[R] Two envelopes problem
Farley, Robert
FarleyR at metro.net
Tue Aug 26 19:13:59 CEST 2008
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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