RE: Observation selection effects

From: Brent Meeker <meekerdb.domain.name.hidden>
Date: Mon, 4 Oct 2004 22:19:42 -0000

>-----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 - 02:16:15 PDT