results achieved. If we assume that at least one of the parties does not know about the behavior of the opponent, then the idea of equilibrium is violated, and the game is played haphazardly.
Consider the matrix game G(3x4)
In this example, the lower price of the game is equal to the upper one: ==9, i.e. the game has a saddle point.
It turns out that in this case the maximin strategies A 2 and IN 2 will sustainable in relation to information about the behavior of the enemy.
Indeed, let the player A learned that the enemy is using a strategy IN 2. But in this case, the player A will continue to follow the strategy A 2 because any deviation from the strategy A 2 will only reduce the gain. Likewise, the information received by the player IN, will not make him deviate from his strategy IN 2 .
Pair of strategies A 2 and IN 2 has the property of stability, and the payoff (in the example under consideration it is equal to 9) achieved with this pair of strategies turns out to be the saddle point of the payoff matrix.
A sign of stability (equilibrium) of a strategy pair is the equality of the lower and upper prices of the game.
Strategies A i And IN j(in this example A 2 , IN 2), under which the equality of the lower and upper prices of the game is satisfied, are called optimal pure strategies, and their combination is the solution of the game. In this case, the game itself is said to be solved in pure strategies.
The value is called the price of the game.
If 0, then the game is beneficial for player A, if 0 - for player B; for =0 the game is fair, i.e. is equally beneficial for both participants.
However, the presence of a saddle point in the game is far from being the rule; rather, it is an exception. Most matrix games do not have a saddle point and therefore do not have optimal pure strategies. However, there is a variety of games that always have a saddle point and, therefore, are solved in pure strategies. These are games with complete information.
Theorem 2. Every game with perfect information has a saddle point and, therefore, is solved in pure strategies, i.e. there is a pair of optimal pure strategies that gives a stable payoff equal to.
If such a game consists only of personal moves, then when each player applies his own optimal pure strategy, it must end with a payoff equal to the price of the game. Let's say a chess game, as a game with complete information, either always ends in a win for White, or always in a win for Black, or always in a draw (we don't know exactly what, because the number of possible strategies in a chess game is huge).
If the game matrix contains a saddle point, then its solution is immediately found using the maximin principle.
The question arises: how to find a solution to the game, payment matrix which has no saddle point? The application of the maximin principle by each of the players provides player A with a gain no less, and a loss no more for the player. Given that, it is natural for player A to want to increase the gain, and for player B to reduce the loss. The search for such a solution leads to the need to apply mixed strategies: to alternate pure strategies with some frequencies.
Definition. A random variable whose values are pure strategies of the player is called his mixed strategy .
Thus, the task of a player's mixed strategy is to indicate the probabilities with which his pure strategies are chosen.
We will denote the mixed strategies of the players A And IN respectively
S A =||p 1 , p 2 , ..., p m ||,
S B =||q 1 , q 2 , ..., q n ||,
where p i is the probability of the player using A clean with strategy A i ; ; q j - the probability of using by the player In a pure strategy B j ; .
In a particular case, when all probabilities, except for one, are equal to zero, and this one is equal to one, the mixed strategy turns into a pure one.
Application mixed strategies is carried out, for example, in the following way: the game is repeated many times, but in each game the player uses different pure strategies with relative frequencies of their use equal to p i And q j .
Mixed strategies in game theory are a model of changeable, flexible tactics, when neither player knows which pure strategy the opponent will choose in a given game.
If the player A applies a mixed strategy S A =||p 1 , p 2 , ..., p m ||, and the player IN mixed strategy S B =||q 1 , q 2 , ..., q n ||, then the average payoff ( expected value) player A is determined by the ratio
Naturally, the expected loss of the player IN is equal to the same value.
So, if the matrix game does not have a saddle point, then the player must use the optimal mixed strategy that will provide the maximum payoff.
The question naturally arises: what considerations should be taken into account when choosing mixed strategies? It turns out that the maximin principle retains its value in this case as well. In addition, the basic theorems of game theory play an important role in understanding the solution of games.
Among the finite games that have practical value, games with a saddle point are relatively rare; more typical is the case when the lower and upper price - games are different. Analyzing the matrices of such games, we came to the conclusion that if each player is given a choice
one - the only strategy., then, based on a reasonably acting opponent, this choice should be determined by the minimax principle. Adhering to our maximin strategy, we certainly guarantee ourselves a payoff equal to the lower price of the game, a, for any behavior of the opponent. A natural question arises: is it possible to guarantee yourself an average payoff greater than a if you apply not just one "pure" strategy, but alternate several strategies at random?
Such combined strategies, consisting in the application of several pure strategies alternating according to a random law with a certain ratio of frequencies, are called mixed strategies in game theory.
Obviously, each pure strategy is a special case of a mixed one, in which all strategies except one are applied with zero frequencies, and this one - with a frequency of 1.
