/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 5 Expected value of conjecture. Su... [FREE SOLUTION] | 91Ó°ÊÓ

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Expected value of conjecture. Suppose both players play your ideal strategy in the Matching Pennies game, what should the expected value of the game be? We could use our previous graphical method to determine the expected value of the game (you might quickly try this just to verify your prediction). However, as we have noted, a major drawback of the graphical solution is that if our players have 3 (or more) options, then we would need to graph an equation in 3 (or more!) variables; which, I hope you agree, we don't want to do. Although we will continue to focus on \(2 \times 2\) games, we will develop a new method which can more easily be used to solve to larger games. We will need a little notation. Let \(\begin{aligned} P_{1}(H) &=\text { the probability that Player } 1 \text { plays } \mathrm{H} \\ P_{1}(T) &=\text { the probability that Player } 1 \text { plays } \mathrm{T} ; \\ P_{2}(H) &=\text { the probability that Player } 2 \text { plays } \mathrm{H} ; \\ P_{2}(T) &=\text { the probability that Player } 2 \text { plays } \mathrm{T} \end{aligned}\) Also, we will let \(E_{1}(H)\) be the expected value for Player 1 playing pure strategy H against a given strategy for Player 2. Similarly, \(E_{2}(H)\) will be Player 2's expected value for playing pure strategy \(\mathrm{H}\).

Short Answer

Expert verified
The expected value of the game is 0 when both players use the ideal strategy.

Step by step solution

01

Define the Game Matrix

The Matching Pennies game can be represented by a payoff matrix for Player 1, where row headings are Player 1's choices (Heads, Tails) and column headings are Player 2's choices (Heads, Tails):\[\begin{bmatrix}1 & -1 \-1 & 1 \\end{bmatrix}\]Here, the values represent Player 1's payoffs, and Player 2 gets the opposite payoff. If they match, Player 1 wins (+1) and Player 2 loses (-1), and vice versa if they don't match.
02

Assign Probabilities

Since both players are using mixed strategies, let's assign probabilities to each action: - Player 1 plays Heads with probability \( P_1(H) \)- Player 1 plays Tails with probability \( P_1(T) = 1 - P_1(H) \)- Player 2 plays Heads with probability \( P_2(H) \)- Player 2 plays Tails with probability \( P_2(T) = 1 - P_2(H) \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Game Matrix
The concept of a game matrix is foundational in game theory. A game matrix represents the possible outcomes and strategies of players in a structured way, usually in a tabular form. In the game of Matching Pennies, the game matrix is defined by a grid where each player's choices are laid out along the rows and columns. This structuring helps in visualizing the interaction between the players.

For example, the Matching Pennies game can be represented as:
  • Rows denote Player 1's strategy choices: Heads (H) and Tails (T).
  • Columns denote Player 2's strategy choices: Heads (H) and Tails (T).
  • The entries in the matrix display Player 1's payoffs with corresponding opposite signs for Player 2.
Understanding a game matrix involves recognizing each cell's value, which indicates the payoff to the players depending on the combination of strategies they choose. In this game, the matrix reflects that if both players choose the same strategy (both Heads or both Tails), Player 1 wins (+1) while Player 2 loses (-1). Conversely, if they choose different strategies, Player 1 loses (-1) and Player 2 wins (+1). This concise setup of a game matrix allows players to clearly see outcomes and plan their strategies accordingly.
Payoff Matrix
A payoff matrix is a specific type of game matrix that assigns numerical values to the outcomes of a player’s strategy choices. It helps players determine their most profitable strategies by offering a clear representation of possible outcomes.

The Matching Pennies payoff matrix for Player 1 looks like this:\[\begin{bmatrix}1 & -1 \-1 & 1\end{bmatrix}\]
  • Here, the entries correspond to Player 1's payoffs.
  • If heads match, Player 1 receives a payoff of +1; otherwise, it’s -1.
The opposite payoffs are true for Player 2. The payoff matrix reflects these results:
  • Matching choices lead to a gain for Player 1 and a loss for Player 2.
  • Non-matching choices result in the reverse.
Understanding the payoff matrix is crucial as it directly impacts decision-making. Each player uses their knowledge of the matrix to anticipate the possible outcomes, which guide them in optimizing their strategies. Players often use the matrix to calculate expected payoffs and make informed decisions about their next move.
Mixed Strategies
Mixed strategies involve players choosing among their possible actions with certain probabilities instead of sticking to one pure strategy. This approach is especially useful in games like Matching Pennies where no single strategy offers a definite winning path.

