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Comment by yanowitz | Hacker News Reader

Isn't the expected value of both envelopes, prior to picking either, 0.5 * X, where X is the sum of the envelopes' value?

So one way to argue this is a fallacious problem is that you can't take an unknown (the actual value of the envelope that you picked) and pretend it's a known. The value of the envelope you are holding, prior to opening it, is 0.5 * X. The value of the other envelope is also (0.5 * (.66 * X)) + (0.5 * (.33 * X)) == 0.5 * X.

This may be too ill-expressed -- but there's something wrong with treating the value of an unopened envelope as anything other than completely probabilistic.

Put differently, don't you have to say

0.5 chance I picked the bigger envelope (call that A). In which case, the other envelope is 1/2A. So if I switch, I lose 1/2A.

0.5 chance I picked the smaller envelope (A/2). So if I switch, I gain 1/2A.

Which means if I switch, half the time, I gain 1/2A and half the time I lose 1/2A, for an expected gain of 0 from switching.

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That is the right answer. The flaw is that the "A"s in the formula are different actual values depending on which envelope you have chosen, but the analysis has them as the same.

In other words the paradoxical analysis is done relatively, which is wrong. You have to analyze probability from an absolute frame of reference.

Exactly, A doesn't equal A for all parts of the formula, so it's silly to add it up.

Well, of course you're right, as is yanowitz. But there is something unsatisfying about this explanation. Intuitively it is obvious, but why is the relative reasoning incorrect?

There is no formal distinction between absolute and relative quantities, and no theorem that says that expected values can only be taken from absolute quantities. There are just random variables, and these have some distributions, and they can be independent or not. Nothing prevents you from taking an expectation of a random variable that is a ratio of two other random variables.

Another angle - I could say that our definition of expected value, based on weighted arithmetic average, is completely arbitrary, and instead define my own expected value G[X], based on the geometric average. Suddenly, the relative approach becomes correct: sqrt(2 * 0.5) = 1, so the expected relative improvement from switching is 1. What the hell is going on?

> Intuitively it is obvious, but why is the relative reasoning incorrect?

Because in the reasoning symbol A is used to denote "expected value of a random variable representing the amount in the envelope you picked" but later on "expected value of a random variable representing the amount in the envelope you picked provided that you picked envelope with more money" and "expected value of a random variable representing the amount in the envelope you picked provided that you picked envelope with less money".

Error comes from splitting reasoning for two cases and failing to factor in the condition on which you split in your further calculation of the cases.

If you solve some equation and you have to split your reasoning in two (or more) cases, the while reasoning those cases you have to remember the condition that you assumed for given case and factor it in (possibly toss away some solutions).

I'd like to see some day less one dimensional way of writing down mathematic reasoning so one can see how information flows through through the course of a proof and errors like this would show up more easily.

OK, now I see there are two slightly different possible formalizations of the problem:

1. You are told that the two envelopes contain amounts A and 2A, but you aren't told what A is. After you pick one envelope, you are allowed to open it, and then you're given the choice to switch. Here the optimal move depends on the distribution of A, and if you don't know it, you can't do much other than pick randomly. After some googling, this is the more common formalization, and it is analyzed in several math papers and blogs.

2. (The version I was assuming.) You are told that the envelopes have, say, $100 and $200. You pick one and you aren't allowed to open it yet. Now you're given the option to switch one last time. There is no problem with undefined priors and weird conditional probabilities in this version. However, the freaking paradox still holds! The expected value you get by switching is $150, no question about that. But the expected relative gain you get by switching is 1.25, there's also no question about that! This is the real paradox to me. Taking an expectation of a relative quantity is intuitively wrong, but why exactly?

Ad.1 If in the reasoning A is a fixed number then you can't assume that you have 1/2 probability that the other envelope contains more money. Probability depends on distribution and if you don't know the distribution it cannot be calculated. Fact that it cannot be calculated does not entitle you to assuming its 1/2.