It turns out that by applying not only pure but also mixed strategies, it is possible to obtain a solution for each finite game, i.e., a pair of (generally mixed) strategies such that when both players apply them, the payoff will be equal to the price game, and in case of any one-sided deviation from the optimal strategy, the payoff can only change in the direction unfavorable for the deviant.
The stated statement is the content of the so-called main theorem of game theory. This theorem was first proved by von Neumann in 1928. The known proofs of the theorem are relatively complex; therefore, we present only its formulation.
Every finite game has at least one solution (perhaps in the realm of mixed strategies).
The payoff resulting from the decision is called the price of the game. It follows from the main theorem that every finite game has a price. It is obvious that the price of the game v always lies between the lower price of the game a and the upper price of the game :
Indeed, there is a maximum guaranteed payoff that we can secure for ourselves using only our own pure strategies. Since mixed strategies include, as a special case, all pure ones, then, allowing, in addition to pure ones, also mixed
strategy, we, in any case, do not worsen our capabilities; hence,
Similarly, considering the capabilities of the opponent, we show that
whence the required inequality (3.1) follows.
Let us introduce a special notation for mixed strategies. If, for example, our mixed strategy consists in applying the AL strategies, with frequencies and we will denote this strategy
Similarly, the adversary's mixed strategy will be denoted by:
where are the frequencies at which strategies are mixed
Suppose that we have found a solution to the game consisting of two optimal mixed strategies S, S. In the general case, not all pure strategies available to a given player are included in his optimal mixed strategy, but only some of them. We will call the strategies included in the player's optimal mixed strategy his "useful" strategies.
It turns out that the solution to the game has another remarkable property: if one of the players adheres to his optimal mixed strategy 5 (5). then the payoff remains unchanged and equal to the price of the game v, regardless of what the other player does, if he. only does not go beyond its "useful" strategies. He, for example, can use any of his "useful" strategies in pure form, and can also mix them in any proportions.
Let's prove this statement. Let there be a solution to the game . For concreteness, we will assume that the optimal mixed strategy consists of a mixture of three
"useful" strategies respectively consists of a mixture of three "useful" strategies
and It is stated that if we stick to strategy S, then the opponent can apply strategies in any proportions, and the payoff will remain unchanged and will still be equal to the price of the game
The player's choice of an action is called move. There are moves personal(the player consciously makes a decision) and random(the outcome of the game does not depend on the will of the player). The set of rules that determine which move a player must make is called strategy. There are strategies clean(nonrandom player decisions) and mixed(the strategy can be considered as a random variable).
saddle point
IN game theory S. t. ( saddle element) - This largest element column game matrices, which is also the smallest element of the corresponding row (in two-person zero-sum game). At this point, therefore, the maximin of one player is equal to the minimax of the other; S. t. there is a point equilibrium.
Minimax theorem
The minimax strategy is called minimax strategy.
The principle that dictates to players the choice of the most "cautious" maximin and minimax strategies is called minimax principle. This principle follows from the reasonable assumption that each player seeks to achieve the opposite goal of the opponent.
The player chooses his actions, assuming that the opponent will act in an unfavorable way, i.e. will try to harm.
Loss function
Loss function is a function that, in the theory of statistical decisions, characterizes the losses due to incorrect decision making based on the observed data. If the problem of estimating the signal parameter against the background of interference is being solved, then the loss function is a measure of the discrepancy between true value estimated parameter and parameter estimate
Player's Optimal Mixed Strategy is a complete set of applications of his pure strategies in multiple repetitions of the game under the same conditions with given probabilities.
A player's mixed strategy is a complete set of application of his pure strategies in the case of multiple repetitions of the game under the same conditions with given probabilities.
1. If all elements of a row are not greater than the corresponding elements of another row, then the original row can be deleted from the payoff matrix. Likewise for columns.
2. The price of the game is unique.
Doc-in: let's say there are 2 game prices v and , which are achieved on the pair and respectively, then
3. If the same number is added to all elements of the payoff matrix, then the optimal mixed strategies will not change, and the price of the game will increase by this number.
Doc-in:
, Where 
4. If all elements of the payoff matrix are multiplied by the same number that is not equal to zero, the price of the game will be multiplied by this number, and the optimal strategies will not change.
Pure strategy player I is to choose one of the n rows of the payoff matrix A, and player II's pure strategy is to choose one of the columns of the same matrix.Optimal pure strategies of players differ from mixed strategies by the presence of a mandatory unit p i = 1, q i = 1. For example: P(1,0), Q(1,0). Here p 1 = 1, q 1 = 1.
Task 1
Based on the payoff matrix, find the optimal pure strategies using the principle of strict dominance. As an answer, write down the vectors P*, Q*.
