The envelope problem, and a possible solution

[jason1990]

[ QUOTE ]
Bayesian probability is a measure of the degree of belief a person has in some proposition. Several attempts have been made to operationalize the intuitive notion of a "degree of belief". The most common approach is based on betting: a degree of belief is reflected in the odds and stakes that the subject is willing to bet on the proposition in question.

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Here's an interesting thought experiment. You choose a number between 1 and 6, write it down, and seal it in an envelope. You are somehow assured that there is no trickery and I have no idea what number you chose. Now, your number is fixed. Suppose, for example, that it's 1 (but I don't know that). I now offer you a wager. I will name a number between 1 and 6. If I name your number, I will give you $60. How much would you be willing to wager on this? What do you think is the probability that I will name 1? I'm not going to roll a die or use any other device. I'm simply going to choose a number.

[AaronBrown]

Bayesians define probability as subjective belief, frequentist by repeated experiment. Both are equally rigorous mathematically, but both have severe technical problems. These seldom lead to practical difficulties, but things like the envelope paradox highlight them.

A Bayesian would say the probability that you have the higher amount in your envelope is 50% before you open it, but since the amount you see will change your subjective estimate, it will be different from 50% after you look. That is consistent and leads to the commonsense result, but it's hard to deal with probabilities that change based on unobservable mental states.

A frequentist would note that if you repeat the experiment many times, you end up with the larger amount 50% of the time, whether you look at your amount or not. Opening the envelope does not change the probability.

Before we open our envelope, call the amount inside X. We know that the other envelope has a 50% chance of holding 0.5*X and a 50% chance of holding 2*X. Its expected holding is 1.25*X. A frequentist has to deny this, because otherwise once we open our envelope and discover X, it would always make sense to switch. So frequentists deny that a 50% chance of losing $X/2 and a 50% chance of winning $X has positive expected value. Again this is mathematically consistent, but hard to accept.

The envelope paradox is stripped down and designed so you can see both arguments at once. It's easy to come up with ad hoc principles that get to the commonsense results. But in real analysis, both statisticians and lay people make both of these assumptions all the time. When there's no reason to think A is better or worse than B, we assume there's a 50% chance that A is better. Then if told we have a 50% chance of losing 0.5*$X and a 50% chance of gaining $X, we calculate our expected gain is 0.25*$X. It's hard to come up with general principles that prevent us from making errors in real problems.

[jason1990]

[ QUOTE ]
Bayesians define probability as subjective belief, frequentist by repeated experiment. Both are equally rigorous mathematically, but both have severe technical problems.

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All statisticians, whether they call themselves Bayesians or Frequentists, use the same formal mathematics, which is simply measure theoretic probability theory. I don't know of any severe technical problems with that branch of mathematics. Perhaps you are talking about philosophical problems.

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A Bayesian would say the probability that you have the higher amount in your envelope is 50% before you open it, but since the amount you see will change your subjective estimate, it will be different from 50% after you look. That is consistent and leads to the commonsense result, but it's hard to deal with probabilities that change based on unobservable mental states.

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The probabilities change based on concrete observable information. To the Bayesian, the sum total of money in both envelopes is a random variable with whatever distribution he decided to use to model his beliefs. If he knew the total amount, the decision of whether or not to switch would be trivial. Once he sees the amount in one envelope, it is possible for that to give him information which would affect the probability distribution of that sum total. It is not difficult to deal with this change in probability, since it is given by the formulas for conditional probability. The only objection that can be made to this is to claim that the Bayesian had no right in the first place to declare the sum total to be a random variable with such-and-such a distribution.

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A frequentist would note that if you repeat the experiment many times, you end up with the larger amount 50% of the time

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What, exactly, is the frequentist repeating? Is he resealing the envelopes, shuffling them up, and picking them again? Then the contents are fixed, say n and 2n, and it is his choice which is random. Or is he being supplied with new envelopes every time? In this case, it is both the contents and his choice which may be regarded as random, and the frequentist cannot proceed with his analysis until he either determines or assumes something about the distribution of the random quantities inside those envelopes. However, once something is known about that, even the frequentist would agree that it is entirely possible for his observation of the first amount to persuade him to switch.

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Before we open our envelope, call the amount inside X.

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Do you mean for X to denote a number or a random variable? Even if you regard the amounts in the envelopes as fixed (say n and 2n), the amount in your particular chosen envelope is random. It is the random variable which is n with probability 0.5 and 2n with probability 0.5. So it seems that you are suggesting X is a random variable.

