The envelope problem, and a possible solution

[PairTheBoard]

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Practicing Bayesian statisticians use the same rigorous mathematics (in particular, the same definition of a probability space) as we probabilists do. Here's a toy example inspired by poker.

Let m be my winrate in BB/100 and s my standard deviation. I could model the outcome of a single 100 hand session by

X = m + sN,

where N is a Normal(0,1). Suppose I know that s=15, but I don't know m. However, I have "beliefs" about m. I believe that m=2. Of course, I'm not certain of it, it could be higher or lower. But I am fairly certain that I am at least a winning player. I might decide to model these beliefs by saying that

m = 2 + M,

where M is a Normal(0,1), independent of N. In "reality", m is a fixed, deterministic number, but I am representing it as a random variable in order to model my beliefs. Similarly, in the envelope paradox, the amounts in the envelopes are fixed, deterministic values, but we may choose to represent them as random variables in order to model our beliefs about the contents of those envelopes.

Now suppose I play a session of poker and lose 100 BB. I now need to adjust my "beliefs". This, of course, is analogous to the act of opening one of those envelopes. The adjustment is done by computing the conditional law of m, given X, and plugging in the observed value X=-100. We get the conditional CDF by computing

P(m <= x | X).

Upon computing this expression and differentiating, we find that the conditional law of m is again normal. According to my hasty calculations, its mean is

(225/226)*2 + (1/226)*X

and its variance is simply 225/226 (regardless of the value of X). Plugging in X=-100, our updated "belief" about our winrate is that it is 350/226=1.55 and our "belief" has a standard deviation of about 0.998.

Obviously, the process gets more complicated when we try to apply it to a sequence of sessions, and we might find it useful to apply the mathematical tools of filtering theory. One aspect of this that I find interesting, as it relates to poker, is that you can get analogs of confidence intervals which appear much more realistic over a smaller sample size. Of course, those estimates always contain some residue of our original "beliefs". For what it's worth, that residue goes away in the limit as the sample size increases.

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Great explanation. Thanks Jason. That's something I can understand. So, is there some controversy about that technique? What is the nature of the Bayesian justification for the technique?


PairTheBoard

[PairTheBoard]

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Dear, PTB. You state that the place where argument 1 breaks down is at the part where it assumes 50% probablility. However, unless the probabilities now take on specific values, switching will still be either +EV or -EV.


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If you pick one of two closed envelopes at random, you have a 50-50 chance of picking the envelope with the larger amount. Once you open the envelope and see the amount A, chances are no longer 50-50 that A is the larger amount.

This is intuitively difficult in the 2E problem. But consider what I'm going to call the 2X problem. You have two X envelopes, one is empty and one contains a dollar. Now consider the same statement as above for the 2X problem.

If you pick one of two closed X envelopes at random, you have a 50-50 chance of picking the envelope with the dollar. Once you open the envelope and see that it's for example empty, chances are no longer 50-50 that you have a dollar in hand. Or if the amount you see is a dollar, chances are no longer 50-50 that you have a dollar in hand.

It's easy for you to accept the statement in the 2X case but difficult for you to accept it in the 2E case. The reason is that you intuitively believe that all 2E envelope amounts are equally likely. But we know that's impossible. Once you open the envelope and see the amount A you now face a new conditional probability for whether it's the larger amount. Just like in the 2X problem, when you see the envelope's empty you now face a new conditional probabiilty that it contains a dollar.

Suppose for example that the Benefactor in 2E picks $1/$2 90% of the time and $2/$4 10% of the time. You don't know this though. Now the game is, He choses his 2E amounts and lets you pick an envelope. You have a 50-50 chance of picking the envelope with the larger amount. Now you look inside and see the amount $2. Are your chances still 50-50 that you have the larger amount? No. Now your chances are 90%. You just don't know it. If you are using an always switch strategy you will be facing -EV doing so when you are looking at your $2 envelope.

PairTheBoard

[BBB]

[ QUOTE ]
Dear, PTB. You state that the place where argument 1 breaks down is at the part where it assumes 50% probablility. However, unless the probabilities now take on specific values, switching will still be either +EV or -EV.

If this is true, then that leaves a bit of a strange situation. I could switch, and then after I did, switch again on the basis of the exact same reasoning, and then again, and again, and in the end I'd have to be rich.

This is basically stating what has been stated before, argument 1 would create +EV situations from both ends of the switch.