R1 | R2 | R3 | R4 |
|
S1 | 3 | 1 | 2 | 5 |
S2 | 2 | 0 | 0 | 3 |
S3 | -3 | -5 | -5 | -2 |
S4 | 0 | -2 | -2 | 1 |
Solution:
We solve all problems using the Matrix Game calculator.
We assume that player I chooses his strategy in such a way as to maximize his payoff, and player II chooses his strategy in such a way as to minimize the payoff of player I.
| Players | B1 | B2 | B3 | B4 | a = min(Ai) |
| A 1 | 3 | 1 | 2 | 5 | 1 |
| A2 | 2 | 0 | 0 | 3 | 0 |
| A 3 | -3 | -5 | -5 | -2 | -5 |
| A4 | 0 | -2 | -2 | 1 | -2 |
| b = max(Bi) | 3 | 1 | 2 | 5 |
The upper price of the game b = min(b j) = 1.
The saddle point (1, 2) indicates a solution to a pair of alternatives (A1,B2). The value of the game is 1.
2. Check the payoff matrix for dominant rows and dominant columns.
Sometimes, based on a simple consideration of the game matrix, it can be said that some pure strategies can enter the optimal mixed strategy only with zero probability.
They say that i-th 1st player's strategy dominates him k-th strategy if a ij ≥ a kj for all j E N and for at least one j a ij > a kj . In this case it is also said that i-th strategy (or line) is dominant, k-th- dominated.
They say that j-th 2nd player's strategy dominates him l-th strategy if for all j E M a ij ≤ a il and for at least one i a ij< a il . В этом случае j-th strategy (column) is called dominant, l-th- dominated.
Strategy A 1 dominates strategy A 2 (all elements of row 1 are greater than or equal to the values of the 2nd row), therefore we exclude the 2nd row of the matrix. Probability p 2 = 0.
Strategy A 1 dominates strategy A 3 (all elements of row 1 are greater than or equal to the values of the 3rd row), therefore we exclude the 3rd row of the matrix. Probability p 3 = 0.
| 3 | 1 | 2 | 5 |
| 0 | -2 | -2 | 1 |
From the standpoint of losses of player B, strategy B 1 dominates strategy B 2 (all elements of column 1 are greater than elements of column 2), therefore, we exclude the 1st column of the matrix. Probability q 1 = 0.
From the standpoint of player B's losses, strategy B 4 dominates strategy B 1 (all elements of column 4 are greater than elements of column 1), therefore, we exclude the 4th column of the matrix. Probability q 4 = 0.
| 1 | 2 |
| -2 | -2 |
We reduced the 4 x 4 game to a 2 x 2 game.
Game solution ( 2 x n
p 1 = 1
p2 = 0
Game price, y = 1
Now we can find the minimax strategy of player B by writing the corresponding system of equations
q 1 = 1
q1 +q2 = 1
Solving this system, we find:
q1 = 1.
Answer:
Game price: y = 1, players strategy vectors:
Q(1, 0), P(1, 0)
∑a ij q j ≤ v
∑a ij p i ≥ v
M(P 1 ;Q) = (1 1) + (2 0) = 1 = v
M(P 2 ;Q) = (-2 1) + (-2 0) = -2 ≤ v
M(P;Q 1) = (1 1) + (-2 0) = 1 = v
M(P;Q 2) = (2 1) + (-2 0) = 2 ≥ v
Since rows and columns were removed from the original matrix, the found probability vectors can be written as:
P(1,0,0,0)
Q(0,1,0,0)
Task 2
Using the payoff matrix, find the lower and upper price of the game. In the presence of a saddle point, write down the vectors of optimal pure strategies P*, Q*.
R1 | R2 | R3 |
|
S1 | -6 | -5 | 0 |
S2 | -8 | -3 | -2 |
S3 | -3 | -2 | 3 |
Solution:
1. Check if the payoff matrix has a saddle point. If yes, then we write out the solution of the game in pure strategies.
| Players | B1 | B2 | B3 | a = min(Ai) |
| A 1 | -6 | -5 | 0 | -6 |
| A2 | -8 | -3 | -2 | -8 |
| A 3 | -3 | -2 | 3 | -3 |
| b = max(Bi) | -3 | -2 | 3 |
We find the guaranteed payoff determined by the lower price of the game a = max(a i) = -3, which indicates the maximum pure strategy A 3 .
The upper price of the game b = min(b j) = -3.
The saddle point (3, 1) indicates a solution to a pair of alternatives (A3,B1). The price of the game is -3.
Answer: P(0,0,1), Q(1,0,0)
Task 3
Using the payoff matrix, find the vectors of optimal strategies P*, Q* and the price of the game. Which player is the winner?
