The envelope problem, and a possible solution

[AaronBrown]

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I am stubbornly sticking to my prior my belief about you, Aaron, which is that you are an educated and intelligent man who can communicate effectively via the written word. That's why I've got to believe that there is some gross misunderstanding going on here.

In both the Paradox and in PTB's Prop Bets, the only techniques we need to use in order to arrive at the "sensible" conclusions are the techniques of elementary probability, such as assumptions [1]-[4] of my previous post. These are consistent, they are certainly not ad hoc, and they are extremely helpful in a wide array of realistic problems. Are you denying this? What am I missing here?

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Thank you for your kind words, although you could have left out "stubbornly." At least you didn't say "against all rational evidence."

I think you keep repeating one side of the paradox, and wondering why I can't see something so simple. I do see it. But there's the other side as well, the one that says the probability that you have the smaller amount is 50% so the expected value of switching is positive. You can't refute a paradox by strengthening one side, that just makes the paradox more puzzling. You have to show why one side is wrong.

I maintain the ways people deny the force of the always switch argument result in restrictions that are commonly ignored, and if not ignored would make statisticians useless.

[jason1990]

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Thank you for your kind words, although you could have left out "stubbornly." At least you didn't say "against all rational evidence."

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[img]/images/graemlins/smile.gif[/img] Sorry, I think I'm a little frustrated over this apparent miscommunication.

Okay, here's the assumptions:
<ul type="square">
[1] if A and B are events, then P(A | B) = P(A and B)/P(B),
[2] if A and B are independent events, then P(A | B) = P(A),
[3] if A and B are mutually exclusive events, then P(A or B) = P(A) + P(B),
[4] if A is an event, then 0 &lt;= P(A) &lt;= 1.[/list]Here's the model we agreed on:

T = total amount in the envelopes
H = 1, if you picked the larger; 0, otherwise
X = (1 + H)T/3 = amount in chosen envelope
Y = (2 - H)T/3 = amount in other envelope

We never said it, but we both should agree that in this setup, T and H should be defined so that they are independent, and H should be defined so that P(H=0)=0.5. (And if someone wants to call T a parameter, that's fine too, just give it a point-mass distribution.) Let us also assume, for simplicity of exposition, that the distribution of T is supported on a countable set, i.e. that T is a discrete random variable. We will take it for granted that we can construct the underlying probability space without any difficulty. The Claim in question is

P(Y = 2X | X = k) = 0.5 for all k.

In this setup, Y=2X iff H=0. So the Claim is simply

(*) P(H = 0 | X = k) = 0.5 for all k.

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I think you keep repeating one side of the paradox, and wondering why I can't see something so simple. I do see it. But there's the other side as well, the one that says the probability that you have the smaller amount is 50% so the expected value of switching is positive. You can't refute a paradox by strengthening one side, that just makes the paradox more puzzling. You have to show why one side is wrong.

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Okay, one side says that Claim (*) is true. The argument which is typically given is some variant of the following: the event {X=k} gives no meaningful information regarding the event {H=0}. These events are therefore independent (for any k). Hence, by Assumption [2],

P(H = 0 | X = k) = P(H = 0) = 0.5.

The problem with this is that the above argument uses a heuristic, not a mathematical, definition of independence. There are many examples of events whose mathematical dependence or independence does not match up with our heuristic notion of this concept. Moreover, not everyone's heuristic understanding of independence would allow them to conclude this.

In order for the argument on this side to carry any weight, we would need to mathematically verify that these events are independent (for all k). Thus, we need to verify that

(**) P(H = 0 and X = k) = P(H = 0)P(X = k) = 0.5P(X = k)

for all k. But according to Assumption [1], this is exactly equivalent to Claim (*). So this side of the argument does nothing but try to prove Claim (*) by applying heuristics to an equivalent formulation of it.

Now, one might argue that these heuristics "ought" to be true. But our mathematical definition of independence stands firm: A and B are independent if P(A and B)=P(A)P(B). This definition is intimately tied to Assumptions [1] and [2]. If you want to "force" these heuristics to be valid, then you will at some point need to deny one of these Assumptions.

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I accept all four of your assumptions, and don't know anyone who does not. But how do they lead to the conclusion?

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Okay, so now we have the other side of the argument in which we try to show Claim (*) is false. We first compute

P(H = 0 and X = k) = P(H = 0 and (1 + H)T/3 = k)
= P(H = 0 and T = 3k)
= P(H = 0)P(T = 3k)
= 0.5P(T = 3k).

Similarly,

P(H = 1 and X = k) = P(H = 1 and (1 + H)T/3 = k)
= P(H = 1 and T = 3k/2)
= P(H = 1)P(T = 3k/2)
= 0.5P(T = 3k/2).