If you have problem with that imagine stack of cards. You and your opponent pick one card. Whoever picks higher card wins. Until you see your card you have 1/2 probability of winning. But after you see you just picked 3 you see your probability of winning sharply changes (depending on what cards are left in the stack).

Ad.2 When you don't open your envelope you must treat amount in it (A) as random variable. Then all I wrote in post above applies.

It doesn't matter that you did not looked into an envelope. Because whether you got higher or lower amount is significant and you don't know which occurred you must split your reasoning and consider both cases. But upon splitting you must remember that condition you assume for each given branch of your reasoning must be taken into account. You can do this by pruning your random variable A with the appropriate condition and that influences its expected value. So by all means you can sum up expected values but not before appropriately adjusting them with the conditions you assumed for branches of your reasoning.

Consider solving equation x^2 + yx +1 = 0 While solving your may split you reasoning to three cases depending on whether y^2-4 is larger, smaller or equal to zero (even though you don't know what is the value of y). But when summing up solution you must remember that you assumed something about y in your branches so for the x you've found y can no longer be any real number.

Ad 1 - Not sure what you're trying to say, all I said is that you don't know the distribution of A - if you knew it, you could make an informed decision on whether to switch.

Ad 2 - If you carry out the analysis you're trying to teach me, you'll find that the expected relative gain from switching is 1.25. Now explain that. (You'll also find that the expected absolute gain is 0 - that's not a paradox.)

Let's denote expected value of money in envelope I picked with A.

If I want to switch I have to consider two cases. Let's label them e1 and e2. e1 means I picked envelope with smaller amount of money. e2 means I picked envelope with larger amount of money. Probability of each case is 1/2.

What is the expected value of money I'll have after switch?

Let's consider first case:

S(e1) = 2 * A(e1)

which mens two times the amount of money I had PROVIDED THAT I PICKED ENVELOPE WITH LESS MONEY.

A(e1) is not equal A If I denote exact amount of dollars in envelope with less money by X then A = 1/2 * X + 1/2 * 2 * X = 3/2 * X but A(e1) = X

Similarly considering case e2 I get expected value of money after switch equal

S(e2) = 1/2 * A(e2)

which means half of the amount of money I had PROVIDED THAT I PICKED ENVELOPE WITH MORE MONEY.

Again A(e2) is not equal A but 2 * X, if you prefer not to use X then A(e1) = 2/3 * A and A(e2) = 4/3 * A

To sum up probability of each case is 1/2 so S that I use to donate expected value of money I could have after switch is equal to:

S = 1/2 * 2 * X + 1/2 * 1/2 * 2 * X = 3/2 * X = A

or if you want without using X

S = 1/2 * 2 * 2/3 * A + 1/2 * 1/2 * 4/3 * A = A

So there is no gain or loss from switching. You can switch zero or more times without changing expected value of money you will get after opening your envelope.

Similar reasoning can be conducted for any amount of envelopes and any probability distribution and I guess it will to same conclusion. I'm not sure if observable symmetry of situation can be taken as proof of this conjecture but I sincerely hope so.

Paradox comes from using same symbol A to denote three different expected values of random variables which I denote above as A, A(e1) and A(e2)

If you have further doubts about above "relative" reasoning I'll gladly try to clear them up.

You just painstakingly carried out the absolute analysis, which I know works.

Let X be the value in the envelope you have, and Y in the other one (X and Y are both random variables with well-known distributions). Then E[Y/X] = 1.25. That's what I wanted you to explain. You just keep saying that E[X] = E[Y], which I know.

Note that this paradox would not arise if X and Y were independent, since then E[Y/X] = E[Y] / E[X] = 1.

What reasoning leads to claim that E[Y/X] = 1.25 ?

It can't be the reasoning from wikipedia about two envelopes problem because it has error in steps 4 and 5 so results in step 7 and further are nonsensical.

I painstakingly shown how this reasoning should look like to be right and that in fact E[Y/X] = E[Y] / E[X] = 1

You're combining two worlds and then adding up probabilities. But there's no world where all three possible values are in play at the same time. There's only two, and you either picked high or low. It's a 50/50 shot.

