Two-Envelope Paradox II

This post continues a discussion of the Two-Envelope Paradox which has been going on at 
Res Cogitans blog.

Suppose you enter a game show where two envelopes are on display.  You are told that envelopes contain gameshow points - points are translated into currency at a constant (linear) rate c at the end of the show.  You are initially asked whether you prefer one envelope to the other.  They are blank and identical, so you do not.  You are then shown the value in one envelope -- 2000 gameshow points.  At this point, you might reason:

1. as you don't know c, there is still no point in swapping.
2. if there is no point in swapping, then the expectation value of swapping is 2000.
3. if the expectation value is 2000 then 2000 = p.1000 + (1-p).4000
4. the probability of swapping and getting 1000 is 2/3.
5. this calculation holds whatever value was in the first chosen envelope.

  i.e., p(the other envelope is worse | your first envelope's value is X) = 2/3 for all conditions x assignign a value X to the first envelope.

6. when given the initial choice of envelopes you have a 2/3 probability of choosing the higher amount.

0 and 6 clearly contradict.  Where did the reasoning go wrong?

I believe it went wronge here: 6 does not follow from 12345.

From p(y|x) for various x we want to infer p(y).  For let y be the case "the other envelope is better," and let x be the condition "the value in the envelope is X."  If we had an integrable prior distribution of envelope values, we could reason using the "law of total probability" that p(y) = sum_x p(y|x) dX.  When the distribution of X is not integrable, can we reason according to a law such as the following?

Law?: if p(y|x) is constant as x ranges over all possible conditions, then the total probability p(y) takes that same value.

When X is not integrable, we can gather conditions x into pairs and find that the probability p(y|pairs of x) is twice as good; then p(y) would take on a larger value by that same law.  That law is not valid.

Arguments 1..5 tease out the consequences of "indifference." I don't think 1..5 are logically necessary at all.  They make a compelling argument about the implications of indifference: what to do if you don't know what gameshow points are worth to you, and you only know that more points are worth more to you. 
I think that once you make the assumption that utility is linear in gameshow points and that p(y|x) is constant , then I think the chain of reasoning 1..5 is correct, leading inexorably to the conclusion 5. But one could go on reasoning after 5, reaching problematic conclusions about small amounts of money and for large amounts of money.  I will try to do that now.

7. Arbitrarily small values exist in the game.  Whatever values we see, we should have expected smaller ones.

What is the domain of the gameshow points?  Is there a minimal value?  1...5 constrains the distribution of envelopes so that for all x a value of x/2 is always twice as likely as 2x.  So, there is no minimal amount of money in this game, and the probability of tiny amounts outweighs the probability of large amounts.  If the first envelope contains 0.01 gameshow points, then I might still switch down without losing expected value because in fact the smaller values are more likely.  There is no "gameshow cent."  1 is rarer than 0.1, which is rarer than 0.01, which is rare than 0.001, etc., so the treatment of small values is not going to come up only rarely.  Perhaps part of the paradox is being pushed into the gap between 0 and arbitrarily small values.  As a result, gameshow points cannot translate into real money.  We could suppose that there is a constant c, such that gameshow points times c = currency in your hometown.  But currency in your hometown doesn't exist in arbitrarily small amounts.  So there has to be a minimal gameshow value.  But by arguments 1..5 there is none.

8. The game show is obtaining its values in some way other than choosing them from a distribution. 

In order to compute p(y) from  p(y|x) using the law of total probability and in order to assign probabilities, we should form an integrable prior distribution.  This will simply violate 1: the notion that you don't know what gameshow cash is worth, so you will treat all values the same.  Of course, you can resist forming a prior distribution, and still reason... but then an "implied distribution" is still present behind your actions.  The implied distribution seems to be x^(-0.5), since for that distribution the chance of switching down is twice the chance of switching up.  This distribution is not integrable.  Not only does it put tremendous weight on small values, it effectively puts nonzero weight at very large values (or at infinity).  if we would cut this distribution off at any arbitrary top and bottom value, we would find a very interesting fact: you *should* switch up from small values, but for top values you should not.  How is the game show supposed to generate its random envelope values?  It must have, in fact, an integrable distribution.  More likely, it has a bucket full of envelopes -- a discrete distribution.  The customer may well not know it, but some such distribution must exist in order to create the game show.