Daily Math Puzzle: 2026-06-17

Approximately 99.9% — This is a classic conditional probability puzzle, often solved using Bayesian reasoning. Let's break it down: 1. **Define Events:** * Let B be the event that Mr. Black is the thief. * Let W be the event that Mr. White is the thief. * Let J be the event that the thief wore the unique jacket. 2. **Initial Probabilities (Prior Beliefs):** * Anya initially considers both suspects equally likely: P(B) = 0.5 and P(W) = 0.5. 3. **Likelihoods (Probability of Evidence given a Suspect):** * If Mr. Black is the thief, he would definitely be wearing the jacket (since he owns it): P(J | B) = 1. * If Mr. White is the thief, the probability that he would be wearing the jacket depends on whether he owns it. Since he's a random individual, the probability he owns the jacket is 1 in 1,000: P(J | W) = 0.001. 4. **Calculate the Overall Probability of the Evidence (P(J)):** The evidence (the jacket being worn by the thief) could have come from either suspect. So, we sum the probabilities of the evidence given each suspect, weighted by their initial likelihood: P(J) = P(J | B) * P(B) + P(J | W) * P(W) P(J) = (1 * 0.5) + (0.001 * 0.5) P(J) = 0.5 + 0.0005 P(J) = 0.5005 5. **Calculate the Posterior Probability (Probability of Mr. Black being the thief, given the evidence P(B | J)):** We use the formula: P(B | J) = [P(J | B) * P(B)] / P(J) P(B | J) = (1 * 0.5) / 0.5005 P(B | J) = 0.5 / 0.5005 P(B | J) = 5000 / 5005 P(B | J) = 1000 / 1001 As a decimal, 1000 / 1001 ≈ 0.999000999... As a percentage, this is approximately 99.90%. Therefore, the probability that Mr. Black is the thief is approximately 99.9%. The trick is to remember that while Mr. Black definitely matches the evidence, there's still a tiny, non-zero chance that Mr. White (as a random person) could also own and be wearing the rare jacket.