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From: John M <jamikes.domain.name.hidden>

Date: Tue, 5 Oct 2004 07:53:53 -0400

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

John Mikes

PS: to excuse my lingo: my 1st Ph.D. was Chemistry-Physics-Math. J

----- Original Message -----

From: "Brent Meeker" <meekerdb.domain.name.hidden>

To: "Everything-List" <everything-list.domain.name.hidden>

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
*

*>
*

Received on Tue Oct 05 2004 - 08:07:43 PDT

Date: Tue, 5 Oct 2004 07:53:53 -0400

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

John Mikes

PS: to excuse my lingo: my 1st Ph.D. was Chemistry-Physics-Math. J

----- Original Message -----

From: "Brent Meeker" <meekerdb.domain.name.hidden>

To: "Everything-List" <everything-list.domain.name.hidden>

Sent: Monday, October 04, 2004 6:19 PM

Subject: RE: Observation selection effects

Received on Tue Oct 05 2004 - 08:07:43 PDT

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