Re: Observation selection effects

From: Jesse Mazer <lasermazer.domain.name.hidden>
Date: Thu, 07 Oct 2004 20:34:55 -0400

Stathis Papaioannou wrote:

>Sorry Jesse, I can see in retrospect that I was insulting your intelligence
>as a rhetorical ploy, and we >shouldn't stoop to that level of debate on
>this list.

No problem, I wasn't insulted...

>You say that you "must incorporate whatever information you have, but no
>more" in the >envelopes/money example. The point I was trying to make with
>my envelope/drug example is that you >need to take into account the fact
>that the amount in each envelope is fixed

Well, I think that's like saying that in the videotaped coinflip example,
you need to take into account the fact that the outcome of the flip is
already fixed. I don't think it matters whether it's "really" fixed or not,
since probabilities are about your knowledge rather than objective reality
(they're epistemological, not ontological), and since you are equally
ignorant of the outcome regardless of whether the flip happens in realtime
or on video, your probabilistic reasoning should be the same.

>But you have passed over the final point in my last post, which I now
>restate:
>
>(1) The original game: envelope A and B, you know one has double the amount
>of the other, but you >don't know which. You open A and find $100. Should
>you switch to B, which may have either $50 or >$200?
>
>(2) A variation: everything is the same, up to the point where you are
>pondering whether to switch to >envelope B, when the millionaire walks in,
>and hidden from view, flips a coin to decide whether to >replace whatever
>was originally in envelope B with either double or half the sum in envelope
>A, i.e. >either $50 or $200.

Well, my argument about the two-envelope paradox all along has been that you
need to think about the probability distribution the millionaire uses to
stuff the two envelopes, and that once you do that the apparent paradox
disappears. So we need to consider what probability distributions the
millionaire used in these examples. Let's say, for example, that the
millionaire flips a coin to decide whether to put $50 or $100 in one
envelope, and then puts double that amount in the other. In that case, if I
pick an envelope randomly, there is a 1/4 chance I'll find $50 inside, a 1/2
chance I'll find $100 inside, and a 1/4 chance I'll find $200 inside. If I
find either $50 or $200, I know with complete certainty how much the other
envelope contains; but if I find $100, then from my point of view there's a
1/2 chance the other envelope contains $50 and a 1/2 chance the other
envelope contains $200.

Now, if you assume that in game (2) the millionaire *only* replaces the
amount in the second envelope with a new amount based on a coinflip *if* I
found $100 in the first envelope, but doesn't mess with the second envelope
if I found $50 or $200 in the first one, then both games are exactly equal,
from my point of view. In both cases, whenever I find $100 in the first
envelope, my average expected return from switching would be (0.5)($50) +
(0.5)($200) = $125, so it's better to switch.

On the other hand, if in game (2) the millionaire replaced the amount in the
second envelope with a new amount based on a coinflip *regardless* of how
much I found in the first envelope, this would change my strategy if I found
either $50 or $200 in the first envelope. In this case, it will be to my
advantage to switch no matter how much I find in the first envelope, since
my average expected return from switching will always be 1.25 times however
much I found in that envelope.

In contrast, in game (1) my average expected return from switching would be
$100 if I found $50 in the first envelope and $100 if I found $200 in the
first envelope, while my average expected return from switching if I found
$100 would still be $125, so my total average expected return from switching
regardless of what I find in the first envelope is (1/4)($100) + (1/2)($125)
+ (1/4)($100) = $112.50, while my average expected return from sticking with
my first choice regardless of how much I find is (1/4)($50) + (1/2)($100) +
(1/4)($200) = $112.50 as well. Again, in game (1) the resolution of the
apparent "paradox" must be that for any possible probability distribution
the millionaire uses to pick the amounts in the envelopes, your average
expected return from sticking with your first choice if you don't open it to
see how much is inside must always be equal to your average expected
winnings if you decide to switch without first checking how much was inside
your first choice.

>Now, which game would you prefer to play, (1) or (2)? They are not the
>same.

With the conditions I mentioned--the millionaire flips a coin to decide
whether to put $50 or $100 in one envelope, then puts double in the other,
and in game (2) he only replaces the amount in the second envelope if you
find $100 in the first envelope you choose--then the two games actually are
exactly the same, in terms of probabilities and average expected returns. If
you want to suggest some different conditions, though, go ahead.

>Look at it another way: game (2) is actually asymmetrical. The amount you
>win if you play it many times >will be different if you switch, because you
>really do have more to gain than to lose by switching (and >the millionaire
>will have to pay out more on average). On the other hand, intuitively, you
>can see that >your expected gains in game (1) should be the same whether
>you switch or not. The paradox comes >from reasoning as if you are playing
>game (2) when you are really playing game (1).

If we say that in game (2) the millionaire replaces the amount in the second
envelope with a new amount based on a coinflip *regardless* of how much you
found in the first envelope, then I agree that the paradox comes from
reasoning as if you're playing game (2) when you're really playing game (1).
But I would say the reason for that is that there's no possible probability
distribution the millionaire could use to stuff the envelopes in game (1)
that would lead to all the same probabilities and average expected returns
as in game (2) -- the reason is that there's no possible distribution where
you should conclude there's a 50% chance of the second envelope containing
double what you found in the first and a 50% chance of it containing half,
regardless of what amount you actually found in the first envelope. For
every real probability distribution the millionaire could use in game (1),
your estimate of the probability of the second envelope containing double or
half will vary depending on the amount you found in the first.

Jesse
Received on Thu Oct 07 2004 - 20:36:44 PDT