The envelope problem, and a possible solution

[PairTheBoard]

[ QUOTE ]
Recently I read a post on the 2 envelopes paradox ( Here).

I will not elaborate on what the problem is, because it is stated and discussed clearly in the provided link and I ask you to please read the OP. I want to discuss a solution put forward by a friend of mine.

A friend of mine argued the following:
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If we were te simulate the envelope problem, it would show that looking at the amount of money in your envelope is not going to affect the EV of always switchin envelopes. Which would be in line with the argument that switching envelopes is neutral EV. He asked me, "if we were to play this game, do you expect that looking at the amount would change anything to you EV?" And I answered: "I do not see how always switching can possibly become anything other than neutral EV, when the only change you make is that you are merely passivily looking at the value now". After which he followed:

Most likely the argument that switching is EV neutral is the correct one. The other argument then cannot be correct, so somewhere in the argument there must be flaw. He argued that the point of determination is key. The amounts in the envelope are determined before the start of the game, not after you learn the value.

If the value of the second envelope was determined after you had learned the value of the first (for example by flipping a coin) then idd switching would be +EV. This is comparable to saying, I give you $100, you can keep it, or I flip a coin; heads you double it, tails you lose half.

However, this is not the case in the envelope problem he said. The values are assigned to the envelopes on beforehand. Even though intuitively it may feel like a switch might double your money or cut it in half after you learn the value; in reality there is only one possible outcome. For example, if you look at the value in envelope 1, and it is $100. Then intuitively it might feel like you have a 50% shot at $50 or $200, but you don't have it. Say the ammount in the second envelope was $50, it would not be possible to get $200 on that trial. The flaw would boil down to using the $100 as new information in determining your EV when it doesn't provide any new information really. So, even though the method of determining your EV is right in argument 1, it is not applicable in that situation.

What I am wondering is, how is this line of reasoning? He is fully convinced that it is the solution to the problem, and a colleague of mine basically suggests the same solution.
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I still am not sure, I'd appreciate some imput.

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I agree with your friend and his colleague under the assumption that the amounts in the envelopes are Fixed. That is, it is NOT the case that the game can be repeated by picking NEW amounts in the envelopes according to some unknown probability distribution. Instead, the game can only be repeated by offering the SAME envelopes to a number of NEW people, none of which know their contents. Under this assumption, each time a Person opens an envelope and sees an amount, the conditional probability of the second envelope being double the amount he sees is either 100% or 0% - he just doesn't know which.

PairTheBoard

[AaronBrown]

[ QUOTE ]
I got frustrated with Aaron in this dialogue before. Sure am glad you started it up again. [img]/images/graemlins/smile.gif[/img]

Here's how my dialogue would go when I hear the problem.

Paradoxer: "I put a random amount X in one envelope and 2X in the other."
Me: "Wait, how was X chosen?"
P: "I'm not telling you. Now, you open and find $100. Should you switch?"
M: "Man, I dunno. I can tell you the correct move if you describe the opponent."
P: "I'm not going to tell you. So, what's the best move?!"

It's retarded. Of course you and I can arrive at different answers if we try to fill-in the gaping hole differently. That doesn't show that the problem's clever; all that shows is that the problem has gaping holes.
-Sam

[/ QUOTE ]
I understand your frustration because I refuse to answer the question. I think the problem is people assume there must be a right answer, and then use that to reason, usually to the idea that you shouldn't switch. My objection is that there is only a right answer if you assume a lot of mathematical machinery, perhaps consistent finite prior beliefs or the existence of an expected value.

It's not just theoretical hair-splitting, as the FX example and years in Purgatory example show. You can make reasonable assumptions that make it +EV to switch. The useful thing is to understand why both the switching and indifference arguments have force, and when to follow one versus the other.

I also understand it's frustrating that life forces us to play a game without clearly specified rules or odds. But don't blame me for that one. The point of this example is not to prove it's a clever question, it's to help you think about real decisions in which you don't how amounts were selected for the envelopes.

Don't argue with me, argue with the Clash:

Clash: "Should I stay or should I go? If I go there will be trouble."
SamIam: "Wait, how was the amount of trouble chosen?"
Clash: "If I stay there will be double."
SamIAM: "Man, I dunno. I can tell you the correct move if you describe the opponent."
Clash: "So you gotta let me know. Should I stay or should I go?"

Sorry SamIAM, you have to choose.