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We know that the other envelope has a 50% chance of holding 0.5*X and a 50% chance of holding 2*X.

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This sentence clearly indicates that the amount in the other envelope is being regarded as a random variable. So I am now even more convinced that you meant for X to be random. Moreover, if I assume that you meant for X to be a fixed number, then you just told me that there is a 50% chance the sum total of both envelopes is 1.5X and a 50% chance it is 3X. Hence, if X is a fixed number, then you are regarding the overall contents of the envelopes as random. However, you have not stated your assumed distribution on that random quantity, nor have you explained how that distribution implies that 1.5X and 3X are equally likely. So I seem to have no choice but to conclude that X is meant to be a random variable.

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Its expected holding is 1.25*X.

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Now wait just a minute. Expected value is a number, not a random variable. So it looks like you're regarding the contents of the chosen envelope as a fixed number. So, as above, you must be regarding the sum total of both envelopes as a random quantity. But, as stated above, you did not tell us your assumed distribution on that random quantity, nor did you explain how that distribution implies that 1.5X and 3X are equally likely. Perhaps you implicitly assumed that the sum total of both envelopes is an unknown quantity which is equally likely to be any positive number. If that's the case, then you have committed exactly the Bayesian fallacy I talked about in my other post.

Or perhaps you really did mean for X to be a random variable, and what you called "expected holding" was really supposed to be the conditional expectation. In that case, when you said "the other envelope has a 50% chance of holding 0.5*X", you were really talking about conditional probability, saying that P(Y=0.5X|X)=0.5, where Y is the contents of the other envelope. But if the sum of the envelopes is not random, then this statement is false, since P(Y=0.5X|X) is the random variable which is 1 if X=2n and 0 if X=n. If this is what you meant, then the problem lies in the fact that you used unconditional probabilities to compute a conditional expectation. In this case, the fallacy in what you wrote has nothing to do with Bayesians vs. Frequentists. It has to do with the problems that arise when we do mathematics informally, without proper regard for rigor. I suppose this is another lesson the Paradox can teach us. I hadn't thought about that.

I would very much like to see you write this down rigorously -- i.e., formally define your probability space, your probability measure(s), and your random variables. You claim to be demonstrating fundamental flaws in probability theory that stem from the Bayesian/Frequentist debate. But without a rigorous presentation, I fail to see how you've demonstrated anything other than (a) the uniform distribution fallacy, or (b) the fact that being inconsistent about what is random and what is fixed, and about what is conditional and what is not, can lead to the wrong answer.

[PairTheBoard]

[ QUOTE ]
Jason -
It has to do with the problems that arise when we do mathematics informally, without proper regard for rigor. I suppose this is another lesson the Paradox can teach us. I hadn't thought about that.


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I think I have you on that one Jason.

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PTB -
I would say it's designed to show the need to be careful when applying mathematical tools like expected value and probabilties. When you speak in those terms you need to have a clear mathematical model in mind to which you are applying them.


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I am now going to reread your post because I enjoyed it so much the first time.

PairTheBoard

[AaronBrown]

[ QUOTE ]
I would very much like to see you write this down rigorously -- i.e., formally define your probability space, your probability measure(s), and your random variables. You claim to be demonstrating fundamental flaws in probability theory that stem from the Bayesian/Frequentist debate. But without a rigorous presentation, I fail to see how you've demonstrated anything other than (a) the uniform distribution fallacy, or (b) the fact that being inconsistent about what is random and what is fixed, and about what is conditional and what is not, can lead to the wrong answer.

[/ QUOTE ]
To answer your specific question, X is a variable, but not a random one. It is the amount in one envelope. The amount in the other envelope is a random variable, with expectation 1.25*X.

You have inverted my meaning. I claim the envelope paradox is a true paradox, two strong arguments that lead to opposite conclusions. Yes, people invent rigorous ways to resolve it, but only by torturing commonsense ideas of uncertainty. They end up with mathematically consistent theories that give absurd results in some cases.

I don't dispute the rigor of either the Bayesain or Frequentist approaches, merely their relevance to practical decision making. The Bayesian says the odds are not 50/50, the Frequentist says the expected value of a 50% chance of getting 0.5*X and a 50% chance of getting 2*X is not 1.25*X. Both say you shouldn't always switch, which agrees with common sense, but only by invoking complex technicalities. In practical decision making, you'd never see these technicalities, so you'd fall into error. Moreover, there are situations in which it does always make sense to switch, the technicalities give no clue as to when those situations occur.