R1 | R2 | R3 | R4 |
|
S1 | -6 | -6 | 2 | 4 |
S2 | 2 | -2 | 7 | -1 |
Solution:
1. Check if the payoff matrix has a saddle point. If yes, then we write out the solution of the game in pure strategies.
We assume that player I chooses his strategy in such a way as to maximize his payoff, and player II chooses his strategy in such a way as to minimize the payoff of player I.
| Players | B1 | B2 | B3 | B4 | a = min(Ai) |
| A 1 | -6 | -6 | 2 | 4 | -6 |
| A2 | 2 | -2 | 7 | -1 | -2 |
| b = max(Bi) | 2 | -2 | 7 | 4 |
We find the guaranteed payoff determined by the lower price of the game a = max(a i) = -2, which indicates the maximum pure strategy A 2 .
The upper price of the game b = min(b j) = -2.
The saddle point (2, 2) indicates a solution to a pair of alternatives (A2,B2). The price of the game is -2.
3. We find a solution to the game in mixed strategies.
We solve the problem by a geometric method, which includes the following steps:
1. In the Cartesian coordinate system, a segment is plotted along the abscissa axis, the length of which is equal to 1. The left end of the segment (point x = 0) corresponds to strategy A 1, the right end - to strategy A 2 (x = 1). Intermediate points x correspond to the probabilities of some mixed strategies S 1 = (p 1 ,p 2).
2. The payoffs of strategy A 1 are plotted on the left y-axis. On the line parallel to the y-axis, the payoffs of strategy A 2 are plotted from point 1.
Game solution ( 2 x n) we draw from the position of player A, who adheres to the maximin strategy. None of the players has dominating and duplicating strategies.
The maximin optimal strategy of player A corresponds to the point N, for which we can write next system equations:
p1 = 0
p2 = 1
Game price, y = -2
Now we can find the minimax strategy of player B by writing the corresponding system of equations, eliminating the strategy B 1 ,B 3 ,B 4 , which gives a clearly greater loss to player B, and hence q 1 = 0,q 3 = 0,q 4 = 0 .
-2q 2 = -2
q2 = 1
Solving this system, we find:
q2 = 1.
Answer:
Game price: y = -2, players strategy vectors:
Q(0, 1, 0, 0), P(0, 1)
4. Check the correctness of the game solution using the strategy optimality criterion.
∑a ij q j ≤ v
∑a ij p i ≥ v
M(P 1 ;Q) = (-6 0) + (-6 1) + (2 0) + (4 0) = -6 ≤ v
M(P 2 ;Q) = (2 0) + (-2 1) + (7 0) + (-1 0) = -2 = v
M(P;Q 1) = (-6 0) + (2 1) = 2 ≥ v
M(P;Q 2) = (-6 0) + (-2 1) = -2 = v
M(P;Q 3) = (2 0) + (7 1) = 7 ≥ v
M(P;Q 4) = (4 0) + (-1 1) = -1 ≥ v
All inequalities are satisfied as equalities or strict inequalities, therefore, the solution of the game is found correctly.
Task 4
Give a detailed answer to the question
"Pure" strategies
We are already familiar with jambs. However, what will happen if jambs are removed from the chain of any strategy? We will get a "pure strategy". Pure strategies are those in the chain of actions of which, from the very root to the effective part, there are no ineffective sub-strategies (jambs), and this can often be evidenced only by the presence of all links in the mind.
Of course, from the point of view of all possible outcomes of applying the strategy, it is difficult for us to talk about the most effective one, since we simply may not have certain experience, and therefore certain intermediate strategies, but it is from our experience that the strategy should be as effective as possible.
The concept of pure strategies is also one of the key concepts in these materials, so I will give an example:
Evening. You are in your native area, hurry home. The milk is running out. As you fly past a "suspicious type of some-many" you hear yourself say "Hey you, [censored]. You don’t go here, the snow will fall on your head!
What will you do? There may be many options. Someone will go to sort things out, someone will get scared and quicken their pace, someone will shout something in response. However, let's think, what is the pure strategy of behavior in this case?
A stranger is shouting something to you on the street. You have your own affairs, on which you actually go. Judging by the text, positive benefits for you from communicating with this person are unlikely. The logical conclusion: calmly go about your business. I draw attention to the fact that it is “calmly”, without a shadow negative emotions but with a healthy indifference to what is happening. How many people will do this? I guess the vast minority. Why?
Because most people have a whole layer of subconscious strategies tied in the lower layers to self-preservation, in particular, these can be: “Always respond to rudeness with rudeness”, “If someone says nasty things, then you need to run”, “If someone rude - you need to fill his face”, “If someone is rude, then there is a danger”, and the like in different variations. Of course, not everyone will take some kind of active action, but emotionally it will hurt almost everyone. And this is a joint.
Pure strategies are always emotionally neutral or positive, and this is embedded in your brain, you just have to use it.
You can read a bit about pure strategies in the notes “Why pure strategies?” and "House, Hopkins, et cetera."
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