It's truly remarkable that a question which is of no philosophical significance or interest could generate so many responses on a philosophy forum! — Janus

It does have some philosophical implications. Some of @andrewk's replies raised good philosophical points regarding the status and significance of probability distributions, which are being involved in the analysis of this apparent paradox.

Yours isn't really a decision tree that the player must make use of since there is no decision for the player to make at the first node. — Pierre-Normand

I might also have pointed out that when I first started doing this a couple days ago I said

This is in fact only a "tree" in a charitable sense. — Srap Tasmaner

The point of the tree is to show that the last decision you make in every case is the exact same as the first decision you made, and whatever decisions you were offered in between.

I'll admit I probably would not understand the philosophical significance of probability distributions even if I had read the relevant posts. I would have thought this is more math than philosophy. What area of philosophy do you think the significance would obtain?

I'll admit I probably would not understand the philosophical significance of probability distributions even if I had read the relevant posts. — Janus

Some might unkindly note that it hasn't stopped me.

You seem to have made a much better fist of it than I ever could! :smile:

I guess it's just not relevant to my area of interest in philosophy. I still don't believe you were really offended, though. :grin:

Sorry, I'm not following this. This sounds like you think I said your expected gain when you have the smaller envelope is zero, which is insane. — Srap Tasmaner

No, that's not what I was saying. I was rather suggesting that, assuming there is some determinate albeit unknown probability distribution of possible envelope pairs, then, conditional on some one specific pair having been selected from this initial range, and consistently with $5 being one of the two amounts within this pair (because $5 is the amount that you have seen in your envelope, say), then, the expected gain of switching appears to be zero, according to your decision tree analysis. According to @JeffJo, it could be $2.5, $6.25, $10, or something else, and not necessarily zero. What it actually is, is unknown to the player. What is known to the player only is the average gain from the unconditional switching strategy. And that is zero.

It's truly remarkable that a question which is of no philosophical significance or interest could generate so many responses on a philosophy forum! — Janus

It's something new. I've been posting here (and on the old place) for 12 years. I've argued over all the typical stuff countless times.

According to the post counter, this will be the 999th post in this thread.

I'm rooting* for it to hit the big 1K, as it has been a most interesting discussion when tempers have stayed within bounds.

* As an Australian resident, I need to point out that I am using the American sense of this word, which is 'hoping for the best for'. In Australia 'rooting' means 'having sex with', which is not what I am doing in relation to this thread.

I wonder if what I've been talking about (made more explicit here), is subjective expected utility.

That's fair enough; I wasn't actually criticising those who are participating in this thread, just expressing my surprise that the OP could be thought to be of any more interest than any kind of logical or intellectual puzzle or conundrum. I acknowledge that I may be missing something due to ignorance, though.

If I hadn't gone inside for coffee, I would have had the 1000th post. I feel bad now, but I have coffee.

What area of philosophy do you think the significance would obtain? — Janus

Probability is a big philosophical topic. It is quite tightly enmeshed with both metaphysics and epistemology. Michael Ayers wrote a lovely book, The Refutation of Determinism, which explores some of the philosophical problems associated with the concepts of probability, necessity and possibility, and also pursues some implications for the problem of free will and determinism. There is also a close connection with epistemology, Gettier examples, and some of the most puzzling paradoxes of Barn Facade County, which arise, it seems to me, from assumptions regarding the grounding of knowledge that are closely related to some of the assumption that give rise to the two-envelope paradox. Maybe I'll create a new topic about this when time permits.

A first step, which is being pursued in this thread, is to get clear on (what should be) the uncontroversial steps in the mathematical reasoning.

I'm not sure which of @JeffJo's examples you're referring to.

As for my "tree" and what it predicts -- You face a choice at the beginning between two values, and the same choice at the end between those same two values. If you flip a coin each time, then your expectation is the average of those two values both times and it is unchanged.

Opening an envelope changes things somewhat, but only somewhat. It gives more substance to the word "switch", because having opened one envelope you will never be allowed to open another. You are now choosing not between two envelopes that can be treated as having equal values inside, although they do not, but between one that has a known value and another that cannot be treated as equal in value.