I'm not anti-statistician, I just think people internalize the mathematical argument and treat it as necessary for reality. It's useful in some cases, not in others. It's good for predicting the chance of getting dealt pocket Aces, bad for telling you when to call an all-in bet on the river.

I don't have a rigorous framework, if I did I'd publish and get the Nobel prize. My claim is no one can give clear, commonsense answers to both of the following questions without leading to absurd conclusions:

(1) Is the probability that you have the higher amount 50%?

(2) If you have a 50% chance of losing some amount and a 50% chance of winning twice that amount, is it a positive expected value bet?

Imagine you have an envelope in your hand, with $10,000 in it. You wonder whether or not to switch. The Bayesian tells you to look inside yourself to figure out what probability you assigned to every possible total amount in the two envelopes before you opened yours. Doesn't that sound a little bit unhelpful? Your only question now is how likely is a $15,000 total versus a $30,000 total? That's what you should be weighing, using your current state of mind, not your prior one.

The frequentist tells you that based on the design of the experiment, he believes that if it were repeated many times, you would end up with the higher amount 50% of the time, therefore, the probability you have the greater amount is 50%. You then say, "Great, you're telling me that the expected value of switching is $12,500." "No," he replies, "that computation is disallowed." Aren't you going to want to take a swing at him?

[PairTheBoard]

[ QUOTE ]
Aaron -
I don't have a rigorous framework, if I did I'd publish and get the Nobel prize. My claim is no one can give clear, commonsense answers to both of the following questions without leading to absurd conclusions:

(1) Is the probability that you have the higher amount 50%?

(2) If you have a 50% chance of losing some amount and a 50% chance of winning twice that amount, is it a positive expected value bet?


[/ QUOTE ]

If there's something to what you say, maybe we should have a group effort to develop such a rigorous framework for your proposition and have it published by 2+2. We could all share in the Nobel prize - I know DS would be up for it.

Can your challenge be met? Let me take a stab at it. I'm not sure how much math language is allowed for "common sense" so I'll try to keep it at a minimum.

For #2. "If you have a 50% chance of losing some amount and a 50% chance of winning twice that amount, is it a positive expected value bet?"

If "the amount" is fixed the answer is yes. However, if you have a 50% chance of winning but "the amount" is $1 when you win and "the amount" is $2 when you lose then your EV is 0.

How's that? Does that agree with common sense?

For #1. "Is the probability that you have the higher amount 50%?"

Where are we at when this question is being asked? Are we where ...

a) Envelopes have been sealed and shuffled. You have just picked one at random. You are taking what's in that envelope sight unseen.

b) Envelopes have been sealed and shuffled, you picked one, opened it. If you repeated the process, Half the time you would see amount A and half the time you would see amount B. Let's say this first time you see amount A and you keep it.

If we're at a) then it it makes sense to say there is a 50% probability your envelope contains the larger amount. It makes common sense because you can repeat this process and over the long run about half the time your envelope will contain the larger amount.

But in b) it doesn't make sense to say that there's a 50% probabilty that the amount A is the larger amount. Either it is or it isn't. If it is, then 100% of the time that you repeat the process to the point in b) where you see amount A, A will still be the larger amount. If you repeat the process, every time you come to the point b) where you have amount A, you will certainly not be claiming you have a 50% chance of A being the larger amount. Of course the first time you do the process your intuition might tell you that you have a 50% chance for A to be the larger amount because you don't know. But faulty intuition does not equate to common sense. Common Sense would insist that you look at things more closely and apply some logic. Common sense would look at the math with a rigorous foundation.


PairTheBoard

[jason1990]

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To answer your specific question, X is a variable, but not a random one. It is the amount in one envelope. The amount in the other envelope is a random variable, with expectation 1.25*X.

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Then you are presenting a Bayesian argument. All of your comments about what a frequentist might say in this argument are out of place. A frequentist would regard X as a random variable.

Let's be clear. There are 3 related quantities here: the amount in the chosen envelope, X, the amount in the other envelope, Y, and the total amount T=X+Y. To the Bayesian, T is a random variable, with whatever distribution he decides to assign to it. To the frequentist, T is fixed. It is an unknown, deterministic quantity. As you put it, it is a variable, but not a random one. To the Bayesian, the random act is the stuffing of the envelopes. To the frequentist, the random act is the choosing. So when you apply the formal machinery of probability theory to this Paradox, you have a choice. You can build a probability space in which T is random -- this is the Bayesian perspective. Or you can build a probability space in which T is fixed -- this is the frequentist perspective. You say that in the argument you presented, X is non-random and Y is random. That means that T is random, so you are presenting an argument from a Bayesian perspective.

