Rationalisation - Solutions

  1. Give the definition of a dominated strategy.

    Bookwork: https://vknight.org/gt/chapters/03/#Definition-of-a-strictly-dominated-strategy

  2. Give the definition of a weakly dominated strategy.

    Bookwork: https://vknight.org/gt/chapters/03/#Definition-of-a-weakly-dominated-strategy

  3. Give the defininition of common knowledge of rationality.

    Bookwork: https://vknight.org/gt/chapters/03/#Definition-of-a-weakly-dominated-strategy

  4. For the following games predict rational behaviour or explain why this cannot be done:

    1. $ A = \begin{pmatrix} 2 & 1\\ 1 & 1\end{pmatrix} \qquad B = \begin{pmatrix} 1 & 1\\ 1 & 3\end{pmatrix} $

    We see that $r_1$ weakly dominates $r_2$ so we have:

    $$A=(2,1)\qquad B =(1,1)$$

    There are no further strategies that can be eliminated.

    We see however that $c_2$ weakly dominates $c_1$ which would give:

    $$\begin{pmatrix}

     (1,1)\\
     (1,3)\\
     \end{pmatrix}$$
    
    

    Again, there are no further strategies that can be eliminated.

    1. $ A = \begin{pmatrix} 2 & 1 & 3 & 17\\ 27 & 3 & 1 & 1\\ 4 & 6 & 7 & 18 \end{pmatrix} \qquad B = \begin{pmatrix} 11 & 9 & 10 & 22\\ 0 & 1 & 1 & 0\\ 2 & 10 & 12 & 0 \end{pmatrix} $

    We see that $c_2$ is weakly dominated by $c_3$ so we have:

    $$ A = \begin{pmatrix} 2 & 3 & 17\\ 27 & 1 & 1\\ 4 & 7 & 18 \end{pmatrix} \qquad B = \begin{pmatrix} 11 & 10 & 22\\ 0 & 1 & 0\\ 2 & 12 & 0 \end{pmatrix} $$

    Now $r_3$ strictly dominates $r_1$ so we have:

    $$ A = \begin{pmatrix} 27 & 1 & 1\\ 4 & 7 & 18 \end{pmatrix} \qquad B = \begin{pmatrix} 0 & 1 & 0\\ 2 & 12 & 0 \end{pmatrix} $$

    Now $c_3$ stricly dominates $c_1$ and $c_4$ so we have:

$$ A = \begin{pmatrix} 1\\ 7 \end{pmatrix} \qquad B = \begin{pmatrix} 1 \\ 12 \end{pmatrix} $$

Thus the predicted rational behaviour is $(r_3, c_3)$.

  1. $ A = \begin{pmatrix} 3 & 3 & 2 \\ 2 & 1 & 3 \end{pmatrix} \qquad B = \begin{pmatrix} 2 & 1 & 3 \\ 2 & 3 & 2 \end{pmatrix} $

    $c_1$ is weakly dominated by $c_3$:

    $$A = \begin{pmatrix} 3 & 2 \\ 1 & 3 \end{pmatrix} \qquad B = \begin{pmatrix} 1 & 3 \\ 3 & 2 \end{pmatrix}$$

    There are no further dominated strategies.

  2. $ A = \begin{pmatrix} 3 & -1\\ 2 & 7\end{pmatrix} \qquad B = \begin{pmatrix} -3 & 1\\ 1 & -6\end{pmatrix} $

    There are no dominated strategies.

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