But it is, for all that, exactly the same choice over again, and however many steps there are between beginning and end, your expected take is the average of the values of the two envelopes. If there's an example in which that is not true, I would be surprised.

I have come, in broad terms, to see probability as a generalization of logic, or logic as a special case of probability, take your pick. I would credit Frank Ramsey for convincing me to begin thinking this way. As my principle interest is the nature of rationality, what's of interest here -- decision theory, broadly -- is still the nature of inference.

What we have been arguing about for fifty pages is what inferences are justified and what aren't.

It's truly remarkable that a question which is of no philosophical significance or interest could generate so many responses on a philosophy forum! — Janus

The OP asks, "What should you do?"

Think of the problem as being in the same family as Pascal's Wager, involving decision theory and epistemology.

The OP asks, "What should you do?"

Think of the problem as being in the same family as Pascal's Wager, involving decision theory and epistemology. — Andrew M

Indeed. And just as is the case with Newcomb's problem, with the two-envelope paradox also, the dominance principle and the (maximum) expected utility principle appear to recommend inconsistent strategies when carelessly applied. Newcomb's problem is more controversial, even, than the two-envelope paradox. It is also quite rich in philosophical implications.

Newcomb's problem is more controversial, even, than the two-envelope paradox. It is also quite rich in philosophical implications. — Pierre-Normand

Definitely a good candidate for a discussion sometime!

I'm not sure which of JeffJo's examples you're referring to. — Srap Tasmaner

I was not referring to a specific example but rather to his general resolution of the apparent problem. It both justifies the zero expected gain for the unconditional switching strategy and explains why the indifference principle can't be applied for inferring that the expected utility of switching is 1.25v, in the case where your envelope contains v.

As for my "tree" and what it predicts -- You face a choice at the beginning between two values, and the same choice at the end between those same two values. If you flip a coin each time, then your expectation is the average of those two values both times and it is unchanged.

I am not sure why you are saying that I am facing a choice rather than saying that I simply don't know whether my envelope is smallest or largest (within the pair that was picked). I am not facing a choice between two values. I am holding v, and I am offered to trade v for another envelope which, for all I know, might contain v/2 or 2v. One wrong inference that one might make is that just because I don't know whether the other envelope contains v/2 or 2v, and don't have any determinate means to estimate it, therefore the indifference principle applies and I can assign 1/2 to the probability of either cases. That would indeed yield an expected value of 1.25v for the second envelope. But that is a wrong inference from the fact that I don't know either P(v, v/2) or P(v, 2v). I only know that those two probabilities add up to one and their exact values are dependent on the prior distribution of possible envelope pairs. The only way for such an initial distribution to be such as to guarantee that P(v, v/2) = P(v, 2v) = 1/2 for any v would be for the initial distribution to be uniform and unbounded. For any other feasible way to randomly determine amounts of value pairs in accordance with some well behaved probability distribution, what the player can infer only is that, on average, after repeated trials, the expected value for switching tends towards zero.

Opening an envelope changes things somewhat, but only somewhat. It gives more substance to the word "switch", because having opened one envelope you will never be allowed to open another. You are now choosing not between two envelopes that can be treated as having equal values inside, although they do not, but between one that has a known value and another that cannot be treated as equal in value.

But it is, for all that, exactly the same choice over again, and however many steps there are between beginning and end, your expected take is the average of the values of the two envelopes. If there's an example in which that is not true, I would be surprised.

I disagree. Suppose the initial distribution is, unbeknownst to you, ((5,10), (10,20), (20,40)). In that case, if you are being dealt 5, the expected value of sticking is 5. You don't know what the expected gain of switching is. But it's not the case that it is therefore zero. That would only be zero if you knew for a fact that (5, 10) is half as likely as (5, 2.5) in the prior distribution.

But it is, for all that, exactly the same choice over again, and however many steps there are between beginning and end, your expected take is the average of the values of the two envelopes. If there's an example in which that is not true, I would be surprised. — Srap Tasmaner

The way I look at the problem is that the expected gain calculation depends on an agent's knowledge. Some agents may have more information than others which they should factor into their calculation.

