The envelope problem, and a possible solution

[PairTheBoard]

Look at this from a Gambler's point of view. You are told about the two envelopes and asked to choose one. You are offered these Proposition Bets.

1. Given 3-2 odds, would you be willing to bet $10 that your envelope contains the smaller amount?

2. Given 3-2 odds, would you be willing to bet the amount in your Envelope that it is the smaller amount?


1 is a good bet for you while 2 is not. Do you see why?


PairTheBoard

[jason1990]

[ QUOTE ]
Look at this from a Gambler's point of view. You are told about the two envelopes and asked to choose one. You are offered these Proposition Bets.

1. Given 3-2 odds, would you be willing to bet $10 that your envelope contains the smaller amount?

2. Given 3-2 odds, would you be willing to bet the amount in your Envelope that it is the smaller amount?


1 is a good bet for you while 2 is not. Do you see why?


PairTheBoard

[/ QUOTE ]

How about...

3. You open the envelope. It contains $10. Given 3-2 odds, would you be willing to bet this $10 that your envelope contains the smaller amount?

[PairTheBoard]

[ QUOTE ]
[ QUOTE ]
Look at this from a Gambler's point of view. You are told about the two envelopes and asked to choose one. You are offered these Proposition Bets.

1. Given 3-2 odds, would you be willing to bet $10 that your envelope contains the smaller amount?

2. Given 3-2 odds, would you be willing to bet the amount in your Envelope that it is the smaller amount?


1 is a good bet for you while 2 is not. Do you see why?


PairTheBoard

[/ QUOTE ]

How about...

3. You open the envelope. It contains $10. Given 3-2 odds, would you be willing to bet this $10 that your envelope contains the smaller amount?

[/ QUOTE ]

Sure, as long as I would have also been allowed to bet $10 had I chosen the other envelope.


PairTheBoard

[AaronBrown]

[ QUOTE ]
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?


[/ QUOTE ]
So many different points are flying around that I think it's clearer if I try to answer one per post.

This is a very good statement of the frequentist position. You're not allowed to compute the expected value and have a variable in the result. So you can't say having 0.5 chance of getting $X/2 and 0.5 chance of getting 2*$X has an expected value of 1.25*$X.

The trouble is for any problem outside a textbook, you have to say this. Even inside a textbook you'll find loads of formulae showing expected value and other parameters computed in terms of variables. A statistical test shows you how to plug your measurements into a formula to get a significance level, that formula is not allowed.

It leads to impossible hair-splitting, which is why Bayesians reject the rule. For a simple example, three baseball fans are arguing over who has the best true underlying batting average in the majors. One says Twin Joe Mauer (102 hits, 260 at bats, 299 plate appearances for a 0.392 average); one says Marlin Miguel Cabrera (94 hits, 274 at bats, 337 plate appearances for a 0.343 average); one says Red Sox (Sock?) Mark Loretta (100 hits, 312 at bats, 339 plate appearances, for a 0.321 average).

Mauer's fan says the other two are clearly wrong, that the probability of two hitters with the same true underlying batting average having this much difference over this many at-bats is less than 5%. This has nothing to do with baseball, it's a pure mathematical question. All agree to accept standard assumptions, that for one player each trial has a fixed probability of success, and all trials for all players are independent. All agree that the data above are correct.

A frequentist statistician from Florida is hired to settle the question. He starts by computing the single batting average that is most likely to produce both Mauer's and Cabrera's results: 0.367. He computes that the probability a true 0.367 hitter will get 102 or more hits out of 260 at bats is 0.181, and the probability of 94 or fewer hits out of 274 at bats is 0.224. Two times the product of these probabilities (since either one could have had the higher actual average) is 0.081. This is more than 0.05 so he cannot reject at the 5% level that Mauer and Cabrera have the same true underlying batting average. The same calculation for Loretta shows a 0.021 significance, so we can reject that Loretta has the same true batting average as Mauer at the 5% level. This is more accurate than the standard t-test, because it uses the exact binomial distribution, but the results are similar, the t-test give 0.119 significance for Cabrera and 0.037 for Loretta.

The Boston fan doesn't like the result so he hires a frequentist statistician from Harvard. "Aha," he says in a thick Harvard accent, "that Florida idiot was computing in terms of a variable. Some plate appearances do not count as at bats (walks, errors, hit by pitch and sacrifices) so the number of at bats for a player was not fixed. We have to do a trinomial computation, assuming the most likely values for joint probability of hit per plate appearance and most likely individual values for probability of no at bat per plate appearance. Now the significance level for Cabrera is 0.026, so we can reject him as Mauer's equal, but Loretta gets 0.070, so his batting cannot be distinguished at the 5% level from Mauer's."

