Re: Two-envelope problem

[Tuan V Nguyen <t.nguyen@garvan.unsw.edu.au>]

Hello folks, 

	I have been waiting for further discussions from math 
gurus on the two-envelope problem, but they seem quiet 
lately. Let me now now summarise and put my view on this 
interesting problem. 

	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. 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. 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 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. 

	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

which would make sense.

	Any further comment? What about Bayes' theorem? 


	Tuan

	


At 12:42 AM 4/18/97 -0500, Dung Trong Nguyen wrote:
>1) The expectation value of keeping the $100
>
>	1 x $100 = $100
>
>2) The expectation value of choosing the other
>
>	1/2 x $50 + 1/2 x $200 = $125
>
>So it is better to take the risk.
>
>d~
>

At 01:06 AM 4/18/97 -0500, Ha Anh Vu wrote:
>I would disagree with this solution. This problem is a classical paradox
>in statistical decision theory, often being stated to fool novices.  The
>punchline is that if you follow this line of reasoning (meaning willing
>to exchange the envelopes), then you would deem it wise to exchange the
>envelopes one more time and indeed, exchange the envelopes infinitely
>many times, an apparent antinomy!


At 04:56 PM 4/20/97 -0500, Paul Pham wrote:
>choose other one because if you're to lose, you lose
>50, otherwise you win 100.