The Puzzle

You have 1,000 coins in a bag:

  • 1 is double-headed (always lands heads)
  • 999 are fair (heads with probability \( \frac{1}{2} \))

You draw one coin uniformly at random and flip it 10 times. It lands heads every time.

Question:

What is the probability that the coin you picked is the double-headed one?

Solution Using Bayes’ Theorem

We define:

  • \( D \): the event that you picked the double-headed coin
  • \( F \): the event that you picked a fair coin
  • \( H^{10} \): the event that you observed 10 heads in a row

We want to compute:

\[ P(D \mid H^{10}) = \frac{P(H^{10} \mid D) \cdot P(D)}{P(H^{10})} \]

Step 1: Prior Probabilities

  • \( P(D) = \frac{1}{1000} \)
  • \( P(F) = \frac{999}{1000} \)

Step 2: Likelihoods

  • \( P(H^{10} \mid D) = 1 \) (always heads)
  • \( P(H^{10} \mid F) = \left(\frac{1}{2}\right)^{10} = \frac{1}{1024} \)

Step 3: Total Probability of 10 Heads

\[ P(H^{10}) = P(H^{10} \mid D) \cdot P(D) + P(H^{10} \mid F) \cdot P(F) = 1 \cdot \frac{1}{1000} + \frac{1}{1024} \cdot \frac{999}{1000} \]

\[ P(H^{10}) = \frac{1}{1000} + \frac{999}{1024000} = \frac{1024 + 999}{1024000} = \frac{2023}{1024000} \]

Step 4: Apply Bayes’ Theorem

\[ P(D \mid H^{10}) = \frac{1 \cdot \frac{1}{1000}}{\frac{2023}{1024000}} = \frac{1024}{2023} \]

Final Answer

\[ \boxed{\frac{1024}{2023}} \approx 0.506 \]

So, given 10 consecutive heads, there’s slightly more than a 50% chance that you picked the double-headed coin.

Reference