[extropy-chat] Re: monty hall paradox again: reds and green gorfs

[Rafal Smigrodzki]

----- Original Message ----- 
From: "Spike" <spike66 at comcast.net>
To: "'ExI chat list'" <extropy-chat at lists.extropy.org>
Sent: Saturday, May 22, 2004 10:48 PM
Subject: RE: [extropy-chat] Re: monty hall paradox again: reds and green
gorfs


>
> > Eliezer Yudkowsky
> >
> > >  ...The probability that
> > > the other envelope is larger must go to 1/3, it doesn't
> > > matter how it gets there."
> >
> > Your mathematician friend is flat wrong, and needs to study Bayesian
> > probability theory. --  Eliezer S. Yudkowsky
>
> OK cool, I was hoping someone would say that.  So where
> is the error?  Are you saying that Bayesian reasoning
> predicts that there *is* a profit in swapping?  Even
> if you have no idea how much a zorg is?  How do you
> set up the Bayesian priors?  How do you set up a
> sim to prove it?
>
### You have two envelopes, one with n zorgs, the other with 2n zorgs. The
probability of choosing either one is 1/2.

You want to calculate the reward/loss associated with staying with the
chosen envelope, vs. swapping after seeing its contents.

The reward for sticking after choosing the 2n zorg envelope is 1 zorgs.

The loss from swapping after choosing the 2n zorg envelope is 1 zorgs.

The loss for sticking with the n envelope is 1 zorgs.

The reward for swapping the n envelope is 1 zorgs.

There are no other courses of action, given the two envelopes. Since you do
not know "n" a priori (yes, you do not have a prior, by the definition of
the problem), you cannot tell whether you have the n or the 2n envelope even
after seeing its contents (obvious, right?). This is why the probability
after opening doesn't "go" anywhere - the contents of the envelope do not
provide you with any information that could allow you to adjust the priors,
or relate the contents to your desires. Since the desires cannot be
consulted (as in the problem with actual dollar amounts), the emotional
reward for swapping is exactly the same as the reward for sticking, always.

I think that a lot of people approach the problem as if there were 3
envelopes: 1/2n, n and 2n, and opening one of them gave clues to the
contents of others (which is similar to the three-door opening problem),
then start calculating erroneous probabilities.

Rafal