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
Have a good day

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

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