Example 21: Speaks the truth 3 times in 4: reports a six, $P = \frac38$

Basics you need

Solution

1. Name the events

   Let

   • $E$⁠: the man reports that a six occurs;
   • $S_1$⁠: a six occurs on the die;
   • $S_2$⁠: a six does not occur.

   $S_1$ and $S_2$ are complements, so they form a partition of the sample space.

   What the die really shows is the hidden cause; his report is what we observe. We want the chance of the cause "six" given the report.
   $\{S_1, S_2\}$ is a partition. Lesson: Bayes' theorem

2. The probabilities

   $$P(S_1) = \frac16, \qquad P(S_2) = \frac56.$$ $$\begin{aligned} P(E|S_1) &= \frac34, \\ P(E|S_2) &= 1 - \frac34 = \frac14. \end{aligned}$$
   If a six did occur, he reports a six exactly when he tells the truth: probability $\frac34$⁠. If a six did not occur, a report of "six" is a lie. Like the book, we assume that his report is just "six" or "not six", so every lie about a non-six is the report "six" and $P(E|S_2)$ is the whole chance of a lie, $\frac14$⁠. This is an assumption of the model: if a lie could name any of the five wrong numbers equally, $P(E|S_2)$ would be $\frac14 \times \frac15 = \frac{1}{20}$ and the answer would be $\frac34$ instead.
   $P(\text{lie}) = 1 - P(\text{truth})$⁠. Basics: rules of probability

3. The chance of the report: total probability

   By the theorem of total probability,

   $$\begin{aligned} P(E) &= P(S_1)P(E|S_1) \\ &\quad + P(S_2)P(E|S_2) \\ &= \frac16\cdot\frac34 + \frac56\cdot\frac14 \\ &= \frac{3}{24} + \frac{5}{24} = \frac{8}{24}. \end{aligned}$$
   The two ways of hearing "six" cannot both happen, so their probabilities add. Both products already have the denominator 24.
   Multiply along a path, add the paths. Basics: tree diagrams

4. Actually a six, given the report: Bayes' theorem

   By Bayes' theorem,

   $$\begin{aligned} P(S_1|E) &= \frac{P(S_1)P(E|S_1)}{P(E)} \\ &= \frac{3}{24} \div \frac{8}{24} = \frac38. \end{aligned}$$
   The top is the true-report path, the bottom is every report of six; the 24 cancels. This is the book's $\frac{1}{8} \times \frac{24}{8} = \frac{3}{8}$⁠, written with a common denominator.
   Bayes' theorem: one path over all the paths. Lesson: Bayes' theorem

5. What the answer means

   Before the report the chance of a six was $\frac16 \approx 0.17$⁠; after it, $\frac38 = 0.375$⁠. The report helps, but it is still more likely ($\frac58$⁠) that the die is not a six.

   A man who is right 3 times in 4 sounds reliable, but here he is reporting something rare. Five times as many throws are non-sixes, and a quarter of those turn into false reports of "six".
   $P(S_1|E) + P(S_2|E) = 1$⁠. Lesson: Bayes' theorem

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