A man has 5 coins in his pocket. Two are double-headed, one is double-tailed, and two are normal. The coins cannot be distinguished unless one looks at them.
a) The man shuts his eyes, chooses a coin at random and tosses it. What is the probability that the lower face of the coin is heads?
b) He opens his eyes and sees that the upper face of the coin is a head. What is the probability that the lower face is a head?
Best Answer
This is a variant of the Bertrand's box "paradox".
a) The probability of getting heads down is the probability of choosing a double headed coin ($\frac{2}{5} \times 1$) plus the probability of getting a normal coin and heads landing down ($\frac{2}{5} \times \frac{1}{2}$). So the total probability is $\frac{3}{5}$.
b) The man has 6 "heads" in his pocket, four of which have heads on the other side, and two of which have tails on the other side. So if the upper face is a head, there is a $\frac{4}{6} = \frac{2}{3}$ probability the lower face is a head also. You can also work this out by conditional probabilities if you are thus inclined.