01 Strategies and utilities

Objectives

  • Define mixed strategies
  • Understand utility calculation for mixed strategies
  • Understand utility calculation as a linear algebraic construct
  • Be able to identify dominated strategies
  • Understand notion of Common knowledge of granularity

Notes

Utility calculations

Use matching pennies form have students play in pairs. Following each game:

  • Ask how many people won?
  • Ask why they won?

Mixed strategies

Look at definition for mixed strategies.

Consider:

\[\begin{split}A = \begin{pmatrix} 2 & -2\\ -1 & 1 \end{pmatrix}\qquad B = \begin{pmatrix} -2 & 2\\ 1 & -1 \end{pmatrix}\end{split}\]

Let us assume we have \(\sigma_r=(.3, .7)\) and \(\sigma_c=(.1, .9)\):

\[u_r(\sigma_r, \sigma_c) = 0.3 \times 0.1 \times 2 + 0.3 \times 0.9 \times (-2) + 0.7 \times 0.1 \times (-1) + 0.7 \times 0.9 \times 1 = 0.08\]

because the game is zero sum we immediately know:

\[u_c(\sigma_r, \sigma_c) = -0.08\]

This corresponds to the linear algebraic multiplication:

\[u_r(\sigma_r, \sigma_c) = \sigma_r A \sigma_c^T\]
\[u_c(\sigma_r, \sigma_c) = \sigma_r B \sigma_c^T\]

(Go through this on the board, make sure students are comfortable.)

This can be done straightforwardly using numpy:

>>> import numpy as np
>>> A = np.array([[2, -2], [-1, 1]])
>>> B = np.array([[-2, 2], [1, -1]])
>>> sigma_r = np.array([.3, .7])
>>> sigma_c = np.array([.1, .9])
>>> np.dot(sigma_r, np.dot(A, sigma_c)), np.dot(sigma_r, np.dot(B, sigma_c))
(0.079..., -0.079...)

Strategy profiles as coordinates on a game

One way to thing of any game \((A, B)\in{\mathbb{R}^{m \times n}}^2\) is as a mapping from the set of strategies \([0,1]_{\mathbb{R}}^{m}\times [0,1]_{\mathbb{R}}^{m}\) to \(\mathbb{R}^2\): the utility space.

Equivalently, if \(S_r, S_c\) are the strategy spaces of the row/column player:

\[(A, B): S_r\times S_c \to \mathbb{R} ^2\]

We can use games defined in nashpy in that way:

>>> import nash
>>> game = nash.Game(A, B)
>>> game[sigma_r, sigma_c]
array([ 0.08, -0.08])

Rationalisation of strategies

Identify two volunteers and play a sequence of zero sum games where they play as a team against me. The group is the row player.

\[\begin{split}A = \begin{pmatrix} 1 & -1\\ -1 & 2 \end{pmatrix}\end{split}\]

(No dominated strategy)

\[\begin{split}A = \begin{pmatrix} 1 & 2\\ -1 & 2 \end{pmatrix}\end{split}\]

(First column weakly dominates second column)

\[\begin{split}A = \begin{pmatrix} 1 & -1\\ -1 & -3 \end{pmatrix}\end{split}\]

(First row strictly dominates second row) (Second column strictly dominates first column)

\[\begin{split}A = \begin{pmatrix} 2 & 2\\ -1 & 2 \end{pmatrix}\end{split}\]

(First row weakly dominates second row) (First column weakly dominates second column)

\[\begin{split}A = \begin{pmatrix} -1 & 2 & 1\\ -2 & -2 & 1\\ 1 & 1 & -1\\ \end{pmatrix}\end{split}\]

(First row dominates second row) (First column dominates second column)

Now pit the two players against each other, the utilities represent the share of the total amount of chocolates/sweets gathered so far:

\[\begin{split}A = \begin{pmatrix} 1 & 0\\ 1.5 & .5 \end{pmatrix}\qquad B = \begin{pmatrix} 1 & 1.5\\ 0 & .5 \end{pmatrix}\end{split}\]

Capture all of the above (on the white board) and discuss each action and why they were taken.

Iterated elimination of dominated strategies

As a class work through the following example.

\[\begin{split}A = \begin{pmatrix} 2 & 5 \\ 1 & 2 \\ 7 & 3 \end{pmatrix}\qquad B = \begin{pmatrix} 0 & 3 \\ 6 & 1 \\ 0 & 1 \end{pmatrix}\end{split}\]

  1. First row dominates second row

    \[\begin{split}A = \begin{pmatrix} 2 & 5 \\ 7 & 3 \end{pmatrix}\qquad B = \begin{pmatrix} 0 & 3 \\ 0 & 1 \end{pmatrix}\end{split}\]

  2. Second column dominates first column

    \[\begin{split}A = \begin{pmatrix} 2\\ 7 \end{pmatrix}\qquad B = \begin{pmatrix} 0\\ 0 \end{pmatrix}\end{split}\]

  3. Second (third) row dominates first row. Thus the rationalised behaviour is \((r_3, c_1)\).

Now return to the last example played as a pair:

\[\begin{split}A = \begin{pmatrix} -1 & 2 & 1\\ -2 & -2 & 1\\ 1 & 1 & -1\\ \end{pmatrix}\qquad B = \begin{pmatrix} 1 & -2 & -1\\ 2 & 2 & -1\\ -1 & -1 & 1\\ \end{pmatrix}\end{split}\]

  1. The first row/column weakly dominate the second row/column:

    \[\begin{split}A = \begin{pmatrix} -1 & 1 \\ 1 & -1 \end{pmatrix}\qquad B = \begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}\end{split}\]

There is nothing further that we can do here.