The problem is the frequentist approach is only rigorous before the experiment is performed, and only if every eventuality is defined (including the significance level; if you wait until after the experiment is performed to select your level, you have erred). Once you have real data, you have to invent a hypothetical repeatable experiment. The Florida statistician assumes each batter got the actual number of at-bats, the Harvard guy assumes each batter got the actual number of plate appearances. We could equally well make assumptions like each batter hit until he got the actual number of hits, or each batter played in the actual number of games and plate appearances were selected from some distribution, or lots of other things. Each assumption gives a different answer.

Some people devote a lot of energy to defining rules so that there is one "natural" answer to common questions. But most people have given up. The outcome of a frequentist significance test on existing data is a matter of opinion.

Bayesians throw all this away and use only confidence intervals, not significance tests. So they have no hair-splitting or inconsistencies. Unfortunately, the probability is still a matter of opinion. The only difference is the two statisticians don't have to criticize each other's work, they just have opposite conclusions because they have different priors.

[AaronBrown]

[ QUOTE ]
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.

[/ QUOTE ]
Here you start with an excellent description of the Bayesian position. The probability changes when you open the envelope. But that is not the common sense idea of probability. Looking at something shouldn't change the probability that it's bigger than something else.

Of course, we all know looking at something changes your subjective belief, but subjective belief does not follow the formal mathematical laws laid down by Bayesians to make everything consistent. A Bayesian takes prior belief as something unexplainable and totally reliable (and known to be consistent), then computes what posterior belief should be, even though it isn't. If you trust subjective belief, just trust the posterior belief, and forget the calculation. If you don't trust subjective belief, don't use it for a prior.

At the end you make the Bayesian reply to this (after an excursion in Frequentistland): mathematical probability doesn't compute anything objective, it helps us train our minds to form more consistent posterior beliefs. I happen to agree with that, but it is far from the common sense idea of a statistician who can weigh evidence in mathematically rigorous ways. I don't see people in this thread saying "long experience with rigorous statistical reasoning has trained my mind to realize always switching is silly, it cannot increase expected value for both people at once." I read "mathematics dictates my answer, anyone who disagrees doesn't understand the math."

For the record, I consider myself an empirical Bayesian, but I'm not dogmatic. That means I use Bayesian methods, but I use empirical evidence and assumptions to form priors rather than my personal subjective belief. I know it's not consistent, but it leads to reliable answers most of the time without Frequentist hair-splitting or Bayesian rejection of useful tools. Empirical Bayesians like resampling and similar techniques. But we don't have a good answer for the envelope paradox.

In the middle you veer to the Frequentist position. the probability is 50% before you pick. Once you pick an envelope the probability goes to zero or one and doesn't change. The only thing that changes when you look at the envelope is your state of knowledge about the probability. That's common sense, that's why Frequentists don't accept the Bayesian approach. The randomness is in the deal, once the cards are dealt the only thing that changes is our knowledge of what happened. Frequentists claim there exists a measureable, objective probability, the same for everyone, but no one knows what it is. Bayesians claim there is only subjective probability, and it's different for everyone, and you can only know your own personal value.

But now you want to repeat the experiment. You dance around it a little bit, but you understand the problem. If you use the same amounts A and B in the envelopes every time, the subject will quickly figure out which is the higher amount. You could use three amounts, like $1, $2 and $4; but this just slows things down a bit. Whatever distribution of amounts you choose to define your hypothetical repeated experiment takes on the same mathematical role as the prior distribution of the Bayesian.

Now we have a different interpretation on the result that always switching implies a prior with infinite expected value (like 1/3 chance of $3 total between the envelopes, 2/9 chance of $6, 4/27 chance of $12 and so on, multiplying the probability by 2/3 each time and the amount by 2; now if I get $1 I know switching gets me $2, anything else means I have a 0.6 chance of $X/2 and a 0.4 chance of 2*$X for an expected value of 1.1*$X).

To a Bayesian it's simple, if you expect infinity, any finite amount is a disappointment and you should switch, but unless we're talking about bets with God or the Devil, priors with infinite expectation are hard to rationalize.

Frequentists are perfectly comfortable imagining infinite hypothetical experiments, in fact their theory depends on them. So they cannot rule out the idea of doing the experiment this way, so that always switching increases expected value. They can't do the experiment with a finite mean distribution, in fact, because the subject would eventually figure it out (not necessarily the entire distribution, if it were continuous, but enough to know when to switch). So they just outlaw the question.

[AaronBrown]

[ QUOTE ]
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.

[/ QUOTE ]
I disagree (what a surprise).

Let me simplify your three variables a bit. There is T, the total amount in the two envelopes and H which is 1 if you have the higher amount (2T/3) and 0 if you have the lower (T/3). These completely define the situation in fewer bits. My X is simply (H+1)*T/3.