Which three worlds? After you have picked one envelope, there are two possible relative outcomes - 2 and 0.5. The average is 1.25 - there is nothing wrong with the math, the only problem is that taking an average of two relative quantities is intuitively not useful for decision making. Putting the intuition on a formal grounding is not something you succeeded with.

Your math is right, but there's more to the story here, specifically to understand why the switching argument is wrong - this is commonly misunderstood, and has more to do with the implied "fairness" of the putting-numbers-in-the-envelope process than the calculation of the odds afterwards.

The super quick version: the number drawing process can't be fair, and that makes those .5/.5 hi/low probabilities wrong in general.

The real impossibility is the idea that we can "pick a random number" such that all numbers are equally probable, with no constraints on the size of that number. If this isn't intuitively obvious, then just try to normalize a uniform probability distribution over the positive real line, you'll see the problem.

In order to select the two amounts, the original problem statement implies that we need to randomly pick A (which we can take as the bigger envelope) in exactly such a way, so basically, the "paradox" is dead-on-arrival, because the true prior distribution that A was picked from must be different - for example, there might be an upper bound, say $100, in which case if we see that our envelope has any number over $50 on it, we are 100% sure that we hold the larger envelope. Otherwise, without some bounds or tapering of the distribution, we can't normalize it, and we can't use it to draw numbers from, so we have no numbers in envelopes to even discuss.

tl;dr version: If you can write me an algorithm that "randomly selects a number" (let's even make it an integer to make it even simpler - of course I mean an Actual Integer, not those short bit sequences that we pretend are integers when we program), such that no integer is more likely to be chosen than any other, then I've got two envelopes that I'd like to show you. :)

On another note, the truly insane might notice that in the same way that we can formally sum series like 1+2+3+... = -1/12 (http://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%C2%B7%...), and 1+1+1+... = -1/2 (http://en.wikipedia.org/wiki/1_%2B_1_%2B_1_%2B_1_%2B_%C2%B7_...), we might consider the probabilistic analogues of these formulas, and "formally normalize" a PDF that is uniform over the positive integers, with a constant probability of -2 for every integer (whatever that means). Similar things could be tried over the real numbers, though it gets messier because the integral analogues of those formal sums tend to come out to zero (for instance, for any integer n, the integral of x^n from 0 to infinity, in a very specific sense, is zero - perhaps I'll do a layman's post on this phenomenon at some point, it's quite useful in some cases), so normalization doesn't really work out so well.

Of course, probabilities can't be negative under the usual rule set, so all of that would be pointless, wouldn't it? :)

“You can't take an unknown and pretend it‘s a known.” is reasonable. However, the paradox still stands if you are allowed to open your chosen envelope and count the money inside before deciding whether to switch.

The problem genuinely does have to do with the use of an impossible probability distribution. It’s not just a straightforward mistake (as in the wrong answer to the Monty Hall problem, say).

Let's say you open the envelope. And there's $20. That means the set of two envelopes is either $20 and $10 or $20 and $40. In one case, the amount of money in play is $30, in the other, $60.

Now, you have no idea which it is. And the fact that you're holding a $20 doesn't tell you anything.

The $20 is fake information. It hasn't revealed any real info, as you knew you'd open something. The key is not to combine a world where the total envelope value is $30 and where the total envelope value is $60 which is what happens if you add up probabilities.

Instead, there's a 50% chance you're holding 1/3 of the total value, 50% chance you are holding 2/3. So if you switch, half the time, you add 1/3, half the time, you lose 1/3. net expected gain from switching: 0.

"use of an impossible probability distribution" - this sounds interesting. Which exact distribution do you mean? To me it seems (though I might be wrong) that the distributions are perfectly valid.

You have a pair of random variables (X, Y) that take values (100, 200) or (200, 100) with equal probability. Then E[X] = E[Y] = 150, there is no question about that. Also, E[X/Y] = 1.25, there is also no question. The only question is why E[X] is useful and E[X/Y] is not useful for our decision making - and I honestly don't know why.

