lew@ihuxr.UUCP (08/25/83)
In Chris Torek's problem, the probability that a gold coin is in the bottom drawer of the second cabinet, given that we found a gold coin in the bottom drawer of the chosen cabinet is 2/3, not 1/2. Chris has turned the problem into a 3-stage decision instead of a 2-stage decision, but this is no problem for Bayes' theorem! We let the events A and B be the "gold in bottom of 2" and "silver in bottom of 2" then A1,A2,A3,B1,B2,B3 are the events "A & we chose cabinet 1", "A & we chose cabinet 2", etc. Now we have P(A1) = P(A2) = ... = P(B3) = 1/6. Letting E be "found gold in bottom of chosen cabinet", we have: P(E|A1) = 0 P(E|B1) = 0 P(E|A2) = 1 P(E|B2) = 0 P(E|A3) = 1 P(E|B3) = 1 ... so Bayes' theorem gives P(A2|E) = P(A3|E) = P(B3|E) = 1/3 and P(A|E) = P(3|E) = 2/3. Lew Mammel, Jr. ihuxr!lew