The envelope problem, and a possible solution

[NaobisDad]

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.

[econophile]

there was a link to a decent explanation to the problem in the original thread. the set up of the problem is that one envelope has $N and the other envelope has $2N in it. the trick is that the value of N is unknown.

but the participant must have some belief about how likely different values of N are. this belief just describes how likely the participant thinks different values for the smaller envelope are. and if the player thinks some values are more likely than others (maybe he believes extremely high values are unlikely) then it is not true that after learning the contents of one envelope the other envelope has a 50/50 chance of being the big envelope.

for example, if you open the envelope and discover a large amount of money, you might think that it is probably the 2N envelope. if you open the envelope and discover a small amount of money, you might think it is probably the N evelope.

so, after opening the first envelope, switching will sometimes have +EV and sometimes -EV. but before you learn the contents of the first envelope, you are indifferent between switching and not switching.

[Paul2432]

[ QUOTE ]
If we were to simulate the envelope problem . . .

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How?

Paul

[SamIAm]

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

P.S. Assuming that X is chosen equally likely from the set of all positive integers is flat wrong. You're not allowed to assume that, because it doesn't make sense. [img]/images/graemlins/smile.gif[/img]

[NaobisDad]

thanks for the reply guys.

@econ: Depending on the amount of money in the envelope, someone might indeed use this information for switching which I expect to influence EVj. However, as I understand it, stated in the original problem, it's not a matter of psychology. If you are indeed allowed to estimate EV according to the method of argument 1, switching would always be +EV. And if you believe arguemtn 2 to be true, it wouldn't matter. It's therefore that I simplified the problem to always switching.

@Paul: we could simply play it, or write a program to play it out. Note that as stated you could play this game without ever really having to actively participate. So you could run a lot of trials. From your response it seems a bit like you believe that how you simulate the problem is going to matter. If this is the case, I'd like to hear those thoughts.

@Sam: Sam, you refer to the sampling method. But at this point I can't really deduce why you believe the manner of sampling is important. You said: "Assuming that X is chosen equally likely from the set of all positive integers is flat wrong. You're not allowed to assume that, because it doesn't make sense." Alright, if we then do not make that assumption, how would that impact the results?

[pzhon]

I don't like the idea that explaining why always switching is equivalent to always not switching is a solution. It's not complete. As bigpooch mentioned in the other thread (and others have mentioned in past threads), you can do better than either always switching or always not switching. This should be added to the statement that you can't assume the conditional expectation is 1/2.

Choose a value t. If you see an amount less than t, switch. If you see an amount greater than t, don't switch. Conditioned on the case that both amounts are less than t, you break even compared with either always switching or never switching. Conditioned on the case that both amounts are greater than t, you break even again. However, if one amount is greater than t, and one amount is less than t, you always end up with the greater amount. If this happens with positive probability, switching with threshold t beats both the strategy of always switching and the strategy of never switching.

Instead of letting t be constant, you can let t be a positive real random variable with positive probability on each interval [x,2x]. Another way to describe this is that you switch probabalistically, with the probability of switching strictly lower at 2x than x, with limits of 0 and 1 as x goes to infinity and 0, respectively. This random threshold will have a positive probability of being between the two amounts no matter what distribution is chosen, and this will beat the strategy of always switching or never switching, even though you can't be sure by how much it will be better.

In fact, you can describe the strategy of always switching as using a threshold of +infinity, and you can describe the strategy of never switching as using a threshold of 0. You can see these thresholds have 0 probability of being between the amounts.

Mike Caro got this wrong, and the Wikipedia article doesn't mention that you can do better than either.

[econophile]

[ QUOTE ]
@econ: Depending on the amount of money in the envelope, someone might indeed use this information for switching which I expect to influence EVj. However, as I understand it, stated in the original problem, it's not a matter of psychology. If you are indeed allowed to estimate EV according to the method of argument 1, switching would always be +EV. And if you believe arguemtn 2 to be true, it wouldn't matter. It's therefore that I simplified the problem to always switching.


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My argument isn't based on psychology, it's based on Bayesian statistics. According to Bayesian statistics, statisticians can have believes about random variables before they gather any data on them. These beliefs are called "priors."

The only prior that allows arguement 1 to be correct is believing that the value of the smaller envelope is distributed uniformly from 0 to infinity. But this distribution doesn't make sense . . . it means you think $5 Billion is as likely a value as $5.

For any other "prior," argument 1 fails.

[TomCollins]

[ QUOTE ]

@Sam: Sam, you refer to the sampling method. But at this point I can't really deduce why you believe the manner of sampling is important. You said: "Assuming that X is chosen equally likely from the set of all positive integers is flat wrong. You're not allowed to assume that, because it doesn't make sense." Alright, if we then do not make that assumption, how would that impact the results?

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If you make the assumption that it is a positive number randomly chosen, and you get an odd amount of money in your envelope (1/4 of the time you will), you know to switch.

[NaobisDad]

thanks for the replies again guys. You'll have to bear with me, I'm no mathmatician or statistician (is that a word?) so I'm slow here.

@econ: high econ I don't know anything about priors, so I'll need some further explanations on that. You said:
"but the participant must have some belief about how likely different values of N are." and went on to describe how the value might influence the way in which he determines ev.
In the original post it is stated that:
"Argument 1: It's +EV to switch. You had a 50/50 chance of picking the high or low envelope so there's a 50% chance that the other envelope is the high and a 50% chance it's the low. Therefore, EV of switch = 0.5*(+100) + 0.5*(-50) = +25."
nowhere here do I read anything about how the size of the value influences the decision to switch. Rather, I read that upon learning the value, switching is EV.

@Phzon: you said: "I don't like the idea that explaining why always switching is equivalent to always not switching is a solution. It's not complete. As bigpooch mentioned in the other thread (and others have mentioned in past threads), you can do better than either always switching or always not switching."

I merely refer to the fact that argument 1 as stated implies that switching will always be positive EV upon learning the value. Argument 2 says it will never matter. You might be able to do better by sometimes switching, but imo that was not stated in the original problem. For that reason I simplified to always switching. Similar to my respons to Econ, choosing a switch strategy might well influence your EV, however as I understand it, that was not waht was stated in the original problem. Correct me if I'm wrong.

@Tom: Are you saying that the sampling method will unintentionally provide information on the value of envelope 2 and that in this manner the sampling method is significant to the problem?

Again, my question really comes down to it that my friend believes it's not a paradox, rather that argument 1 is based on a fallacy. How did he do? is he right?

[jason1990]

The question in the OP is: Is switching +EV? The answer is, it depends. It depends on (i) the amount you see and (ii) the way in which those amounts were chosen by the person who put the money in there.

Both arguments are flawed. Argument 2 is flawed because it doesn't even answer the question. It answers the question, is always switching +EV? Argument 1 is flawed because, after seeing the first amount, there is not a 50% chance you have the higher amount. The probability depends on (i) and (ii) above. If you make assumptions about (ii), then you can calculate the EV. Sometimes it will be positive, sometimes it will be negative.

Your friend is wrong. He did not identify the flaw in Argument 1. Specifically, your friend thinks that the fact that the money was determined before you chose has something to do with the EV. But we know from our poker experience that this is a silly argument. The deck is shuffled and the order of the cards is determined prior to any of our bets. But that does not affect our EV calculations, which are based on the fact that we don't know this order, not that the order is yet to be determined.