For example, the host may know what envelope pair has been selected (rendering the initial distribution irrelevant) but not know the amount in the player's chosen but unopened envelope. For the host (as for the player), the expected gain from switching is zero. However the host calculates this by averaging the two known amounts instead of averaging the unknown X and 2X.

When the player opens the envelope revealing the amount, the host will now know the actual gain (or loss) were the player to switch. So, from the host's perspective, the expected gain just equals the actual gain. The host now has all the relevant information.

Conversely, the expected gain that the player calculates will still be the unconditional gain of zero since she doesn't know the initial distribution or both amounts in the selected envelope pair.

A further perspective is held by those who know the chosen amount and also know the initial distribution but not which envelope pair was initially selected. So for the equiprobable ({$5,$10},{$10,$20}} distribution, the expected gains from switching for $5, $10 and $20 respectively are $5, $2.50 and -$10. The $10 amount is the only amount where not all the relevant information is known (i.e., which envelope pair it belongs to). But there is enough information to recommend switching should the player see the $10 amount.

Each expected gain calculation is justified given the information the agent has. But since that information can be different, each agent can (justifiably) reach different conclusions regarding whether the player should stick, switch or be indifferent.

Suppose the initial distribution is, unbeknownst to you, ((5,10), (10,20), (20,40)). In that case, if you are being dealt 5, the expected value of sticking is 5. You don't know what the expected gain of switching is. But it's not the case that it is therefore zero. That would only be zero if you knew for a fact that (5, 10) is half as likely as (5, 2.5) in the prior distribution. — Pierre-Normand

This makes no sense to me. Initial distribution of what? If these are pairs of envelopes from which will be chosen the pair that the player confronts, then not only is this sample space unknown to the player, she never interacts with it. She will face the pair chosen and no other.

If that pair is {5, 10}, and she draws the 5, then she has the low value and she can only gain by switching with no risk of loss. The only case in which switching does not produce an actual gain or loss is when, contrary to the rules, the envelopes have the same value. By switching she gets the 10; but the 10 was there all along and she might have chosen it at the start. What doesn't change is your overall expectation from playing. Always Stick and Always Switch are not strategies that increase or decrease your expected take from playing.

I am not sure why you are saying that I am facing a choice rather than saying that I simply don't know whether my envelope is smallest or largest (within the pair that was picked). — Pierre-Normand

Why would your ignorance preclude you from facing a choice and making a decision? In the OP, you make at least two choices: which envelope to claim, and whether to keep it or trade it for the other. Whether you end up with the most or the least you could get depends on those two decisions and nothing else. What the amounts are depends on someone else.

This makes no sense to me. Initial distribution of what? If these are pairs of envelopes from which will be chosen the pair that the player confronts, then not only is this sample space unknown to the player, she never interacts with it. She will face the pair chosen and no other. — Srap Tasmaner

Yes, it is tempting to say that if the game is only being played once then the shape of the initial distribution(*) isn't relevant to the definition of the player's known problem space. That's a fair point, and I may not have been sufficiently sensitive to it. This is broadly the argument @Michael and @Jeremiah are making, although they are reaching opposite conclusions regarding the expected gain from switching. (And they're, in two different ways, both right; hence the paradox). The player doesn't know what this distribution is, and, since she only is playing once, and hence only is being dealt one single envelope pair, how is it relevant what the other unrealized envelope pairs (and their probabilistic frequencies) were within this specific distribution? But it is actually relevant what the distribution might look like since there are logical constraints on the shape of that distribution that can be inferred from the assumptions that ground either strategies (i.e. switching or being indifferent to switching).

The advocate of the switching strategy is actually right in saying that if, even after she has seen that her envelope contains v, she still has no reason to assign an at least twice higher probability to the other envelope being v/2 rather than 2v, then she is justified in switching. The switching strategy can only be justified since, precisely, the prior probability distribution of this game, which is only being played once, doesn't interact with this player's problem space. When she gets to play such a game again, if ever, the prior distribution might be different.