How do you reason your perspective deals with this problem?

@BBB, thank you as well for your elaborate explanation. As far as I can see my reply to PTB is also valid as a response to your post and I ask you the same thing.

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NaobisDad,

I'm not quite clear on what your question is... If your strategy is to always switch, no matter what amount is in your envelope, then this strategy will always be neutral EV. This is because, if the amount that you see will not affect your decision, then switching after looking is the same as switching without looking, and argument 2 applies. Once you see the amount in the envelope, if you can infer any information based on that amount as to the chances that the amount you see is N versus those that it is 2N, then switching may become +EV or -EV, depending on what the amount tells you.

Suppose we change the problem slightly. Let's say you're presented with two envelopes, except that all you're told is that one envelope contains mroe than the other (not necessarily twice as much). You open one and you see $100. Should you switch? What if you see $1 million dollars?

If you apply similar logic to argument 1 in the N-2N case, you could conclude that you should always switch, no matter what amount you see. After all, if you see $100, switching can only cost you up to $100, but it could earn you thousands, millions, or more. If you see $1 million, switching could cost you a million, but it could gain you billions or trillions.

But actually, a strategy of switching no matter what amount we see is clearly neutral EV (prior to opening the envelope), just like in the N-2N problem. But in this case, it's harder to jump to the not-necessarily-true conclusion of argument 1, becuase it's easier to see that automatically figuring that the other envelope will contain more money than ours on average is arbitrary and not necessarily correct.

If you were actually presented with this new problem, and you found $100 in your envelop, then if you think there's even a small chance that the other envelope contains thousands of dollars or more, then of course you should switch. However, if you know that there's no way this guy could afford to give you even $200, then I would keep the $100. Similar reasoning should be used if you find $10 million dollars (although personally, that would be enough money that if I felt that on average, the other envelope would contain $20 million dollars, but if there was a significant chance that it contained much much less than $10 million, then I would probably decide that $10 million would be life-changing money for me, and I wouldn't risk the switch, even though switching would be significantly +EV if looked at strictly in terms of dollar amounts (but if I were already a billionaire, then I would switch)).

[jason1990]

What I presented was a Bayesian analysis, but the exact same mathematics could have been done by a frequentist in a slightly different context. Here is the modified context.

Suppose, by doing a survey or something, I determine that the winrate of a randomly selected 2+2er is a Normal(2,1) random variable. Suppose I also somehow know that every 2+2er has a standard deviation of 15. I now select a random 2+2er and observe him as he plays a single 100 hand session. He happens to lose 100 BB. I now conclude, as did the Bayesian, that his winrate is normal with mean 1.55 and standard deviation 0.998.

When I say this, though, what I mean (as a frequentist) is this: imagine I repeatedly select random 2+2ers and watch one session. If the player loses 100 BB, then I (somehow) record his actual winrate. I will then generate a list of winrates of random players who happened to lose 100 BB. My conclusion is that, in the long run, the resulting histogram will approach a Gaussian curve with the appropriate parameters.

As a frequentist, both my assumption and my conclusion are statements about long term frequencies of repeatable experiments. The Bayesian is saying something different. The Bayesian is not starting with a random player. (In the example, I was the Bayesian and I was analyzing myself.) So his assumption about the winrate is not a statement about a repeatable experiment, but about his beliefs. Mathematically, everything is the same. The difference is philosophical and the debate is over the meaning of the statement "the probability of event A is p."

In my opinion, Bayesian methods should not be summarily dismissed simply because their assumptions cannot be phrased in terms of repeatable experiments. I think that, at least intuitively, probabilities of one-time events are meaningful to the human mind. For example, I know that in 2003 it made sense to me to discuss with others the probability that Bush would be re-elected. Also, in practice, I think that Bayesian assumptions tend to be eduacted guesses based on empirical evidence, and not simply arbitrarily chosen "beliefs".

But when it comes to the actual computations and the rigorous mathematical framework, there is no real difference.

[PairTheBoard]

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The difference is philosophical and the debate is over the meaning of the statement "the probability of event A is p."

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What does a Bayesian say that statement means?


PairTheBoard

[PairTheBoard]

Ok, from this Wikipedia Article on Bayesian Probabilty

"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.

When beliefs have degrees, the theorems of the probability calculus become criteria for the rationality of sets of beliefs in the same way that the theorems of first order logic are criteria for the rationality of sets of beliefs. Many authors regard degrees of belief as extensions of the classical truth values (true and false)."