Now we have to be precise about timing. Before T is selected, it is a random variable to both schools of thought. Once it is selected, but not revealed to you, a Frequentist regards it as a fixed unknown quantity, to a Bayesian it is still subjectively random to you, and you have a prior distribution about its value. Both schools agree H is still random.

When you pick the envelope, a Frequentist says H goes from random to fixed but unknown. Nothing happens for the Bayesian, there is no information since the envelopes are identical from the outside, hence no change in subjective belief.

When you open the envelope to see the amount, nothing happens for the Frequentist (he admits your subjective beliefs might change, but that has nothing to do with objective probability). For the Bayesian, H and T remain random variables, but their probability distributions change.

Now it gets murky for Frequentists. Once you pick the envelope, H and T are fixed, and therefore X is as well. So a Frequentist will disallow any probability statements about it, including computing expected values in terms of it.

But statisticians are supposed to give us decision rules, not tell us everything is fixed but unknown so any decision is either right or wrong. A practical decision rule requires us to let X vary. "Switch for X< $100 but not for larger amounts," explicitly treats X as a variable (or parameter if you prefer, the point is that it's not fixed).

The Frequentist does this by inventing hypothetical infinite repetitions of the experiment, and mentally computing the rule before T and H are determined. Then X is a variable and we can compute the expected value of different switching strategies. But there are many ways to convert a decision problem to realizations of hypothetical infinite experiments, and they lead to different answers.

[PairTheBoard]

[ QUOTE ]
Here you start with an excellent description of the Bayesian position. The probability changes when you open the envelope. But that is not the common sense idea of probability. Looking at something shouldn't change the probability that it's bigger than something else.


[/ QUOTE ]

You asked for something that makes common sense. How much common sense does your statement make? Suppose you have two envelopes. One contains an amount bigger than $100 and one contains an amount smaller than $100. You shuffle and chose. Before looking inside it makes common sense to say there is a 50-50 chance the amount inside is bigger than $100. After you open and look inside it certainly makes common sense to say that probabilty has changed. Yet you would say,

[ QUOTE ]
The probability changes when you open the envelope. But that is not the common sense idea of probability. Looking at something shouldn't change the probability that it's bigger than something else.


[/ QUOTE ]

I don't think that makes common sense at all.


Actually, as Jason pointed out, to be technically correct I should have stated a) this way,

a) Envelopes have been sealed and shuffled. You are about to pick one at random. You will take what is in the envelope you decide to pick.

At that point the probability is 50% that you will chose the envelope with the larger amount. However, after you chose it you have determined the outcome whether you look inside or not and the probabilty is technically no longer 50-50. I was being a bit loose about where the outcome is clearly determined. Simliliar to being dealt a random card. Before the deal the chance is 1/52 you'll be dealt the Ace of Spades. Once the card has been dealt face down the outcome has been determined. Either it is or isn't the Ace of Spades although speaking loosely there's usually no harm in saying the probabilty is 1/52 that you've been dealt the Ace of Spades. However, once you look at the card the probabilty is clearly no longer 1/52. Either it is or isn't the Ace of Spades.

Certainly once you open the envelope the outcome is determined. If the envelope amounts are N/2N and you open to see N then the question whether your envelope has the larger amount has definitely been determined. You may not know the determination yet but it's been determined nevertheless.

PairTheBoard

[PairTheBoard]

[ QUOTE ]
PTB -
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?


[/ QUOTE ]

[ QUOTE ]
Aaron -
This is a very good statement of the frequentist position. You're not allowed to compute the expected value and have a variable in the result. So you can't say having 0.5 chance of getting $X/2 and 0.5 chance of getting 2*$X has an expected value of 1.25*$X.


[/ QUOTE ]

You asked for a common sense answer to your question. I think my answer made very good common sense. I also think it makes very good common sense when doing a calculation involving a term $X that $X should stand for the same thing throughout the calculation, whether you call it a "variable" or anything else.

With envelopes containing N and 2N, "Switching" amounts to betting half your envelope amount at 2-1 odds that it is the smaller amount. You say this is a true paradox and both arguments are equally strong. Do you thing the strength of argument for the "Switchers" would help them win money taking Part 2 in the following Proposition Bet?

[ QUOTE ]
PTB -
Look at this from a Gambler's point of view. You are told about the two envelopes and asked to choose one. You are offered these Proposition Bets.

1. Given 3-2 odds, would you be willing to bet $10 that your envelope contains the smaller amount?

2. Given 3-2 odds, would you be willing to bet the amount in your Envelope that it is the smaller amount?


1 is a good bet for you while 2 is not. Do you see why?


