2014-I-19

$\begin{array}{cl}
= & \dfrac{1}{6} + \dfrac{5}{6} \times \dfrac{5}{6} \times \dfrac{1}{6} + \ldots \\
= & \dfrac{\frac{1}{6}}{1-(\frac{5}{6})^2} \\
= & \dfrac{6}{11}
\end{array}$

1. The probability of getting $10$ tokens

   $\begin{array}{cl}
   = & \dfrac{1}{8}\times \dfrac{1}{8} \times 8 \\
   = & \dfrac{1}{8}
   \end{array}$

   The probability of getting $5$ tokens

   $\begin{array}{cl}
   = & \dfrac{7}{8}\times \dfrac{1}{8} + \dfrac{7}{8}\times \dfrac{1}{8} \\
   = & \dfrac{7}{32}
   \end{array}$

   Therefore, the expected number of tokens got

   $\begin{array}{cl}
   = & 10 \times \dfrac{1}{8} + 5 \times \dfrac{7}{32} + 0 \times (1- \dfrac{1}{8} – \dfrac{7}{32}) \\
   = & \dfrac{75}{32}
   \end{array}$

2. For Option 2. The probability of getting $50$ tokens

   $\begin{array}{cl}
   = & \dfrac{1}{8}\times \dfrac{1}{8} \times \dfrac{1}{8} \times 8 \\
   = & \dfrac{1}{64}
   \end{array}$

   The potability of getting $10$ tokens

   $\begin{array}{cl}
   = & \dfrac{1}{8} \times \dfrac{2}{8} \times \dfrac{1}{8} \times 6 + \dfrac{1}{8} \times \dfrac{2}{8} \times \dfrac{1}{8} \times 6 + \dfrac{1}{8} \times \dfrac{2}{8} \times \dfrac{1}{8} \times 6 \\
   = & \dfrac{9}{128}
   \end{array}$

   The probability of getting $5$ tokens

   $\begin{array}{cl}
   = & \dfrac{C^3_2 \times 7 \times 2}{8^3} \\
   = & \dfrac{21}{256}
   \end{array}$

   Therefore, the expected number of tokens got

   $\begin{array}{cl}
   = & 50 \times \dfrac{1}{64} + 10 \times \dfrac{9}{128} + 5 \times \dfrac{21}{256} + 0 \times (1 – \dfrac{1}{64} – \dfrac{9}{128} – \dfrac{21}{256} )\\
   = & \dfrac{485}{256}
   \end{array}$

   Since $\dfrac{75}{32} > \dfrac{485}{256}$, then the player of the second round should adopt Option 1.

3. The probability of Ada getting no tokens

   $\begin{array}{cl}
   = & 1 – \dfrac{6}{11} \times ( \dfrac{1}{8} + \dfrac{7}{32} ) \\
   = & \dfrac{13}{16} \\
   = & 0.8125 \\
   < & 0.9 \end{array}$

   Therefore, the claim is not correct.