"A quite different interpretation of the term probable has been developed by frequentists. In this interpretation, what are probable are not propositions entertained by believers, but events considered as members of collectives to which the tools of statistical analysis can be applied.

The Bayesian interpretation of probability allows probabilities to be assigned to all propositions (or, in some formulations, to the events signified by those propositions) independently of any reference class within which purported facts can be thought to have a relative frequency. Although Bayesian probability is not relative to a reference class, it is relative to the subject: it is not inconsistent for different persons to assign different Bayesian probabilities to the same proposition. For this reason Bayesian probabilities are sometimes called personal probabilities (although there are theories of personal probability which lack some features that have come to be identified with Bayesianism).

Although there is no reason why different interpretations (senses) of a word cannot be used in different contexts, there is a history of antagonism between Bayesians and frequentists, with the latter often rejecting the Bayesian interpretation as ill-grounded. The groups have also disagreed about which of the two senses reflects what is commonly meant by the term 'probable'.

To illustrate, whereas both a frequency probability and a Bayesian probability (of, e.g., 0.5) could be assigned to the proposition that the next tossed coin will land heads, only a Bayesian probability could be assigned to the proposition, entertained by a particular person, that there was life on Mars a billion years ago—because this assertion is made without reference to any population relative to which the relative frequency could be defined."


I admit I am out of my depth with this Bayesian Probabilty. I'm not sure what to make of it. For a Bayesian to say that a probalility is his degree of belief at first seems unworkably vague. But when he is willing to place bets giving odds based on that degree of belief/aka probability then it seems more meaningful.

It is confusing when terms like probability and expectation are being tossed around in a mish mash of Frequentist and Bayesian meanings. It would be nice if they would differentiate their terms.


PairTheBoard

[PairTheBoard]

So Jason, in your opinion would the Bayesian perspective yield anything essentially different than the Frequentist when looking at the Two Envelope Problem?


PairTheBoard

[NaobisDad]

Here is a link to an article on this problem

http://newbabe.pobox.com/~mjd/blog/math/envelope.html

I'll get back to this thread a little later. Fatigue is taking over and that led to the poorly formulated last post I made in reply to PTB and BBB. I'll get back to this. Thanks guys.

[SamIAm]

I was paging through this pile of posts to see how it could have possibly gone to 48 replies. I was hoping there was a flame-war; maybe Aaron insulted BruceZ's mother, and Bruce countered by questioning Aaron's sexual prowess. (It could happen.)

No luck, but I did catch the quote below.
[ QUOTE ]
Clash: "Should I stay or should I go? If I go there will be trouble."
SamIam: "Wait, how was the amount of trouble chosen?"
Clash: "If I stay there will be double."
SamIAM: "Man, I dunno. I can tell you the correct move if you describe the opponent."
Clash: "So you gotta let me know. Should I stay or should I go?"

[/ QUOTE ]I've always found that song to be hilarious. Listen, TheClash, if you don't want double the trouble, get out now! (Duh.)

The truly unanswerable 80s song questions are "Voulez-vous coucher avec moi (ce soir)?" and "Who's bad?".

Quite the paradox.
-Sam

[jason1990]

In my opinion, the only thing a Frequentist can say about the Two Envelope Paradox is what you yourself have already said.

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the amounts in the envelopes are Fixed...each time a Person opens an envelope and sees an amount, the conditional probability of the second envelope being double the amount he sees is either 100% or 0% - he just doesn't know which.

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There is no repeatable experiment in the Paradox. As you said, there is no sequence of envelopes for you to choose from. But neither is there a sequence of people choosing from those envelopes. There's just you and the envelopes and a one time choice. Introducing any kind of repeatable experiment is adding something to the Paradox that wasn't there.

In my opinion, the Paradox is entirely a lesson for Bayesians. The Bayesian equivalent of "I know nothing" is the uniform distribution. If a Bayesian has no information about an unknown quantity, other than the fact that it is a member of some set, then it is common for the Bayesian to assume that the unknown quantity is a random element of that set with a probability distribution such that all members are equally likely.

In fact, many people who have never even heard the word "Bayesian" have made this Bayesian assumption, without even explicitly realizing they are doing so. Anyone who has assumed this has a little Bayesian in them, and this Paradox is a lesson for the Bayesian in all of us.