# 02 Rationalisation¶

## Corresponding chapters¶

Duration: 50 minutes

## Objectives¶

• Be able to identify dominated strategies
• Understand notion of Common knowledge of rationality

## Notes¶

### 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.

Ask students to repeat the game against each other in groups of two (use “days of doing the dishes perhaps?”).

### 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

$\begin{split}A = \begin{pmatrix} 2\\ 7 \end{pmatrix}\qquad B = \begin{pmatrix} 3\\ 1 \end{pmatrix}\end{split}$
3. First row dominates third row. Thus the rationalised behaviour is $$(r_1, c_2)$$.

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