Hence, by Assumption [3],

P(X = k) = P(H = 0 and X = k) + P(H = 1 and X = k)
= 0.5P(T = 3k) + 0.5P(T = 3k/2).

Now let us assume Claim (*) is true. This is equivalent to Claim (**), so (after a little algebra) the above computations imply

P(T = 3k) = P(T = 3k/2) for all k.

Choose some a&gt;0 such that P(T=a)&gt;0. Let d=P(T=a). If we let k=2a/3, then the above implies that P(T=2a)=P(T=a)=d. By induction, P(T=(2^n)*a)=d for all n. By Assumption [3], for all positive integers M,

P(T = a or T = 2a or ... or T = (2^{M-1})*a) = Md.

If we choose M such that Md&gt;1, then this contradicts Assumption [4], so Claim (*) cannot be true. Or, equivalently, it is not the case that {H=0} and {X=k} are independent for all k.

Again, if you want to deny Assumptions [1] and/or [2] and redefine independence to more closely match the heuristic ideas in the Paradox, then fine. Another possibility is to deny Assumption [4] and try to permit the use of "unnormalizable priors" (e.g. a uniform distribution on the naturals). I really don't know if any attempts have been made to develop such theories. I certainly have never heard of any. But I would consider such theories ad hoc, and I definitely think that any theory which does not assume [1]-[4] is well outside the realm of what is commomly used by practicing statisticians.

[PairTheBoard]

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So there's no mystery with the proposition bet? It doesn't produce a true paradox with equally strong arguments on both sides for why it might be a good bet? All it takes is simple logic to see why it's not a good bet? Yet the Envelope Paradox is essentially the same situation. Only in the Envelope "Paradox" the Proposition asks if you want to bet half the amount in your envelope giving you 2-1 odds.

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I'm not sure why you keep belaboring this. I know the envelope switch is not a good bet, that's not the point. I don't believe that a friend and I can make each other infinitely wealthy by passing two envelopes back and forth all day (I tried, it doesn't work). You can, however, have two people get positive EV from opposite sides of a zero sum bet, if you tradie options with people who keep score in different currencies. But that's not the aspect of the paradox we've been discussing.

I think the hard part is to come up with a rigorous theory that spells out exactly when switching makes sense, that is also flexible enough to use for practical decision problems. You don't think that's hard, because you will accept things that I feel cannot be true and that are devilishly hard to apply. You're not alone in that, many people agree with you (although few are quite as broadminded as you, most limit themselves to one hard-to-swallow pill).

None of this makes me think you don't know basic logic or math, I don't know why you think that of me. If I were to get ad hominem, I'd bet you haven't done a lot of real world statistical analysis. Not because I think you've shown yourself incapable of it, but because the day-to-day experience of trying to apply your principles leads many practitioners to become agnostic like me.

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I was not trying to ridicule you by asking you to address the 2E Proposition Bet Paradox. I think the 2E Proposition Bet Paradox is a better version of the 2E Paradox because it strips away a couple of irrelevant factors that make for psychological muddy waters. In the 2E paradox the fact that you are being given the amount in the envelope tends to hide the fact that you are simply making a bet when you switch. Also, in the 2E paradox, since switching is EV neutral, you tend to feel less psychologically compelled to make a decision about what is best.

The 2E Proposition Bet Paradox is right there in front of you. Even a frequentist will take the first bet.

1) Given 3-2 odds bet $10 your envelope contains the smaller amount.

This is clearly a +EV bet. Even after opening the envelope, assuming you don't know the envelope amounts. Although the frequentist will say the outcome has been determined he will still put his money on what amounts to a 50-50 proposition with 3-2 odds in his favor. He will do this in the same way he will bet that a randomly dealt card having been dealt face down is the Ace of Spades if you give him 80-1 odds.

But as soon as you change the bet to 2)

2) Given 3-2 odds bet the amount in the envelope that it is the smaller amount.

any gambler worth his salt knows the bet is no good. Yet the same Paradox arises. You are being given 3-2 odds on what you just said was a 50-50 bet. That's got to be +EV.

When confronted with the 2E Proposition Bet Paradox you respond,

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But it takes only logic, not probability theory, to see that the bet in (2) can be converted to an obviously poor bet of winning half the total amounts in the two envelopes if you have the smaller amount and paying two thirds of the total amount if you have the larger amount. That's not a mystery to me or anyone who thinks through it clearly.


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But with the original - But equivalent - Two Envelope Paradox you insist

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But there's the other side as well, the one that says the probability that you have the smaller amount is 50% so the expected value of switching is positive. You can't refute a paradox by strengthening one side, that just makes the paradox more puzzling. You have to show why one side is wrong.


