Double Head

Solution

Answer: 20.4%

Let \[ D \] be the event that we picked the double-headed coin, and \[ N \] be the event that we picked a normal coin. Initially, before flipping the coin, the probabilities are \[ P(D) = \frac{1}{500}, \qquad P(N) = \frac{499}{500}. \] Now define the event \[ H \] as observing \(7\) heads in a row. If the coin is double-headed, then heads occurs every time, so \[ P(H|D) = 1. \] If the coin is normal, then each flip has probability \(\frac12\) of heads, so \[ P(H|N) = \left(\frac12\right)^7 = \frac{1}{128}. \] Now apply Bayes' theorem to compute the probability that the coin is double-headed given the observed sequence of heads: \[ P(D|H) = \frac{P(H|D)P(D)} {P(H|D)P(D) + P(H|N)P(N)}. \] \[ P(D|H) = \frac{1 \cdot \frac{1}{500}} {1 \cdot \frac{1}{500} + \frac{1}{128}\cdot \frac{499}{500}}. \] \[ P(D|H) = \frac{1}{627/128} = \frac{128}{627} \approx 0.204. \] So the probability that the coin is double-headed after observing \(7\) heads in a row is approximately 20.4%.