[Masquerade]

Your intuition is correct of course and it's EV neutral - at least for meta-beings in a universe where all amounts of money are equally likely.

It's funny seeing Aaron's avatar here instead of a finance forum where I usually read his wisdom. I assume he's making the point that once you switch to real money in this world and real people with utility functions then it's not so clear-cut. Suppose the amount in the envelope was 1/2 of all the money in the world for example?

[NaobisDad]

Aaron, there is something that puzzles me about your reply regarding argument 1. Under the assumption that the method used for figuring EV in this argument is correct, then it would seem to me, that without any further assumptions that switching would always be EV.

Or am I now implicitly making the assumption that the value (say N) is an element of a set of positive real numbers?

My point is that as stated, argument 1 seems to argue that knowing the ammount will lead to a change in EV irrespective of the size of the ammount in the envelope. That is, the player will change strategy despite the fact that he/she didn't really get new information. Also, as stated, it would seem that unless N was a negative value, switching would always be EV.

What am I missing?

[NaobisDad]

Hey PTB,

you said: "I agree with your friend and his colleague under the assumption that the amounts in the envelopes are Fixed."

COnsider the following. If I made 100 pairs of envelopes with variable ammounts, such that each pair of envelopes contains N and 2N values. We play the game. You choose an envelope at random. And you then switch blind, or you don't. Would you predict that now switching affects EV?

Also would you think that your EV would be affected if now after picking an envelope you are now shown the value that ethe envelope contained?

[NaobisDad]

Hi, Masquerade. From your statement I deduce that you then agree that argument 1 must be flawed somewhere. Where do you think that is?

[NaobisDad]

Hello, Jason. thank you again for an elaborate reply.

As I understand it the crux of your post is that if your know the value in one envelope, then given the sampling method this knowledge will affect the chance that the value in the other envelope will be either double or half the amount you now see.

however, in the original problem I don't believe it is argued that the value is used in that way. Rather, a very straightforward, simple modification of estimating EV is used. And the probability with which the amount in the other envelope is N or 2N seems dependent on the probability with which you pick an envelope.

Then you said: "It is EV neutral to always switch. Argument 2 is correct about that. But the problem did not ask about the EV of always switching. It asks about the EV of switching."

I still don't understand, and others have argued this point also and I didn't understand then either, if argument 1 is correct, why then would you ever NOT switch?

[NaobisDad]

Hi, BBB. You said: "All we are told in the original problem is that one of the envelopes contains twice the other. It tells us nothing about the chances that the other envelope contains twice as much, and to simply assume that these chances are 50% is not correct."

Do you then imply that we choose an envelope in a non random way? Or do you think that those probabilities do not depend on the probability with which you choose the initial envelope? I would think that I have 50% chance op picking up N or 2N.

"The EV of switching no matter what will be -X if we chose the 2X envelope, and X if we chose the 2X envelope. Since there's a 50% chance of chossing either envelope, the net EV of switching no matter what is zero."

I agree here, and as I read it, it is the pointe of your post. Although in the more involved example you do point to the fact that the size of the ammount can matter.

If you conclude that barring other information switching is neutral EV, then you must believe that as stated in the original problem the reasoning in argument 1 is false. However from your post I do not understand how you think this is so.

[lorinda]

Previous envelope 2+2 discussion

Lori

[PairTheBoard]

[ QUOTE ]
Hey PTB,

you said: "I agree with your friend and his colleague under the assumption that the amounts in the envelopes are Fixed."

COnsider the following. If I made 100 pairs of envelopes with variable ammounts, such that each pair of envelopes contains N and 2N values. We play the game. You choose an envelope at random. And you then switch blind, or you don't. Would you predict that now switching affects EV?

Also would you think that your EV would be affected if now after picking an envelope you are now shown the value that ethe envelope contained?

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No and No.

You ask, where does argument 1 break down? It's in the conclusion once seeing the amount A in Env 1, that chances are 50-50 that Env 2 has 2A. With Fixed envelope amounts the chances are either 0% or 100% you just don't know which. That's the principle that negates argument 1. If the amounts come from a population of 100 envelope pairs - or some other unknown probabilty distribution - then the conditional probabilty that Env 2 has 2A is still something other than 50% in general, and just like above you don't know what it is. It's the same principle negating argument 1. The principle is just more apparent in the Fixed Amount case.

PairTheBoard