[R] Two envelopes problem

[Farley, Robert]

So if I put $10 and $20 into the envelopes, then told you that the
values were multiples of $10, it would be wrong for you to assess
probabilities on $100, $1,000,000 and so on? :-)  

But what if you reasoned that there were far more multiples of 10 above
20 than below 20?

What if I was really evil and the multiples of 10 that I always put in
the envelopes were 0x$10 and 0x$10.  When you opened an empty envelope,
what are the odds the other has $1,000,000?  What are the odds it has
$10?

It seems the information given can be misleading. 



 
Robert Farley
Metro
www.Metro.net 
 
-----Original Message-----
From: r-help-bounces at r-project.org [mailto:r-help-bounces at r-project.org]
On Behalf Of Duncan Murdoch
Sent: Tuesday, August 26, 2008 05:15
To: Jim Lemon
Cc: r-help at r-project.org; Mario
Subject: Re: [R] Two envelopes problem

On 26/08/2008 7:54 AM, Jim Lemon wrote:
> Hi again,
> Oops, I meant the expected value of the swap is:
> 
> 5*0.5 + 20*0.5 = 12.5
> 
> Too late, must get to bed.

But that is still wrong.  You want a conditional expectation, 
conditional on the observed value (10 in this case).  The answer depends

on the distribution of the amount X, where the envelopes contain X and 
2X.  For example, if you knew that X was at most 5, you would know you 
had just observed 2X, and switching would be  a bad idea.

The paradox arises because people want to put a nonsensical Unif(0, 
infinity) distribution on X.  The Wikipedia article points out that it 
can also arise in cases where the distribution on X has infinite mean: 
a mathematically valid but still nonsensical possibility.

Duncan Murdoch

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