What rather defines the problem space of this player, in accordance with the general specification of the problem that is provided to her (as described in the OP), is the range of possible envelope pairs (and their probabilistic weighs) that is being merely consistent with the general (and abstract) specification of the problem. There are however (roughly) two ways to generate such a range: bounded or unbounded. If it's unbounded, then the switching strategy is justified since the expected gain from switching is 1,25v conditionally on any v being found in the first envelope. If the range is bounded, because the player assumes that there is some finite amount of money M in the universe, however big M might be, then her problem space is such that the average expected gain from switching is zero. This is because her problem space is built up from some arbitrary weighing of all the possible bounded prior distributions, and all of those individually yield an average expected gain from switching that is zero, and so is any weighted sum of them.

(*) I am defining this initial distribution with reference to the method, unknown to the player, by means of which the initial contents of the two envelopes is effectively being determined, by means of a pseudo-random number generator, quantum device, or whatever.

A further perspective is held by those who know the chosen amount and also know the initial distribution but not which envelope pair was initially selected. — Andrew M

Yes, absolutely, and this is specifically beyond the OP. The distributions we've been talking about have almost always been (or should have been) unknown to the player. The player doesn't even know that there is some selection process. There are values in envelopes. How they got there can be discussed, and that can be interesting when the player has, say, partial knowledge of that process, but it is not the source of the paradox, in my opinion.

There are values in envelopes. How they got there can be discussed, and that can be interesting when the player has, say, partial knowledge of that process, but it is not the source of the paradox, in my opinion. — Srap Tasmaner

Whereas, on my view, it is the source of the paradox ;-)

Whereas, on my view, it is the source of the paradox ;-) — Pierre-Normand

Yes! This is exactly what we disagree on.

-- For a change, I'm going to take a little time and think through my response. --

Conversely, the expected gain that the player calculates will still be the unconditional gain of zero since she doesn't know the initial distribution or both amounts in the selected envelope pair. — Andrew M

On the assumption, of course, that the player takes this initial distribution to be bounded above by M for some (possibly uncknown) M; or that, if unbounded, its sum or integral converges.

Why would your ignorance preclude you from facing a choice and making a decision? In the OP, you make at least two choices: which envelope to claim, and whether to keep it or trade it for the other. Whether you end up with the most or the least you could get depends on those two decisions and nothing else. What the amounts are depends on someone else. — Srap Tasmaner

Your ignorance doesn't preclude you from facing a choice and making a decision. What it precludes you from doing is basing your decision to switch (if you decide to) on a determinate expected gain, since the value of such an expected gain (conditionally on your having seen v in the first envelope) is unknown (and, in particular, is not known to be zero).

Only in the case where the distribution is uniform and unbounded can you know what the expected gain from switching is, and know it to be precisely 1.25v. (And that's because, in the unbounded case, it is precluded that you might ever hit the top of the distribution(*) and hence lose in one fell swoop all the expected gains potentially accrued from the other cases in the average time that it takes for this large loss to randomly occur). In that unbounded case, you would know that switching will earn you either v/2 or 2*v with equal probability. It merely appears paradoxical, in that unbounded case, that you are always entitled to switch but that, nevertheless, if you were to switch blindly the first time, and only open the second envelope to find some value w, then, you would still expect 1.25w from switching back. It is this apparent paradox, occurring with infinite and uniform distributions, that my Hilbert Grand Hotel analogy was meant to illustrate.

(*) I am only considering uniform probability distributions over elements of a single discrete doubling sequence, for simplicity. We can also assume it to be truncated below, at $1, say.

Hilbert's Grand Hotel Revisited

Here is an improvement on my earlier Hilbert Grand Hotel analogy to the two-envelope paradox. The present modification makes it into a much closer analogy.

