Exercise 2.41

Answers

Let $G_{i}$ be the event that the $i$ -th door contains a goat, and let $D_{i}$ be the event that Monty opens door $i$ .

P (G_{1} | D_{2}, G_{2}) = \frac{P (G_{1}) P (D_{2}, G_{2} | G_{1})}{P (D_{2}, G_{2})} .

\begin{array}{l} P (G_{1}) P (D_{2}, G_{2} | G_{1}) & = \frac{2}{3} (P (G_{2} | G_{1}) P (D_{2} | G_{1}, G_{2})) \\ = \frac{2}{3} (\frac{1}{2} (p + (1 - p) \frac{1}{2})) \\ = \frac{2}{3} (\frac{1}{2} p + \frac{1}{4} - \frac{1}{4} p) \\ = \frac{1}{6} (p + 1) . \end{array}

Thus,

\begin{array}{l} P (G_{1} | D_{2}, G_{2}) & = \frac{\frac{1}{6} (p + 1)}{\frac{1}{6} (p + 1) + \frac{1}{3} \times \frac{1}{2}} \\ = \frac{p + 1}{p + 2} . \end{array}

Note that when $p = 1$ , the result matches that of the basic Monty Hall problem.

fifthist

2021-12-05 00:00