If it were (100, 200) or (200, 100) with equal probability, you'd switch iff the first envelope contains 100. Nothing interesting about that scenario.

The problem states that (and is only interesting because) one of the envelopes contains twice as much as the other, but not how much, so that the amount in the first envelope tells you nothing.

This is actually impossible because there is no information about how the amount (let's say the bigger of the two) is distributed, which usually implies uniform distribution. But a uniform distribution is only possible if you assume an upper limit (otherwise, what's the expected value?). If there is an upper limit, the question again becomes quite easy (you switch if the first envelope contains less than half the upper limit).

"Nothing interesting about that scenario." OK, but let's say you're not allowed to see the contents of the envelope you picked.

In this case, the argument that by switching you get 125% on average still holds!

I'd claim that this is even more interesting and paradoxical than the scenario you're talking about, where your explanation is correct (the distribution of the bigger value is crucial for your decision, and you know nothing about it).

Isn't the expected value of both envelopes, prior to picking either, 0.5 X, where X is the sum of the envelopes' value?*

Only if the value of X is determinate. As matters stand for the person picking, it is not. It is simply undefined.

The solution to the "paradox" is simply that expected value doesn't always mean what we'd like it to mean.

Yanowitz defines X as the sum of the value of both envelopes. That amount is not undefined and not indeterminate. The amount inside an individual envelope isn't undefined either: it has a probability distribution.

Sorry, but no.

No probability distribution has been defined for the amount inside an individual envelope, and you can't just pull one out of thin air. That's not how mathematics works. If you had a probability distribution for that, you could make all sorts of arguments from it. But we're not given one, and we're given no information from which we can imply the existence of one. Therefore there is no such distribution. Period. (Interestingly, assuming such a distribution gives useful information even when it is the wrong distribution, but that is a complicated variant on the problem.)

Now you can argue until the cows come home about what probability "really" means, or what is "really" true in a made up problem. Plenty of philosophers are willing to argue that with you. But the way that mathematicians look at this is quite simple. You can only work with the information you are given, and you can't make up information that you weren't given. Arguing about the fine points of reality within a piece of fiction is, not to put too fine a point on it, just plain silly. After all it is a made up example. So don't do that.

We are given certain information. We look in the one envelope if we wish, and we get a dollar value. Based on that we can draw inferences about what might be in the other envelope. We can create an expected value for said envelope. All that is fine.

But our intuition about what the expected value has to say about what we should do is wrong. Nothing is wrong in the math - it is our intuition that is wrong here. Our intuition is based on what happens if you encounter large numbers of similar, but independent, events. However in this case we have a singular, and extremely unlikely to be repeated, event. Therefore the basis for our intuition is lacking here. And we are mislead.

No probability distribution has been defined for the amount inside an individual envelope

I don't understand this assertion. It seems to me that both envelopes have the trivial distribution:

  P(x=A) = 1/2, P(x=2A) = 1/2, P(x) = 0 for all other x.

and that this is clear from the phrasing of the question. Extracting this distribution from the text is no different than solving any other problem not stated in rigorous mathematical terms. With this distribution, with A properly defined (see the next point) upfront, the whole paradox disappears.

--

Nothing is wrong in the math [..]

I think the statement in the article that

  4. If A is the smaller amount the other envelope contains 2A.

  5. If A is the larger amount the other envelope contains A/2.

is wrong, because it tries to define A twice with different concrete values and goes on to treat them as the same value. If your point is that saying 'A is the quantity in the envelop you just chose' makes A ill defined, then I guess you are right, but that doesn't mean that A cannot be defined properly and it doesn't mean the intended problem can't be solved.

> > No probability distribution has been defined for the amount inside an individual envelope

> I don't understand this assertion. It seems to me that both envelopes have the trivial distribution:

What I mean is that there is no a priori distribution in terms of actual numbers. What are the odds that the amount in the envelope before you open it is in the range $5 - $500? We are given no such information. Before you open the envelope you are given, A is a variable, but not a random variable. Its possible values ideally would have a uniform distribution on an infinite set, which is mathematically impossible. There is no such distribution. (There is an alternate approach which is to claim that that the sum of the two envelopes has an unknown distribution, which creates a distribution for the values of each envelope. This leads to a very different mathematical idealization of the problem. But the other distribution, being unknown, is not available to you. I will trace the reasoning for this through as well.)

