From: Jonathan Rosenberg <jxr@thumper.bellcore.com>
Subject: Re: Not the Monty Haul problem
Organization: Bellcore
Date: Sat, 21 Nov 92 16:23:31 GMT

It's true that this discussion really belongs on "something".math, but
since it was asked here & I've seen two incorrect answers, I'm going to
provide an answer here.  This problem is actually much tricker than it
seems.

Jeff Tang Asks:

> Question:

> You're given two envelopes and told that the contents are determined
> randomly, except that one has twice as much money as the other.  You open
> one envelope and find 1 unit.  Now, you're given the opportunity to either
> take the unit you found, or give it up and take the contents of the other
> envelope.  What do you do?

> Answer 1:
> Well, since there wasn't really any information revealed by opening the
> envelope, switching to the other envelope wouldn't be any different than
> choosing that envelope in the first place.  Therefore, it doesn't matter if
> you switch.

> Answer 2:
> Well, the other envelope has either .5 or 2 units.  Since there's a 50%
> chance of each of these, you'll come out with 1.25 units on average as
> opposed to sticking with the single unit, so you should switch.

> What's wrong with one (or both) of these answers?

Believe it or not, BOTH answers are wrong, though Answer #2 is actually
closer to a correct answer.  No, this is not a trick.  Let me explain.

The difficulty is that the problem is not completely specified &, so, is
unanswerable.  Let's look at the problem statement again:

	You're given two envelopes and told that the contents are determined
	randomly, except that one has twice as much money as the other.

The difficulty lies in the phrase "contents are determined randomly". 
We can assume without loss of generality that the contents are
determined as follows: choose a number U & place that many units in an
envelope.  Now place 0.5U or 2U (determined by a flip of a coin, say)
units in the other envelope.

To answer the question correctly requires knowing the distribution of U
(i.e., what is the distribution of the random variable determining the
contents).  Without knowing the distribution, there is no answer.

You're probably thinking that this is nit-picking & that it's obvious
what the distribution is.  Well, I can think of 2 obvious possibilties:

1) U is chosen with a uniform distribution from the interval [0, M]. 
That is, the minimum that can be put in an envelpe is 0 units & the max
is M units.  In this case, there is a simple algorithm for maximizing
the expected value:

	If the enveople you choose has <= M/2 units in it, switch to the other
	envelope.
	Otherwise, stick with you first choice.

Notice that answer #2 is a variation of this for the case M=2.

2) U is chosen with a uniform distribution from the interval [0,
infinity).  That is, the minimum number of units is 0, but there is no
maximum.  I suspect that this is implicitly what everyone assumed the
distribution was.

If this is the correct meaning, then it's easy to deal with.  That's
because the construction specified is impossible: it's not possible to
choose a number uniformly from an infinite sequence.  Thus, the problem
is ill-defined & there is no answer.

------

On rethinking this, I realized that you might argue that the problem is
well-defined & that the chooser just is not told the distribution used. 
That seems to be ok.  In that case, answer #1 is correct.  Answer #2 has
flawed logic because it is calculating an average of repeated trails
that are not like the original trial.  I.e., answer #2 assumes that in
each trail the envelopes will contain either 0.5 & 1 units or 1 & 2
units.  But, that ain't the original problem.  The original problem says
that you have no information about the amounts in the envelopes (other
than 1 is twice the other).

JR

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