Re: Two-envelope problem

[AnHai Doan <anhai@cs.washington.edu>]

Woa folks, I just took a look at this thread. Envelopes! I love
envelopes! Pls allow me to add to the "mess" :-).

On Mon, 21 Apr 1997, Tuan V Nguyen wrote:

> 	The original problem is:
> 
> >        There are two envelopes containing money. One 
> >envelope contains twice as much as the other, however since the 
> >envelopes look identical you cannot tell one apart from the 
> >other. You choose one envelope and you see that it contains 
> >$100. Is it better to keep the $100 or choose the other 
> >envelope?
> 
> 	Firstly, Dung used the expected value concept to work 
> out that the expected value of the 2nd envelope is higher 
> than the first (selected) one, hence it would be wise to 
> take the risk. 

I think his solution is correct.

>Then, Ha Anh Vu disagreed, saying that the 
> idea of expectation can not be applied here since it 
> assumes infinite repeated selection, which is not possible 
> in this case. 

I don't think so. Suppose you have exchanged the first envelope for the
second one. Then if you are to choose between the two actions "keep the
second one" and "exchange the second for the first one", you still have
the same expected monetary value of $125 for the first action, and $100
for the second one. So you will choose the first action, namely, to keep
the second envelope.

As long as you HAVE OPENED at least one envelope, you will never face the
possibility of "infinite switching". This may seems to arise only when you
don't open any envelope, but even in this case, we have no paradox, and
the solution of keeping the picked envelope is the correct and intuitive
one.

>Paul Pham then commented that "choose other 
> one because if you're to lose, you lose 50, otherwise you 
> win 100". I think Paul's reasoning is too simplistic; well, 
> you can lose up to $100, not $50.

I don't understand why you can lose up to $100. I think anh Paul Pham
provided the intuitive explanation of the seemingly counterintuitive
solution of exchanging the first envelope for the second one. If you do
the exchange, you face equal chances of losing $50 and gaining $100, thus
you'd better exchange the envelopes.

> 	I have difficulty with the application of expected 
> value concept in this problem as well. In fact, if you 
> apply this concept, then regardless of what the value in 
> the first envelope (say X1), then the second one would be 
> 1.25*X1. Although the mathematics is correct, it does not 
> make sense to me.

Some time the commonsensical solution is not the correct one. This
envelope problem is very much similar to the (in)famous Monty Hall
problem, which is briefly as follows: One of the three boxes contains
a prize, the other two contains nothing. You choose one box, I will 
open an empty box among the two remaining boxes. Now you can either choose
to keep your initially chosen box, or to exchange it for the unopened
box among the two remaining boxes. The correct solution for this problem
is to exchange the boxes, which is counterintuitive on the first thought.

> 
> 	So why? I think while the problem is stated in terms 
> of ratios of amounts, if one re-stated the question in an 
> arithmetic form, then intuition and expectatioon would co-
> incide, i.e. instead of saying that one envelope contained 
> *twice as much* as the other, one would say that, say, one 
> envelope contained *$50 more* than the other.  Hence, if 
> there was X1 in the first envelope, then the expectation 
> for the second would be:
> 
>    E(X2) = 0.5(X1 - 50) + 0.5(X1 + 50) = X1

I don't understand why it is "X1 + 50" in the above equation. You are
solving a modified version of the original envelope problem.

In summary:

1) Some people would say that there is not enough information provided
in this problem to have a solution. Namely, they object to the fact that
the probabilities in the problem can't be formulated and calculated in the
frequentist interpretation (with repeatable trials and so on). I would say
that this problem is perfectly solvable under the subjective
interpretation of probability, and provides a very reasonable solution.
Subjective (also called Bayesian) probability and Bayesian decision
making have well-grounded (if controversial) theoretical and practical
foundations, and are rapidly gaining acceptance and use in many 
sciences, inclusing statistics, sociology, artificial intelligence,
business decision making, just to name a few.

2) Under the subjective interpretation, with only the information provided
in the problem text, the probability of the remaining envelope containing
$200 is 0.5. As long as the act of picking an envelope includes opening
the envelope, Anh Dung's solution of exchanging the first envelope for the
second one is correct. An intuitive explanation of this solution is
provided by anh Paul Pham.

3) If the act of picking an envelope doesn't include opening it to get to
know the amount inside, then if we are to use the  expected-monetary-value
approach, we seems to get into the paradox of infinite switching. But I
think this is not the case, the intuitive and correct solution is to keep
the first envelope. I don't present my arguments here as I am interested
in knowing what you guys think.

Hope this adds to the mess.
Hai.