Rather than considering an Infinite Hotel where the countably infinitely many rooms are numbered with the natural numbers, this time, they will be numbered with the (also countably infinitely many) strictly positive rational numbers. Hence, for instance, room 3/2 will be located midway between room 1 and room 2, and will be small enough to fit between them. Guests of the Hilbert Rational Hotel who want to go into rooms corresponding to rational numbers that have very large denominators will have to swallow very many magic shrinking pills, let us assume (and may have to leave their luggage behind). Initially, infinitely many guests are being randomly distributed such that there are, on average, about one hundred guests in each room. Every morning, each guest is awarded $Q, where Q is their rational room number. Every night, each guest is being given the opportunity to flip a coin to determine if she will move from room Q to either room Q/2 (if she lands tails up) or to room 2*Q (if she lands heads up). Let us suppose that all the guests are greedy and rational and, hence, all choose to flip the coin (rather than stay) in order to increase their expected earnings to 1.25*Q on the next morning. (If we take a random sample of guests within any given bounded segment of Hilbert's Rational Hotel, it is clear that they fare better, on average, than they would if they had all chosen to stay. On average, they fare 1.25 times better.)

The first thing to notice is that, after the daily reshuffling of guests, the average population of the rooms stays the same since each room is accessible from exactly two other rooms and as many guests, on average, move from room Q to room 2Q, for any Q, than the reverse. So, although any finite random sample of guests improves its fare 1.25 times, on average, the average room occupancy doesn't change and the overall population density (measured as guests per room) on any segment of the hotel doesn't vary at all.

Now, suppose that on some particular nights -- on Sunday nights, say -- all the guests swallow a pill that makes them forget what room it is that they had moved from on the previous night. They are then offered the opportunity to blindly move back to whatever room it is that they previously occupied. If they are rational and greedy, they ought to choose to move back since, if they now are in room Q, it is equally likely that they came from room Q/2 than it is that they came from room 2*Q. Hence, on average, the guests who move back to the room where they came from fare 1.25 times better than they would if they stayed.

This explains why, likewise, in the two-envelope game, with an unbounded and uniform distribution, (if such a thing is intelligible at all), it appears rational to switch envelopes in order to increase one's expected value by 1.25 and, nevertheless, after one has switched blindly, it would appear to be equally rational to switch back in order to increase one's expected value by 1.25. Such distributions, though, can no more be instantiated in real games than Hilbert's Grand Hotel can be built.

Yes, absolutely, and this is specifically beyond the OP. The distributions we've been talking about have almost always been (or should have been) unknown to the player. — Srap Tasmaner

Right, the player doesn't have that perspective but we do when discussing hypotheticals with specific distributions.

The player doesn't even know that there is some selection process. There are values in envelopes. How they got there can be discussed, and that can be interesting when the player has, say, partial knowledge of that process, but it is not the source of the paradox, in my opinion. — Srap Tasmaner

What is the source of the paradox in your view?

As I see it, the source of the paradox is the idea that the player can calculate the expected gain conditional on the observed amount. That would require knowing the specific probabilities for {v/2,v} and {v,2v} which is just what she doesn't know.

On the assumption, of course, that the player takes this initial distribution to be bounded above by M for some (possibly uncknown) M; or [...] — Pierre-Normand

Yes, which I take to be a reasonable (physical) assumption.

n the case of the two envelopes paradox, the case is similar. The player never has the opportunity to chose which branch to take at the first node. So, the player must treat this bifurcation as occurring within a black box, as it were, and assign each branch some probability. But, unlike my example with two equally biased dice, those probabilities are unknown. — Pierre-Normand

I like this explanation. And I thought of a possibly better way explain how "unknown" is used in the TEP, by analogy:

• Say you are given a geometry problem where one angle is represented by the unknown "x", and you are asked to solve for another angle "y."
• Your answer will be expressed in functional form, something like y(x) = 60° + x + arctan(sin(x)/2).
• The unknown "x" can be any angle in some range, and is more properly called an independent variable.
• It is true that the value of "y" is not known. But it is not an "unknown," it is a dependent variable that has a very specific relationship to x. That relationship is determined by the details of the OP and the laws of geometry.

Even if you treat the independent random variable V, for the value v in your envelope, (or the smaller value x of the pair) "as an unknown," the value y in the other envelope is still represented by the dependent random variable Y that has a very specific relationship to X, as determined by the OP and the laws of probability.

The point is that "as an unknown" is a description of your knowledge, not the role played by a value in the problem.