If you open the envelope in front of you, A then becomes becomes a number, and there is indeed now a probability distribution for the contents of the other envelope. And that distribution is now, as you said

  P(x=A) = 1/2, P(x=2A) = 1/2, P(x) = 0 for all other x.

(Note, the alternate approach would now assert that the number in the the envelope provides unknown information on whether it is larger. Therefore the other envelope now has a different unknown distribution that depends on the actual magnitude of A.)

But now we arrive at a different point. Now we have a probability distribution for the other envelope, but our intuition fails for the simple reason that our intuition of the meaning of expected value is based on what happens after repeated opportunities at similar random events. (Insert the strong law of large numbers, etc.)

(The alternate solution finds that the other distribution causes the other envelope, on average over all possible values you could see, to have the same value as your own. However this is of no obvious direct use.)

From the point of view of the standard mathematical idealization, what you guys have been doing is trying to train your intuition to create arbitrary distinctions that avoid conclusions that bother you. But you're drawing a distinction that doesn't make sense. If you've opened the envelope and you have $10 sitting there, then A is $10. Period. It isn't one of two different numbers, it is the number you see in front of you. That has now become a provided fact. At this point you can draw a distribution for the other envelope. But your intuition about expected value fail. Why? Because this scenario, with this set-up, is not something that can become subject to repeated trials.

(The alternate approach comes to a different conclusion. Your intuition, they claim, is thrown by the fact that you do not have the critical information about the actual distribution of the other envelope, and so cannot come up with the correct numbers to decide. Interestingly in a bizarre twist, it turns out that if you make up a distribution and pretend it is the unknown one, as long as that distribution has some probability of answers falling in every possible range, your decision winds up being right better than half the time. However how much more you are right depends on the unknown distribution, and is therefore unknown to you.)

That is the clearest, most succinct explanation I've heard. Thanks.

So basically the fallacy is in counting gains only, without counting opportunity cost [of the switch].

Heard on the street?

Under a frequentist interpretation of probability that's correct (and the paradox doesn't arise), but Bayesian decision theory does allow you to reason in the way the paradox proposes, starting in a situation where you've already picked an envelope, and estimating the utility of switching to the other one by multiplying the probability that you're in the high->low configuration with the value of a high->low switch, and likewise with a low->high switch. So you have 50% chance of halving your money, 50% of doubling, and thus estimated 1.25x return.

Frequentist probability doesn't let you do that, because it doesn't let you say "there's a 50% likelihood I'm in this state, and 50% that I'm in the other". Instead, it says you must be in one or the other state (non-probabilistically), and the probabilities are only attached to a previous process that generated that outcome. But Bayesian probability does let you interpret the probabilities as belief in each current state, so you have a 50% belief you're in one state, and a 50% belief you're in the other, and the value of any decision is thus 0.5 * value_if_I_was_in_situation_A + 0.5 * value_if_I_was_in_situation_B.

(If you don't allow that kind of computation, Bayesian decision theory has to be revised in some other cases as well.)

No, Bayesian decision theory does not allow you to reason in that way.

Suppose the amounts in the envelopes are A and B. A Bayesian would say:

Prob(I have A) = 1/2 and Prob(I have B) = 1/2

The expected gain from switching is:

Prob(I have A)x(what I'd gain from switching = B-A) + Prob(I have B)x(what I'd gain from switching = A-B)

which is: 1/2 x (B-A) + 1/2 x (A-B) = 0

Ah yes, this one can be solved that way; oops. My understanding of the literature is that variants are much more problematic, though, and require more complex restrictions on typical Bayesian probability frameworks to exorcise them.

This is one good review, with a proposed solution (from a Bayesian perspective): http://books.google.com/books?id=14_ykEOAZ6AC&pg=PA49

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