Probability: Envelope Paradox

Factual Questions
41

ultrafilter,

Well, you’ve said that a fixed but unknown quantity is not a variable, which is true. But you’ve not shown any reason for it not to have the same properties as a variable for expected values purposes.

Consider the likelihood function before it is drawn. It is clearly a true probability. Now it is chosen, and the value has a real existence to those who know it. From the perspective of those who don’t, it has the same properties as a probability. If there was a 50% likelihood until now, it continues to have a 50% likelihood, in the sense that in an expected 50% of the cases the results will turn out to be that particular way.

It is true that “If you know that something is fixed but unknown, you have more information about it than if you know nothing” as you say, but that additional information is completely unbiased in either direction, and does not affect the probability.

(BTW, the Monty Hall problem is not the product of information being known - it is the result of the opening of the second door being non-random, thus biasing the remaining group of closed doors, IOW making a higher percentage of them winners).

42

No, I didn’t mean to disagree with you on that. It is true, you must treat a fixed but unknown quantity as a variable to work with it. I think this a classic “theory vs. praxis” type-thing.

I concur.

[quote]
**It is true that “If you know that something is fixed but unknown, you have more information about it than if you know nothing” as you say, but that additional information is completely unbiased in either direction, and does not affect the probability.

True, that was a bad example. But after a long day at the office, it was all I could come up with. I should’ve given the “neighbor has two kids and says one of them’s a girl” example.

43

The most interesting thing about this thread are all the posts that read additional information into the OP or try to answer by analogy with more complicated questions.

The simple answer is that there is no paradox. The expected value of B is 1.25 * A, and the expected value of A is, by similar argument, 1.25 * B. There is clearly no reason to prefer one to the other. Which, of course, is consistent with the idea that there was a 50-50 chance of picking the right envelope in the first place.

I think the problem comes from trying to make physical sense of the concept of expected value. For example, the expected value when a six sided die is tossed is 3.5. Let me know if you ever roll a 3.5 on a standard die.

44

hrm… this is how I reason it. I can be wrong.

if you have envelope A and envelope B and you can choose one, you are not going to know if you will be better off choosing A or B.

The math is this.
A = X
B = 2X or .5X

Okay, the above setup is wrong and that is what people are trying to tell you.
It should be
A(1) = X has 50% chance of being true.
A(2) = 2X has 50% chance of being true.
B(1) = X has 50% chance of being true.
B(2) = 2X has 50% chance of being true.

The reason it does not matter if you choose A or B is that you don’t know which envelope you got in your hand- A(1) or A(2) and same if you chose B.

The reason that you are not getting it is that you assume X is the amount you get when you open your envelope, but that is wrong way to look at it. You should say X is the min or max and go from there.

If X is min:
So let’s say X=10 and you choose A, so you get 10 half the time and 20 the other half the time, you will average 15 if you choose A, but same hold true if you choose B, you will average 15.

If X is max:
So let’s say X=10 and you choose A, so you get 10 half the time and 5 the other half the time, you will average 7.5 if you choose A, but same hold true if you choose B, you will average 7.5.

Is that more clear?
DOes it make sense?

If that does not, then let me add this.
You cannot combine the probability of the amount in A and B, at least not the way you are doing it.
No matter if you chose A or B, from your perspective A can contain three values and B can also contain the same three values- 5, 10, 20. or using variables- 0.5X, X, 2X where X is what you choose. Remember you don’t know what A or B is.

To prove your flaw:
You choose A.
The amount is A=X.
B has a 50% chance of being 2X or .5X.
Okay, you are coming from your perspective, which is not reality.

Reality is that you chose A.
A=X
B is either 2X or 0.5X, but cannot be both. So over 100 times, you would get all 2X or all 0.5X. But not half 2X and half 0.5X, so you cannot average the two. If you want to be able to average the two, you must not used your perspective but use an objective perspective.
Objective perspective is
50% chance that A=X,B=2X
50% chance that A=2X,B=X

I don’t know if this helps.