Two Envelopes Paradox(It's blowing my mind)

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Hopefully this is the right forum for this.

I was looking at some paradoxes on the internet, and trying to "solve" them(usually the paradox is built on a flawed premise, or it isn't paradoxical at all), and I came across one that I simply cannot understand.

It goes like this: suppose there are two envelopes, with money in them, A and B, and you want to pick the one with the most money. You are told that one envelope contains exactly twice as much money as the other. You are allowed to open one of the envelopes, and then given a choice to switch envelopes or stick with the one you opened. You open A, and you see that it contains, say, $10. Now, you know that there is a 50% chance that B will contain $5 and a 50% chance that it contains $20. You can calculate the average expected value of envelope B like so: 0.5*5 + 0.5*20 = 12.5. Therefore, the expected value of B is greater than A and you should switch to B. However, if you had initially picked B, you would come to the conclusion that you should switch to A regardless of the actual value of B.

We can simplify the problem so that you don't even have to open any envelopes. One envelope has twice as much money as the other. The expected value of envelope B would be 0.5*0.5A + 0.5*2A = 1.25 A, thus, B is a better pick than A. However, the same logic can be used to show that A is a better pick than B. So a seemingly valid premise and seemingly valid logical deductions lead us to a seemingly invalid conclusion. That's why it's a paradox.

I just can't figure out the solution to this problem. There has to be some mistake somewhere, yet every logical deduction made seems valid. Isn't it universally true that you can assign an expected value to a choice by multiplying the value of each of it's outcomes by each outcome's probability and summing the results? Can anyone figure this out?

Also, feel free to post your own paradoxes or similar logic problems.

This sounds a bit like the Monty Hall problem.

ruveyn

There does seem to be one way one can come to a profitable conclusion.

If the total amount of cents in the first envelope is an odd number then there is no possible half amount that can occur in the second envelope; therefore the greatest amount of currency is in the second envelope.

If the first envelope has $100.01, the two possible answers are $50.005 and $200.02. Since $50.005 is not a correct amount of currency the second envelope must have $200.02.

This is the only situation I can see that there is 100% probability there will be a greater amount in the second envelope, all other situations there is only a 50% probability.

You're misunderstanding the problem a bit, Ericys. The goal isn't to find how to make the most profitable decision, the goal is to understand why a (seemingly) fully possible premise and valid logical deductions lead to a contradictory conclusion. Usually, with these paradoxes, the initial premise contains some sort of contradiction or impossibility, which isn't the case here. Sometimes, there's a flaw in the logical deductions, and that's why the conclusion is contradictory. However, all of the logic seems valid to me.

@ ruveyn: Yes, it does remind of the Monty Hall problem. However, it's a lot trickier, because while the Monty Hall problem goes against intuition, this paradox seems to defy logic. I think it would take a logician or mathematician to make sense of it.

There's a 50 percent chance you'll get more money from the second envelope, but there's only a 10 percent chance of that.

Nice problem, I hadn't heard of it before.

After some twisting and turning I googled it and it seems like it was first put forward in 1953 and has been befuddling mathematicians and economists ever since.

http://en.wikipedia.org/wiki/Two_envelopes_problem

Sign posted at the neighbourhood swimming pool:
'Non-swimmers may not approach this pool unless they know how to swim.'

It's because 50% of the double amount will always be the same as the envelope you opened, then you add in the other possible value, going over 100%. 50% of 200% is 100%. It's algebra.

As far as 'why' that all is, that's just one of those basic questions, like where did the universe/God first 'come from', why is there gravity, why is there electricity, etc. etc. etc.

Another way of looking at it..

If the other envelope is twice as much, that's an extra $10. If it's half as much, you only lose $5.

Due to the fact the second choice is dependent on the first choice having occurred the probability of the second choice being greater than the first choice can be shown:

P(A then B) = P(A) * P(B)
P(A then B) = 0.5 * 0.5
= 0.25

This means given that one has picked one of two envelopes with a probability 0.5 of being the larger amount, the second choice will reduce the probability to 0.25.

First choice will have a 50 percent chance
Second choice will have a 25 percent chance

This appears to be a disproof of the idea that if something has an average expected value higher than some actual value that it is definitely the better choice.

If there are a million lottery tickets, one worth a million dollars and all the others worth nothing, and I offer to sell you a lottery ticket for one dollar, that is fair according to average expected value, but I'm probably ripping you off.

If the goal is to find out what is in both envelopes you pick both envelopes reguardless of the possibility of netting less money. Either way you walk away with more money then when you started and, the knowledge of what was in both envelopes. Or, maybe I'm missing somethign here.

Speaking vaguely, I think that the problem is that the notion of expected value only makes sense if the probability distribution is "finite".

For example, suppose that we changed the problem so that the money in the smaller envelope is an integer number of dollars from 1 to 100, and has an equal 1/100 chance of being any of these 100 values. Then the mathematics would work out, because if you get a higher amount of money then there is a lower chance that you chose the smaller envelope, if you get an amount of money over 100 then it is certain that you did not choose the smaller envelope, and if you get an odd number then it is certain that you chose the smaller envelope.

However, the "probability distribution" implied by the question is that the smaller envelope could contain any amount of money, no matter how large, and possibly it might not even be an integer, or it might not even be an integer multiple of a cent. But this probability distribution is not "finite".

Think about it like this: if there are an infinite number of quantities which the smaller envelope might contain, and each of these quantities has an equal chance C of being right, then we are in an impossible situation. Why? Because if C = 0, then it is impossible for any particular quantity to be right! But if C > 0, then the total probability is an infinite sum of C's, which is infinity, not 1! There are mathematical tools which can help to deal with this type of situation, but I think that when you read the fine-print for using these tools you will discover the resolution to the paradox.

Here is my favourite paradox. You need to know the idea of "proof by contradiction". If you're not sure, look it up on Wiki.

CLAIM: Each positive integer can be named in under fourteen English words.

PROOF: Suppose for contradiction that not every positive integer can be named in under fourteen English words. Then there is a smallest positive integer that cannot be named in under fourteen English words, call it x. But x can be named in under fourteen English words, since x is "the smallest positive integer that cannot be named in under fourteen English words", giving us our contradiction.

CLAIM: It is impossible that each positive integer can be named in under fourteen English words.

PROOF: There is a finite number of English words, call it N. So there are only N^1 + N^2 + ... + N^13 possible combinations of under fourteen English words. But there are an infinite number of positive integers!