[/ QUOTE ]


PairTheBoard

[jason1990]

Your baseball example brings up very important issues. If "true underlying batting averages" really exist, there is no way for us to determine, with certainty, what they are. The question of how to "best" decide which true value is the highest is a philosophical question (that uses mathematics), but not an intrinsically mathematical question. The pros, cons, consistencies, and inconsistencies of the various approaches to this question are very important and worth discussing and questioning.

But the Paradox asks a mathematical question: Starting with a mathematical model, and applying mathematical steps, we arrive at contradictory mathematical statements. What is wrong with the math?

I can see that you want to use the Paradox as a vehicle for discussing the (important) philosophical issues at the heart of statistical inference. In my opinion, this is an excellent use of the Paradox. But rather than coming right out and saying that is what you are doing, you phrase things in a way that can lead the less informed to believe that the branch of mathematics which we call probability theory is logically inconsistent.

[ QUOTE ]
There is T, the total amount in the two envelopes and H which is 1 if you have the higher amount (2T/3) and 0 if you have the lower (T/3). These completely define the situation in fewer bits. My X is simply (H+1)*T/3.

[/ QUOTE ]
Okay, great. (Now we just need a probability space on which these objects are defined.) Let me add one more thing. The amount in the other envelope is

Y = T - X = (2 - H)T/3.

The Paradox presents a seemingly valid mathematical argument which shows that

(*) E[Y | X] = 1.25X.

But this statement is false, so what is wrong with the math?

[ QUOTE ]
Now we have to be precise about timing. Before T is selected, it is a random variable to both schools of thought.

[/ QUOTE ]
This is not necessarily the case. If your probability space models T as a random variable (at any point in "time"), then it must model the distribution of T. It's okay if it's a distribution involving an unknown parameter or just a completely unknown distribution. But there must be a distribution. There is nothing in the statement of the Paradox that says anything about this distribution, so if it is part of your probability space, then it is based on your own assumptions.

The probability space need not impose any assumptions at all. It is not necessary for the probability space to model what happens before T is selected. The quantity T can exist in the model simply as a parameter.

[ QUOTE ]
Once it is selected, but not revealed to you, a Frequentist regards it as a fixed unknown quantity,

[/ QUOTE ]
As long as we're trying to formalize things, please allow me to clarify your statement. Random variables do not become parameters. Once a random variable, always a random variable. Of course, you can condition on a random variable. For example, if, in your model, T is a random variable, then you can compute P(D|T) for any event D, and you can compute E[Z|T] for any random variable Z. When you do this, you will find that there are functions g and h (that depend on D and Z) such that

P(D | T) = g(T) and E[Z | T] = h(T).

In this sense, you may regard T as a "parameter" in the functions g and h. But it is really still a random variable. In particular, the above equalities are equalities between random variables and represent almost sure equality, i.e.

P(E[Z | T] = h(T)) = 1.

When we speak informally, we say things like, "before you select the amounts, T is random, but after you select it, it is fixed." But it's important for people to realize that this is just informal language. It has no mathematical translation other than the formulas I gave above for conditional probability and conditional expectation. "Time" does not appear in the mathematical models we are talking about.

But the question still remains: why does the argument presented in the Paradox not prove (*)? What are the mathematical principles that it violates? Once this is established, then we can discuss how these principles contradict our intuition and what implications this has on the philosophical foundations of the various methods in statistical inference. But your discussion of the Paradox can lead people to believe that:
<ul type="square">[*]The branch of mathematics called probability theory comes in two varieties, one for Bayesians and one for frequentists.[*]These varieties employ different mathematical principles, which are not consistent with one another, and can therefore arrive at contradictory mathematical conclusions.[/list]What I'm trying to emphasize is that these statements are false. There is only one branch of mathematics called probability theory, and it is used by both Bayesians and frequentists. The "proof" in the Paradox violates the mathematical principles of probability and is a mathematical fallacy, no matter what your philosophical inclination. It is not a genuine mathematical paradox.

Philosophical discussions beyond this are very valuable and, in my opinion, are a very big part of what this Paradox is about. But no one should be so confused as to think that the above two bulleted items are true.

[AaronBrown]

[ QUOTE ]
You asked for something that makes common sense. How much common sense does your statement make?

[/ QUOTE ]
Remember, I've said from the beginning that there is no common sense answer.

At this point, I think we should just agree that your common sense is broader than mine. You can accept probabilities that change when you look at them, as a Bayesian does, and also mathematical formulae that are different depending on whether X is a number or a variable, as a Frequentist does.

I'd rather reject both and live with having no rigorous theory of probability that can also answer useful questions. Many people agree with me, many others choose the less distasteful (to them) of the two alternatives above. But you are rare in being willing to embrace both. For you, the envelope problem is no problem at all.

That probably sounds sarcastic, and maybe I meant it that way a little, but I admit I have no argument against it. You might well be right, I just hope you aren't.