Mixed strategy

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The concept of mixed strategy is used in game theory as a generalization of the concept of (pure) strategy . A strategy is the establishment of a complete plan of action before a game . With a mixed strategy, the player does not make a direct decision, but rather chooses a random mechanism that determines a pure strategy. The decision made in a game for a specific action plan is therefore purely random and is only indirectly subject to strategic considerations, insofar as the player has taken it into account when selecting the random mechanism. Mathematically, a mixed strategy is characterized by a probability distribution over the pure strategies.

The transition from a game in which only pure strategies are considered to the game in which mixed strategies are also permitted is also referred to as mixed expansion .

Mixed strategies were first used by Émile Borel (1921) and John von Neumann (1928).

Existence of a Nash equilibrium under mixed strategies

In some normal form games, there is no Nash equilibrium in the pure strategy area . This means that there is no combination of strategies from which no single player can gain an advantage by changing his strategy on his own. However, every finite game has a Nash equilibrium in mixed strategies. A Nash equilibrium in mixed strategies therefore consists of a mixed strategy for each player, with the property that each player's mixed strategy is the best answer to the mixed strategies of the other players.

Examples

Symmetrical game

2 players each have a black and a white marble. The rules are: Player A wins if the colors of the marbles are the same when drawn (black-black or white-white). Player B wins if the colors of the marbles are different (white-black or black-white). What could player A's strategy look like? If he chooses the black marble, player B will always choose the white marble and player A loses. Even if player A changes his strategy and chooses the white marble, player B also changes his strategy and this time chooses black as the answer - A loses again.

If player B starts, player A will also adjust his strategy. It follows that no player can gain an advantage with the right combination of marbles. If the opponent guesses the strategy, he can always choose a suitable counter-strategy that will secure him victory and vice versa.

Player A / Player B black White
black 1, −1 −1, 1
White −1, 1 1, −1

In this described game, there cannot be a Nash equilibrium if both players choose a pure strategy. The only remedy can be a randomized selection, i.e. a game using a random selection of procedures. Only if both players take the white or black marble purely by chance with a probability of 50% would there be no incentive for anyone to deviate from this random strategy and a Nash equilibrium inevitably arises.

The proof:

In practice, the problem in the example described above can be solved by both players pulling the marbles out of a darkened vessel (urn drawing).

Asymmetrical game

Player A has to park her car and can choose between a very convenient parking space, which is unfortunately illegal, and a legal but distant parking lot. The convenient parking lot secures her a profit of 10 (if she is not caught) and the one further away has no profit (i.e. 0). If she is caught in the convenient parking lot, she has to pay a fine (her loss here is -90). Player B is from the city and can check the parking lots. Since inspecting takes time, the corresponding payout is -1. At the same time, illegal parking in the city causes high losses of -10. These losses are partially offset when the wrongdoer is caught and has to pay a fine, then it is -6 for the city. The situation is shown in the following profit matrix:

Motorist / inspector Check Don't check
Illegal parking -90, -6 10, -10
Legal parking 0, -1 0.0

There is no Nash equilibrium here in pure strategies. The driver chooses legal parking and the inspector inspects. But it would be better if the driver parks legally and doesn't even have to be inspected.

Nevertheless, the strategies can be selected randomly. For this purpose, we assume that the driver is probably parking in the convenient, illegal parking lot. Consequently, with the opposite probability, she chooses the parking lot further away. She wants to choose these probabilities so that the inspector has no incentive to deviate from his strategy. So she has to make his expected profit equal for both of his strategies. If the inspector decides to check, his expected profit is . If the inspector chooses not to control, his expected gain is . By equating these terms, we get and the driver should park wrong with this probability. The other way around we assume that the inspector will probably decide to check. Then we have to solve and maintain .

The two mixed strategies for player A and for player B then form a mixed Nash equilibrium.

literature

supporting documents

  1. Robert S. Pindyck / Daniel L. Rubinfeld: Microeconomics , p 662, Pearson Education, Munich, 2003, ISBN 3-8273-7025-6 .
  2. Manfred J. Holler / Gerhard Illing: Introduction to Game Theory , Springer Verlag, Heidelberg, 2008, ISBN 978-3540693727 .
  3. Jörg Bewersdorff: With luck, logic and bluff: Mathematics in play - methods, results and limits, p. 250, Vieweg + Teubner, Wiesbaden, 2007, ISBN 3-8348-0087-2 .
  4. Harald Wiese: Decision and Game Theory , p. 199, Springer Verlag, Berlin, 2001, ISBN 3540427473 .
  5. Gernot Sieg: Game Theory , p. 17, Oldenbourg Verlag, Munich, 2005, ISBN 978-3486275261 .
  6. ^ Christian Rieck: Game Theory: An Introduction , p. 78, Christian Rieck Verlag, Eschborn, 2008, ISBN 3-924043-91-4 .
  7. ^ Anna R. Karlin and Yuval Peres: Game Theory, Alive . December 13, 2016, p. 76 .