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It's easy to see why that side is wrong in the 2E Proposition Bet Paradox. It's because you're being asked to bet the amount in the envelope. The amount you bet also determines whether you win or lose. Bet a little and win. Bet a lot and lose. Not +EV even with 3-2 odds.

The 2E Paradox is equivalent as a Paradox. In the 2E Paradox you are being asked to bet half the amount in the envelope and are given 2-1 odds. The amount you bet determines whether you win or lose. Bet a little and win. Bet a lot and lose.

In my opinion this Does give the reason "why one side is wrong." And I think it makes common sense.


PairTheBoard

[jason1990]

Hi PTB, I have a new Prop Bet for you. (And, of course, for anyone else who would like to respond.)

I will bet the amount in my Envelope that it is the smaller amount. If I lose, I will pay the amount in my envelope. But if I win, I want you to pay me (3T+6)/(2T+3) times that amount, where T is the total amount in both envelopes. (Note that this factor is always strictly less than 2. And, of course, you are free to choose T.) Moreover, I want to be able to look at the contents of my envelope before agreeing whether or not to wager. That is, after I look at what's inside my envelope, I am free to call off the bet and we both go home with no money changing hands.

Who does this bet favor? Or is that somehow impossible to objectively determine from the given information?

[AaronBrown]

Now I'm going to reverse myself and recombine the threads.

I think both Jason1990 and PairThe Board have redefined the original paradox in opposite ways.

Jason1990 does a thorough job of showing that no consistent prior distribution on T justifies the idea that the probability of switching is always 50%. However, there are consistent prior distributions that justify always switching, but they have infinite expectation for T. That's certainly a strong argument against switching, if it is made before T is selected.

But after T and H have been selected, and we look at X, all of that seems like abstract machinery, irrelevant to the problem at hand. I'm looking at $100, considerations of infinite expected value for T seem silly. I have no idea how T was selected, arguing from the assumption I know its probability distribution recalls the "assume a can opener" joke; the Bayesians get around that by asserting that I must have some subjective belief on the subject, but that has its own problems.

I'm not saying Jason1990 is wrong here, just that he drags into the problem a lot of formalism that is hard to apply to real decisions. If we ask him whether we should switch or not, instead of whether it's rational to always switch, none of the stuff he's added to the problem is useful. This is what I mean by hair-splitting. It offends my common sense that a simple problem like switch or don't switch requires such abstraction to avoid inconsistency. My intuition says that there's something Jason1990 (and I) don't know about the nature of probability, and that's why he wants to fence it in with abstractions and I maintain a mystical faith in common sense.

PairTheBoard, on the other hand, wants to throw out all the prior stuff and just have two envelopes, and he wants me to make a choice without seeing the amount in mine. The envelope paradox is that before I see the amount in my envelope switching is obviously EV neutral, but once I know the amount (that is, once I make PairTheBoard's proposition bet impossible) switching seems to be EV positive.

I think you're both dodging the boundary issue. If you start before T and H are determined, Jason1990's argument seems unassailable. If you ignore how T is selected, determine H but don't tell me what it is, PairTheBoard has the answer. But when precisely does the probability change? Is it when T is determined? When H is determined? When I know X? And what precisely is the nature of the thing "probability" that changes?

[PairTheBoard]

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Hi PTB, I have a new Prop Bet for you. (And, of course, for anyone else who would like to respond.)

I will bet the amount in my Envelope that it is the smaller amount. If I lose, I will pay the amount in my envelope. But if I win, I want you to pay me (3T+6)/(2T+3) times that amount, where T is the total amount in both envelopes. (Note that this factor is always strictly less than 2. And, of course, you are free to choose T.) Moreover, I want to be able to look at the contents of my envelope before agreeing whether or not to wager. That is, after I look at what's inside my envelope, I am free to call off the bet and we both go home with no money changing hands.

Who does this bet favor? Or is that somehow impossible to objectively determine from the given information?

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This is another reason I think the Proposition Bet Paradox is a better version of the Two Envelope Paradox for focusing in on what Aaron (and I) view as the basic conundrum - How can what looks like a 50-50 bet, given better than even money odds, not be +EV?

I really played a hunch that 3-2 odds would be slim enough to defeat any system like SamIAm and others might propose when a prior distribution with finite expected value is assumed for T=total in envelopes. Thereby stripping that issue from the central conundrum mentioned above.

Assuming 3-2 can't be beat and knowing 2-1 can be beat the question naturally arises as to what happens in between. Your function adresses exactly that question, giving close to 2-1 odds for small T and close to 3-2 odds for large T. I assume you've worked it out and can show a system for declining the bet that beats those varying odds. However, can you prove a more general Theorum about it that answers all such questions?

