Randomizer Conditional Probability Problem

Question

• The Randomizer holds the 6-sided die in one fist and the 8-sided die in the other.
• The Roller selects one of the Randomizer’s fists and covertly takes the die.
• The Roller rolls the die in secret and reports the result to the table.

a) If the number rolled is 1-6 then P(six-sided) = ?

My method: $P(6-sided|1-6 rolled)=\frac{|6-sided\cap1-6rolled|}{|1-6 rolled|}=\frac{1}{2}$. The answer is $\frac{4}{7}$. What did I do wrong?

Answer 1

Your method assumes the 14 outcomes are equally likely, but they are not. It is harder to roll a 1 with the 8-sided die than with the 6-sided die.

The correct approach is to use Bayes's rule. $$P(\text{6-sided} \mid \text{rolled 1-6}) = \frac{P(\text{6-sided, rolled 1-6})}{P(\text{rolled 1-6})} = \frac{P(\text{6-sided, rolled 1-6})}{P(\text{6-sided, rolled 1-6}) + P(\text{8-sided, rolled 1-6})} = \frac{\frac{1}{2} \cdot 1}{\frac{1}{2} \cdot 1 + \frac{1}{2} \cdot \frac{6}{8}} = \frac{4}{7}.$$

Answer 2

Let $D_6$ be the event that the six-sided die is selected and $D_8$ be the event that the eight-sided die is selected. Also let $R_6$ be the event that a number between $1$ and $6$ is reported by the roller.

You then wrote

$$ P(D_6 \mid R_6) = \frac{P(D_6 \cap R_6)}{P(R_6)} $$

This is correct. But the value of this is not $1/2$. You may have deduced that $P(D_6 \cap R_6) = P(D_6) = 1/2$. But what is $P(R_6)$, the a priori probability that a number between $1$ and $6$ is reported? It is not $1$.

Can you take it from there?