Arbitrarily label the two envelopes A and B. Assume that no envelope is empty. Let «X» denote the amount of money in envelope X. We are told that either «A» =

Arbitrarily label the two envelopes A and B. Assume that no envelope is empty. Let «X» denote the amount of money in envelope X. We are told that either «A» = 2«B» or «B» = 2«A». Because these propositions are exhaustive and mutually exclusive, we know that P(«A» = 2«B») + P(«B» = 2«A») = 1. (1) Since we have no knowledge that would make either proposition more plausible than the other, we also have (by symmetry) that P(«A» = 2«B») = P(«B» = 2«A»). (2) From (1) and (2) we can solve for the individual probabilities of the propositions: P(«A» = 2«B») = P(«B» = 2«A») = 1/2. (3) Therefore, from the initial conditions, we have no reason to prefer either envelope. Now we are given the information that we have picked one of the envelopes at random (let's say it's A). We are further given the information that A contains $20, that is «A» = $20. How does this new knowledge affect the probabilities? It has no effect because we can't say anything more about either proposition without also knowing «B» as well, and we don't know it. That is, P(«A» = 2«B» | «A» = $20) = P(«A» = 2«B») and P(«B» = 2«A» | «A» = $20) = P(«B» = 2«A»). Therefore, our probability assignments from (3) remain unchanged, and we have no reason to prefer one envelope to the other, let alone swap A for B.