Also, can you show that fixed 3-2 odds can't be beat? What's the cuttoff? Any fixed odds less than 2-1? That would be nice to know because I would like to offer the Proposition Bet to Aaron allowing him to look in the envelope and giving him the option to decline the bet.


PairTheBoard

[PairTheBoard]

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PairTheBoard, on the other hand, wants to throw out all the prior stuff and just have two envelopes, and he wants me to make a choice without seeing the amount in mine. The envelope paradox is that before I see the amount in my envelope switching is obviously EV neutral, but once I know the amount (that is, once I make PairTheBoard's proposition bet impossible) switching seems to be EV positive.


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NO! The Proposition Bet is a one time offer. So looking at the amount in the envelope does not make the bet impossible. This is another reason why the Proposition Bet Paradox is better for focusing on the conundrum - why is this apparent 50-50 bet with better than even money odds not +EV? On a one time basis you can look at your envelope amount and decline the bet if you like. As the maker of the proposition I'm betting that people I offer it too will not improve on a coin flip type decision enough to beat the 3-2 odds I'm offering. You can go into gymnastics about how I might be choosing the amounts and how seeing the envelope amount might help you improve your decision if you want. But in doing so you will be avoiding the central conundrum. After looking at the envelope amount you are happy to bet $10. You see it as a 50-50 proposition getting 3-2 odds. That's +EV. Why aren't you just as happy betting the Envelope amount? Isn't it still 50-50? Aren't you still getting 3-2 odds? Why isn't it still +EV?

The answer is simple. The amount of your bet now determines whether you win or lose - even though you don't know which. That messes with your EV.



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If you ignore how T is selected, determine H but don't tell me what it is, PairTheBoard has the answer. But when precisely does the probability change? ... When H is determined? When I know X? And what precisely is the nature of the thing "probability" that changes?

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I think we are finally getting to your point. What are your best answers to the questions you pose above?


PairTheBoard

[AaronBrown]

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Who does this bet favor? Or is that somehow impossible to objectively determine from the given information?

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I think it's a game, and the better player will win.

If you know the distribution I use to pick T, it's obvious that you cannot lose, because you look at the amount, know the probabilities and take only positive expected value opportunities. I cannot lose, because I can pick T always to be zero (I'm pretty sure any non-trivial T has to have an infinite expected value to avoid giving you a nonzero edge).

However, if you don't know how I pick T then it comes down to a guessing game. For example, suppose I pick T = $3 with probability 1/3, $6 with 2/9, $12 with 4/27; doubling the amount and multiplying the probability by 2/3 each time up to some limit. This gives you a small positive edge, $0.10 (a bit less for some of the smaller T's), to take the bet for every amount you see, up the the maximum possible amount (2/3 of my maximum T). But your expected loss if you take the bet for that amount dwarfs your dime.

Say, for example, that I continue the distribution but won't make T more than $1 million (the extra probability gets added to the maximum T, $786,432). 1 time in 6 you'll see $1 and have a $1.66 EV from taking the bet. 0.833 of the time, you'll see an amount between $2 and $262,144 and have a $0.10 positive expectation from taking the bet. But 1 time out of 2,596 you'll see $524,288 and have a negative $524,288 expectation from taking the bet. The unconditional expected value of that occurrance, assuming you take every bet, is negative $177.

Of course, I can try to be trickier than this. I can always set my distribution to guarantee you never have a big expected EV from any bet, and stick in a few big negative EV bets anywhere I think you're likely to fall for them.

All it all, it sound as if we're getting back to poker.

[AaronBrown]

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If you ignore how T is selected, determine H but don't tell me what it is, PairTheBoard has the answer. But when precisely does the probability change? ... When H is determined? When I know X? And what precisely is the nature of the thing "probability" that changes?

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I think we are finally getting to your point. What are your best answers to the questions you pose above?

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I don't have good answers, and I don't like the answers other people accept as good. But I believe there are good answers and when someone discovers them we will not only be able to do statistics with more rigor, we will improve the practice of statistics and risk-taking.

[PairTheBoard]

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If you ignore how T is selected, determine H but don't tell me what it is, PairTheBoard has the answer. But when precisely does the probability change? ... When H is determined? When I know X? And what precisely is the nature of the thing "probability" that changes?

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I think we are finally getting to your point. What are your best answers to the questions you pose above?

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I don't have good answers, and I don't like the answers other people accept as good. But I believe there are good answers and when someone discovers them we will not only be able to do statistics with more rigor, we will improve the practice of statistics and risk-taking.

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Fair enough. I would say the lesson of the Two Envelope Paradox is this:

Beware of Proposition Bets where the Outcome Determines the Size of your Bet.


PairTheBoard