

The Two Envelopes Problem: the most boring paradox in the world - dools
http://iaindooley.com/post/61292324101/the-two-envelopes-problem-the-most-boring-paradox-in

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ColinWright
The author appears not to understand the problem as presented - certainly the
"solution" as given is inconsistent with my understanding of the problem.

Consider your position. I present you with two envelopes, and tell you that
each contains a number, and one of those numbers is twice the other. You
choose an envelope, and find that it contains the number 8.3467394756297. You
can choose to swap, if you like, and your objective is to maximize the
expected value of the number in the envelope you're holding.

Do you swap? Or not? Here's a line of reasoning.

Call the number I've chosen Z. The other envelope contains either Z/2 or 2Z,
and it would appear to be 50:50 as to which it is. If you swap, my expected
value is then (Z/2) _0.5 + (2Z)_ 0.5 = 5Z/4\. That's bigger than Z, so you
should swap.

But that means you should always swap, which doesn't seem to make sense.

Perhaps this version explains more clearly to you the paradox. Or not, maybe
you still think it's all obvious. In that case, could you revise your
explanation for this context, away from the $20 vs $10 or $40 issues?

~~~
dools
Yep so the way you've stated it there is my understanding of the problem,
except for this bit:

 _You choose an envelope, and find that it contains the number
8.3467394756297_

You're actually given the chance to switch without having first seen the value
in the envelope.

If we forget the notion of switching and expected values for a second, I just
want to focus on the very start of the statement where you say:

 _Call the number I 've chosen Z. The other envelope contains either Z/2 or
2Z_

The issue as I see it is that the probability exists that the other envelope
contains Z/2 or 2Z, but not both - ie. the probability that one of those
exists negates the other from existing, because in order for both to be
possible you're introducing a 3rd value where only 2 existed at the start.

If you ignore all the envelopes and switching and whatnot you can boil it down
to this:

I have two values A and B. I ask you to select one. What is the probability
that the remaining value is C? Obviously it's zero.

So if A and B are envelopes, how does this change? If B is twice A how does it
affect anything?

All the other information provided in this paradox just serves to distract you
from the fact that in order for you to have selected a value, and for the
probability to exist that either half that value or twice that value is in the
remaining envelope, you would have had to have 3 values present at the start
of the exercise.

~~~
dalai
_The issue as I see it is that the probability exists that the other envelope
contains Z /2 or 2Z, but not both - ie. the probability that one of those
exists negates the other from existing, because in order for both to be
possible you're introducing a 3rd value where only 2 existed at the start._

So there is 50% probability that the other one contains Z/2 and 50% that it
has 2Z. Do you disagree with that statement?

 _I have two values A and B. I ask you to select one. What is the probability
that the remaining value is C? Obviously it 's zero._

No it doesn't. According to the problem, both 2Z and Z/2 are possible and you
don't have the information to decide either way. Think of a simple coin toss,
before I show you the result it is 50-50 to be heads or tails, even though it
is just one of the two.

~~~
dools
_So there is 50% probability that the other one contains Z /2 and 50% that it
has 2Z. Do you disagree with that statement?_

Yes! Well, in some circumstances. Unless you know how the values were placed
into the envelope to begin with, how could you possibly assign a probability
to it? I just found this great link today:
[http://lesswrong.com/lw/dy9/solving_the_two_envelopes_proble...](http://lesswrong.com/lw/dy9/solving_the_two_envelopes_problem/)
which basically resolves the problem I'm having by taking into account how the
values get into the envelopes in the first place.

If you have absolutely no knowledge of how the values got into the envelope in
the first place then there's a 50/50 chance you got the higher value on your
first pick, and a 50/50 chance you'll get the higher value on your second
pick. You can't make a more informed decision unless you know how the values
were initially chosen.