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You have inverted my meaning. I claim the envelope paradox is a true paradox, two strong arguments that lead to opposite conclusions. Yes, people invent rigorous ways to resolve it, but only by torturing commonsense ideas of uncertainty.

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It really seems like what you're calling common sense consists merely of some assumptions that we make by habit, which may or may not be justified, and the mistakes we make when we are careless about mathematics. If that's right, then it's probably good for such "common sense" to be tortured every once in a while so we learn to be more careful.

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The Bayesian says the odds are not 50/50,

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Yes, he says the conditional probability depends on X and may not be 50/50.

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the Frequentist says the expected value of a 50% chance of getting 0.5*X and a 50% chance of getting 2*X is not 1.25*X.

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No, to the frequentist, X is a random variable, not an ordinary variable as you have declared. So the frequentist would say that if you switch, then yes, there is a 50% chance you will end up with 2X. But that will be the 50% of the time when X was the smaller amount, so 2X is not as big as you think. And yes, there is a 50% chance you will end up with 0.5X. But that will be the 50% of the time when X was the larger amount. So 0.5X is larger than you think. According to the frequentist, you are essentially committing the fallacy of equivocation, since the X in 2X is something different from the X in 0.5X. If you (correctly) do the math in the frequentist's probability space, you will see that your expected return over those times when you decided to switch is the same as over those times when you didn't. Both are simply T/2.

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Both say you shouldn't always switch, which agrees with common sense, but only by invoking complex technicalities.

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The Bayesian doesn't say this. He simply says the conditional probability isn't necessarily 50/50. The frequentist says that switching will not change your EV, but he is not invoking any complex technicalities. He uses only basic probability and his assumption that T is fixed.

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In practical decision making, you'd never see these technicalities, so you'd fall into error.

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In practical decision making, we might make hasty assumptions without considering all the evidence, and we might perform hasty calculations without carefully considering what we're calculating. If that's the case, then we deserve to fall into error.

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I don't have a rigorous framework, if I did I'd publish and get the Nobel prize. My claim is no one can give clear, commonsense answers to both of the following questions without leading to absurd conclusions:

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Perhaps no one can give answers that are clear enough to satisfy you or that match up with your notion of common sense, but this only shows that the concepts in this Paradox are nontrivial and worth thinking about.

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(1) Is the probability that you have the higher amount 50%?

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Only if you're a Bayesian and only if that happens to be your degree of belief. If you've already chosen the envelope, then to the frequentist, the experiment is over. You have already selected your envelope. There is nothing random left to do, so there are no probabilities to speak of. Either you have the higher amount or you don't. It's an unknown proposition, but it's not random. Only a Bayesian would speak of the probability that the proposition is true, and he would speak of it in terms of his degree of belief.

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(2) If you have a 50% chance of losing some amount and a 50% chance of winning twice that amount, is it a positive expected value bet?

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If "amount" means two different things in the two different clauses of your sentence, then no, it's not necessarily a positive expected value bet. But to a frequentist, this question is irrelevant after you have selected the first envelope. Switching will result in a fixed, deterministic outcome. There is no 50% chance of anything when you switch. You switch and you get what's coming to you. Just because you don't know what's coming, doesn't mean you are free to assign a probability to it. (Unless, of course, you're a Bayesian.)

[ QUOTE ]
Imagine you have an envelope in your hand, with $10,000 in it. You wonder whether or not to switch. The Bayesian tells you to look inside yourself to figure out what probability you assigned to every possible total amount in the two envelopes before you opened yours. Doesn't that sound a little bit unhelpful?

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Actually, the Bayesian would probably tell you to look around you, not inside you. Where are you? Where did the envelopes come from? What do you know about the person who put the money in them? Is there any information at all that can help you decide whether it is more or less likely that there is $15,000 total rather than $30,000 total in these envelopes? Even according to the mathematics of conditional probability, these are the only two events whose probabilities matter. The Bayesian would not tell you to consider what was the prior likelihood of there being $100,000 in the envelopes. He knows that at this point, that is irrelevant.