In our Matching Pennies example, we describe mixed strategies as follows:
  • Player 1 can play Heads with a probability of \(P_1(H)\) and Tails with \(P_1(T) = 1 - P_1(H)\).
  • Similarly, Player 2 can play Heads with a probability of \(P_2(H)\) and Tails with \(P_2(T) = 1 - P_2(H)\).
Mixed strategies offer a randomized element that makes it harder for opponents to predict and counteract a player's decisions. By varying their choices according to these probabilities, players can potentially maximize their expected payoffs while maintaining an element of surprise.

This strategic flexibility is a fundamental aspect of game theory, especially in games with no apparent pure strategy equilibrium. By applying mixed strategies, players balance their actions to avoid predictability and make the game more competitive.

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Most popular questions from this chapter

Consider the system of equations $$\begin{array}{cc} 3 x+2 y-w-v & =0 \\ 2 x-y+3 z+w+5 v & =0 \\ x+2 y+6 z-w & =0 \\ -y+z-3 w+v & =0 \\ x+y+z+w+v & =1 \end{array}$$ a Set up the augmented matrix for this system. b Use row reduction to find the solution. Now we are ready to apply everything we have learned about solving repeated zero-sum games to a much more challenging game in the next section!

Consider the system of equations $$\begin{array}{cl} 2 x+y-4 z & =10 \\ 3 x+5 z & =-5 \\ y+2 z & =7 \end{array}$$ a Set up the augmented matrix for this system. b Use row reduction to find the solution.

Consider the following zero-sum game $$ \left[\begin{array}{cc} -1 & 1 \\ 0 & 2 \end{array}\right] $$ This game has an equilibrium pair. Convince yourself that if this game is played repeatedly, each player should choose to play a pure strategy. Thus, if the game has an equilibrium we know that players will play the pure strategies determined by the equilibrium pairs. So let's get back to thinking about games without equilibrium pairs. If we play such a game once, can we predict the outcome? What about if we repeat the game several times- can we predict the outcome? Think about tossing a coin. If you toss it once, can you predict the outcome? What if you toss it 100 times- can you predict the outcome? Not exactly, but we can say what we expect: if we toss a coin 100 times we expect to have half of the coins turn up heads and half turn up tails. This may not be the actual outcome, but it is a reasonable prediction. Now is a good time to remind yourself about finding the expected value!! Recall the familiar game Rock-Paper-Scissors: ROCK beats SCISSORS, SCISSORS beat PAPER, and PAPER beats ROCK. Using the payoff matrix and experimentation, we will try to determine the best strategy for this game.

It should be easy to see that the maximum of the minimum expected payoffs occurs at the intersection of the two lines. Step 3. Find the intersection of the two lines. Step 3 a. Find the equation for Line \(C\). This is the line passing through the points (0,1) and \((1,-1) .\) It has slope -2 and \(y\) -intercept 1\. Thus, it has equation \(y=-2 x+1\). [Although the \(x\) -axis represents probability \(p\) and the \(y\) -axis represents expected payoff \(m,\) you are probably more comfortable solving equations-at least for the moment-in \(x\) and \(y\). Step \(3 \mathrm{~b}\). Find the equation for Line \(\mathrm{D}\). This is the line passing through the points (0,0) and (1,2) . It has slope 2 and \(y\) -intercept 0 . Thus, it has equation \(y=2 x\). Step \(3 \mathrm{c}\). Use substitution to find the point of intersection. $$\begin{array}{rlr}2 x & = & -2 x+1 \\ 4 x & = & 1 \\ x & = & \frac{1}{4}\end{array}$$ Substituting \(x=\frac{1}{4}\) back in to either original equation, say \(y=2 x,\) gives us \(y=\frac{1}{2}\). Thus, the point of intersection is \((1 / 4,1 / 2)\). Step 4. Determine Player 1's maximin mixed strategy. Recalling that the first coordinate is \(p,\) the probability that Player 1 plays \(B,\) we know that Player 1 will play \(\mathrm{B}\) with probability \(1 / 4\), and thus, play \(\mathrm{A}\) with probability \(3 / 4[1-1 / 4=3 / 4] .\) The expected payoff for Player 1 is \(1 / 2\) It is important to check back to your original intuition about the game from Exercise 3.2.2. Did it seem as though Player 1 should play A more often than B? Let's make a few important observations. First, it should be clear from the graph that Player 1 expects a payoff of \(1 / 2\) NO MATTER WHAT PLAYER 2 DOES. Furthermore, since this is a zero-sum game, we know that Player 2 's expected payoff is \(-1 / 2\). It is important to note that this graph does not give us any information about an optimal strategy for Player \(2 .\) We will see how to find a strategy for Player 2 in the following exercises. Can you think of how you might do this?

Consider the system of equations $$\begin{array}{cc} 4 x+8 y-4 z & =4 \\ 3 x+8 y+5 z & =-11 \\ -2 x+y+12 z & =-17 \end{array}$$ a Set up the augmented matrix for this system. b Use row reduction to find the solution.

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