probability-and-computing-solutions/exercise01.05.tex at master · emsbach/probability-and-computing-solutions · GitHub

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\paragraph{Exercise 1.5}
\begin{enumerate}
  \item[(a)] Assume that Bob chooses die A and Alice chooses die B. In the
    following we list all of the possible events and their probabilities.
    \begin{align*}
    \begin{tabular}{c|c|c|c}
        die A   &die B  &Probability    &Winner \\
        \hline
            1       &2      &1/3 $\cdot$ 1/3 = 1/9 &Alice\\
            1       &4      &1/9        &Alice\\
            1       &9      &1/9        &Alice\\
            6       &2      &1/9        &Bob\\
            6       &4      &1/9        &Bob\\
            6       &9      &1/9        &Alice\\
            8       &2      &1/9        &Bob\\
            8       &4      &1/9        &Bob\\
            8       &9      &1/9        &Alice\\
    \end{tabular}
  \end{align*}
    Therefore the probability that Alice wins is $5 \cdot \frac{1}{9} =
    \frac{5}{9} > \frac{1}{2}$.

  \item[(b)] Assume that Bob chooses die B and Alice chooses die C. In the
    following we list all of the possible events and their probabilities.
    \begin{align*}
    \begin{tabular}{c|c|c|c}
        die B   &die C  &Probability  &Winner \\
        \hline
            2       &3      &1/9      &Alice\\
            2       &5      &1/9      &Alice\\
            2       &7      &1/9      &Alice\\
            4       &3      &1/9      &Bob\\
            4       &5      &1/9      &Alice\\
            4       &7      &1/9      &Alice\\
            9       &3      &1/9      &Bob\\
            9       &5      &1/9      &Bob\\
            9       &7      &1/9      &Bob\\
    \end{tabular}
  \end{align*}
    Again, the probability that Alice wins is $5 \cdot \frac{1}{9} =
    \frac{5}{9} > \frac{1}{2}$.

  \item[(c)] Assume that Bob chooses die C and Alice chooses die A. In the
    following we list all of the possible events and their probabilities.
    \begin{align*}
    \begin{tabular}{c|c|c|c}
        die C   &die A  &Probability  &Winner \\
        \hline
            3       &1      &1/9      &Bob\\
            3       &6      &1/9      &Alice\\
            3       &8      &1/9      &Alice\\
            5       &1      &1/9      &Bob\\
            5       &6      &1/9      &Alice\\
            5       &8      &1/9      &Alice\\
            7       &1      &1/9      &Bob\\
            7       &6      &1/9      &Bob\\
            7       &8      &1/9      &Alice\\
    \end{tabular}
  \end{align*}
    And still, the probability that Alice wins is $5 \cdot \frac{1}{9} =
    \frac{5}{9} > \frac{1}{2}$.

\end{enumerate}