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Your only question now is how likely is a $15,000 total versus a $30,000 total? That's what you should be weighing, using your current state of mind, not your prior one.

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Yes, if you're thinking like a Bayesian, this is the question you would consider. And yes, it is the only question you need to consider. As I stated above, this is in fact implied by the laws of conditional probability. As for states of mind, I think you're talking about degrees of belief. Obviously, you should use your current, updated (a.k.a. posterior) beliefs and not your prior ones. That's the whole point of the Bayesian method. You are spot on here in everything you said, in the sense that you are in complete agreement with the Bayesians. Now we just need to take one more step and realize that one of the points of the Paradox is, when you adopt this Bayesian approach, you should not blindly assign 50% as your degree of belief, regardless of the amount in the first envelope.

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The frequentist tells you that based on the design of the experiment, he believes that if it were repeated many times, you would end up with the higher amount 50% of the time,

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Yes. But to be more precise, he believes that before you select your envelope, there is a 50% chance you will select the higher amount.

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therefore, the probability you have the greater amount is 50%.

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No, no, no. The frequentist doesn't buy into all of this degree of belief business. You've already selected your envelope. To the frequentist, the randomness is over. Probability is no longer applicable. Either you have it or you don't. Just because the frequentist doesn't know the answer, doesn't mean he assigns a probability to it.

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You then say, "Great, you're telling me that the expected value of switching is $12,500." "No," he replies, "that computation is disallowed." Aren't you going to want to take a swing at him?

[/ QUOTE ]
Obviously, he doesn't say anything like that. He tells you that there is no expected value, because there's nothing random anymore. There is some fixed amount of money in that other envelope (i.e. it's a variable, but not a random one). He'll tell you that the amount is either $5,000 or $20,000, but he doesn't know which. If you ask him, what is the probability that the other envelope contains $20,000, he'll tell you that either it does or it doesn't. There is no in between and there is no probability.

And, frankly, yes, I would probably want to take a swing at him, because this is about the most unhelpful answer I can imagine. I would wish that he would get off his frequentist high-horse for once and act a little like a Bayesian. Maybe he could tell me if the guy who gave us the envelopes is wearing a fake Rolex. Or maybe he thinks the other envelope looks a bit thicker. Something, anything, would be more helpful than that.

Anyway, it seems like your point is that, when we apply probability theory to the situation in this paradox, we produce strong arguments that lead to different conclusions. This, therefore, casts doubt on the relevance of probability theory to practical decision making. But you have presented no evidence for this. You merely presented a loose and informal argument, which contains hidden assumptions and fallacies of equivocation. But regardless of what is wrong with this argument, it remains the case that the argument is not the result of applying the formal machinery of probability theory. To do that, you need a rigorous framework. The reason you don't have one with which to present these contradictions, is because it doesn't exist. When you rigorize your probability model, these issues evaporate and you can no longer get away with the kind of sleight of hand that's happening in the argument you presented.

But in the end, the "Paradox" isn't about real life or practical decisions. It's a math puzzle. And the challenge is to find the flaw. All you have done, essentially, is restated the paradox. Granted, you have made it somewhat more convincing with your skill in the use of language. But it still contains all of the same fallacies that were present in it originally. You then declare that it is a genuine paradox. In other words, you answer the challenge of finding the flaw by saying that the flaw is in mathematics itself. To me, that's the absurd conclusion.

[BBB]

[ QUOTE ]
My claim is no one can give clear, commonsense answers to both of the following questions without leading to absurd conclusions:

(1) Is the probability that you have the higher amount 50%?

(2) If you have a 50% chance of losing some amount and a 50% chance of winning twice that amount, is it a positive expected value bet?


[/ QUOTE ]

Well, the answer to (2) is yes, of course.

I think question (1) is what drives the paradox, and a key point to the entire world of probability. Without any extraneous information, it is meaningless, once we see $100 in an envelope, to assign a 50-50 chance, or a 2-1 chance, or any specific probability to the likelihood that N is $50 versus that that N is $200. So any argument which makes such a claim lacks foundation. If we have information (in this case, we can make judgements based on our life experience), then and only then can we assign probabilities. And the better the information we have is, the more likely it is that our probabilties will lead us to the correct envelope.

For example, suppose I tell you that I'm thinking of a ballplayer, and I don't tell you who, what team he plays for, or what type of baseball league he's in (or even if he's in a league), and I ask you to give the probability that he'll hit a grand slam tonight. If you knew absolutely nothing about baseball, and you didn't know what a grand slam was, you would presumably infer whatever you could, and give your best estimate. But to blindly guess 50% (or any other number) without thinking would be meaningless (and presumably someone who did know baseball would come up with a probability lower than 50%).

