Dear Brent, I enjoyed your description which was quite a show for my totally qualitative (and IN-formally intuitive) logic - how you, quanti people substitute common sense for those letters and numbers. Norman's paradox is an orthodox paradox and you made it into a metadox (no paradox). Have a good day

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John Mikes PS: to excuse my lingo: my 1st Ph.D. was Chemistry-Physics-Math. J ----- Original Message ----- From: "Brent Meeker" <[EMAIL PROTECTED]> To: "Everything-List" <[EMAIL PROTECTED]> Sent: Monday, October 04, 2004 6:19 PM Subject: RE: Observation selection effects > >-----Original Message----- > >Norman Samish: > > > >>The "Flip-Flop" game described by Stathis Papaioannou > >strikes me as a > >>version of the old Two-Envelope Paradox. > >> > >>Assume an eccentric millionaire offers you your choice > >of either of two > >>sealed envelopes, A or B, both containing money. One > >envelope contains > >>twice as much as the other. After you choose an > >envelope you will have the > >>option of trading it for the other envelope. > >> > >>Suppose you pick envelope A. You open it and see that > >it contains $100. > >>Now you have to decide if you will keep the $100, or > >will you trade it for > >>whatever is in envelope B? > >> > >>You might reason as follows: since one envelope has > >twice what the other > >>one > >>has, envelope B either has 200 dollars or 50 dollars, with equal > >>probability. If you switch, you stand to either win > >$100 or to lose $50. > >>Since you stand to win more than you stand to lose, you > >should switch. > >> > >>But just before you tell the eccentric millionaire that > >you would like to > >>switch, another thought might occur to you. If you had > >picked envelope B, > >>you would have come to exactly the same conclusion. So > >if the above > >>argument is valid, you should switch no matter which > >envelope you choose. > >> > >>Therefore the argument for always switching is NOT > >valid - but I am unable, > >>at the moment, to tell you why! > > > Of course in the real world you have some idea about how much > money is in play so if you see a very large amount you infer it's > probably the larger amount. But even without this assumption of > realism it's an interesting problem and taken as stated there's > still no paradox. I saw this problem several years ago and here's > my solution. It takes the problem as stated, but I do make one > small additional restrictive assumption: > > Let: s = envelope with smaller amount is selected. > l = envelope with larger amount is selected. > m = the amount in the selected envelope. > > Since any valid resolution of the paradox would have to work for > ratios of money other than two, also define: > > r = the ratio of the larger amount to the smaller. > > Now here comes the restrictive assumption, which can be thought of > as a restrictive rule about how the amounts are chosen which I > hope to generalize away later. Expressed as a rule, it is this: > > The person putting in the money selects, at random (not > necessarily uniformly), the smaller amount from a range (x1, x2) > such that x2 < r*x1. In other words, the range of possible > amounts is such that the larger and smaller amount do not overlap. > Then, for any interval of the range (x,x+dx) for the smaller > amount with probability p, there is a corresponding interval (r*x, > r*x+r*dx) with probability p for the larger amount. Since the > latter interval is longer by a factor of r > > P(l|m)/P(s|m) = r , > > In other words, no matter what m is, it is r-times more likely to > fall in a large-amount interval than in a small-amount interval. > > But since l and s are the only possibilities (and here's where I > need the non-overlap), > > P(1|m) + P(s|m) = 1 > > which implies, > > P(s|m) = 1/(1+r) and P(1|m) = r/(1+r) . > > Then the rest is straightforward algebra. The expected values are: > > E(don't switch) = m > > E(switch) = P(s|m)rm + P(l|m)m/r > = [1/(1+r)]rm + [r/(1+r)]m/r > = m > > and no paradox. > > Brent Meeker >