Re: Observation selection effects

From: Jesse Mazer <lasermazer.domain.name.hidden>
Date: Thu, 07 Oct 2004 05:01:11 -0400

Stathis Papaioannou wrote:

>Jesse Mazer wrote:
>
>>I don't think that's a good counterargument, because the whole concept of
>>probability is based on ignorance...
>
>
>No, I don't agree! Probability is based in a sense on ignorance, but you
>must make full use of such information as you do have.

Of course--I didn't mean it was based *only* on ignorance, you must
incorporate whatever information you have into your estimate of the
probability, but no more. Your argument violates the "but no more" rule,
since it incorporates the knowledge of an observer who has seen how much
money both envelopes contain, while I only know how much money one envelope
contains.

>If you toss a fair coin, is Pr(heads)=0.5? According to your argument, it
>could actually be anything between zero and one, because it is possible I
>am lying about it being a fair coin!

My argument implies nothing of the sort. But your argument would seem to
imply that if I am watching a videotape of a fair coin toss, then if someone
else has already watched the tape, it would be permissible to incorporate
their knowledge of the outcome of the toss into a probability calculation,
even if I myself don't have this knowledge.

>Here is another "two envelope" example:
>
>Two envelopes, A and B, contain two doses of the drug Lifesavium, the
>Correct Dose and the Half Dose. If you give the patient more than 1.5 times
>the Correct Dose you will certainly kill him, while if you give him the
>Half Dose you will save his life, although he won't make an immediate
>recovery as he would if you gave him the Correct Dose. If you don't give
>him any medication at all, again, he will surely die. Once you open an
>envelope, the medication in in such a form that you must give the full
>dose, or nothing.
>
>You are faced with the two envelopes, the above information and the sick
>patient, with no other help, on a desert island. There is one further
>complication: if you open the first envelope, and then decide to open the
>second envelope, you must destroy the contents of the first envelope in
>order to get to the second envelope.
>
>OK: so you open envelope A and find that it contains 10mg of Lifesavium.
>You don't know whether this is The Correct Dose or the Half Dose; so
>envelope B may have either 5mg or 20mg, right? And if 10mg is the Correct
>Dose, then if you discard envelope A in favour of envelope B, there is a
>50% chance that envelope B will have double the Correct Dose and you will
>kill the patient - so you had better stick with envelope A, right?
>
>I think you can see the error in the above argument. You already know that
>the amount in each envelope is fixed, so even though you have no idea of
>the actual dosages involved, or which envelope contains which dose, even
>after opening the first envelope, there is NO WAY you can give the patient
>an overdose. There is no way envelope B can contain 20mg of Lifesavium, but
>even though you cannot know this, you can use the above reasoning to deduce
>that there is no expected benefit from choosing a strategy of switching or
>not switching - as you can also see intuitively from the symmetry of the
>situation, whether you choose envelope A or B first.

This case is not analogous to the two-envelope problem, because in this case
it is part of *my* knowledge that one envelope contains the Correct dose and
the other contains the Half dose, and neither contains a Double dose. In
contrast, your analysis of the two-envelope problem relied on information I
don't have, namely whether the two envelopes contained $50 and $100 or $25
and $50.

Suppose I know that the envelope-stuffer flipped a fair coin to decide
whether to put $25 or $50 in one envelope, then put double that amount in
the other. I randomly choose one envelope and open it, and find $50. Do you
disagree that my average expected return from switching would now be
(0.5)(25) + (0.5)(100) = 62.5? If this experiment was repeated many times
and we looked only at the subset of cases where the first envelope I opened
contained $50 and I chose to switch, wouldn't my average winnings in this
subset of cases be $62.50?

>
>In the game with the envelopes and the money, the analogous error is to
>think that there is a possibility of doubling your money when you have
>actually picked the envelope containing the larger sum first.

But *you* don't know that the envelope you picked was the one with the
larger sum. This is akin to arguing that it is an "error" to think there is
a possibility of winning if you bet $100 on heads in a videotaped coin toss,
since someone who's already watched the tape knows it comes up tails, even
though you don't know that. Would you indeed say it's an error to believe my
average expected return is $50 in this case?

Jesse
Received on Thu Oct 07 2004 - 05:03:03 PDT