Basically, what I'm saying is that any probability is simply an estimate based on some nonzero amount of information. Any random number based on zero information, however, cannot be fairly called a probability.

So my answer to question (1) would be based on factors such as the amount in the envelope, what the envelopes looked like, how they were presented to me, and by whom. It would also be based on my life experience. But without the complete picture, all I can say is that if I am able to learn anything useful by seeing the amount in the envelope, then and only then could I give a meaningful answer to the question. Otherwise, I cannot tell you what the probability is that I have chosen the smaller amount, and consequently I cannot give you a meaningful answer as to what the EV of switching would be. (I can still tell you that applying a strategy of deciding to switch before choosing and opening an envelope would be a neutral EV strategy.)

So, regarding the original statement of the paradox, I would say that neither argument 1 or 2 is correct, and that the answer to the question, "After you view the envelope, is it +EV to switch?" is "it depends". (Argument 2 is correct when it states that always switching is a neutral EV strategy, but that does not necessarily mean that switching is neutral EV once you view the contents of either envelope. Argument 1 is simply bogus, since it arbitrarily assumes a 50-50 probability with no foundation.)

Finally, to relate to poker, clearly, to make better estimates of hand probabilities, it is best to gain as much information as possible during prior deals. Gaining as much information as possible, and then making the proper decisions about how to play a hand (i.e., those that maximize EV) based on this information, is perfect poker, IMO.

[jason1990]

It occurred to me that there is an interesting connection between this Paradox and confidence intervals.

Suppose I am about to play 10,000 hands of poker and that I know my standard deviation is 15 BB/100, so that my standard error over those 10,000 hands will be 1.5. Let m be my true winrate and M the winrate I will observe over those 10,000 hands. The number m is a fixed, but unknown quantity. The number M is a random variable.

Before I play, I can say that

P(M - 3 < m < M + 3) = 0.95.

This is a meaningful statement and it is true. Now, I go play and I observe that M=4. It is tempting to plug this into the above statement and conclude that

P(1 < m < 7) = 0.95.

But I can't do that. That's meaningless. The number m is fixed. Either 1<m<7 is true or it's not. There's no probability involved. This is a well known problem and it is why [1,7] is called a "confidence interval". We don't say that m is in [1,7] with 95% probability. We say that we are 95% confident that m is in [1,7]. This 95% is not a probability and we cannot work with it as though it were a probability. For example, it makes no sense to ask about the expected value of m.

On the other hand, if I were a Bayesian, then I could regard m as a random variable, assign it some probability distribution, and talk about actual probabilities involving m. I could talk about m's expected value, and about its conditional expectation given M.

Okay, now back to the Paradox. Let T be the total amount in the envelopes and X the amount in the one I choose. Suppose I am a frequentist, so that T is a fixed, but unknown quantity, and X is a random variable. Before I open that envelope, I can say that P(X=T/3)=0.5. This is a meaningful and true statement. Now I open the envelope and find that X=100. It is tempting to plug this in and conclude that P(T=300)=0.5. But I can't do that. That's meaningless. The number T is fixed. Either T=300 or it does not. There's no probability involved. I might say that I am 50% "confident" that T=300. Hence, I am 50% "confident" that the other envelope contains 200. Similarly, I am 50% "confident" the other envelope contains 50. But I can't work with "confidences" the way I can with probabilities. That's why we gave them a different name, so we wouldn't get confused and accidentally do things like compute expectation.

But if I were a Bayesian, then I could regard T as a random variable, assign it some probability distribution, and talk about actual probabilities involving T, and hence, involving the other envelope. Then, as a Bayesian, I could actually compute the conditional expectation of that other envelope, given X=100.

[Tater10]

If i offer 2 envelopes with different amounts in each, I would expect to that person to walk away with the average of the two amounts in the long run.

Why is 2.0 and 0.5 so special? Any 2 amounts using (x + 1/x)/2 will have a value greater than 1, this doesnt mean to switch.

I offer 2 envelopes to someone and she opens it and there is $24. I open the other and she sees that there is $3 in it. She then says "Oh, I should have switched if I had known that I had a 50/50 chance of 8x my money or 1/8x my money because I should walk away with (8 + .125)/2 or 4.0625x."