The two envelope paradox... AAARGH!

[Fist and Faith]

The chances that you picked X in the first place are 50/50.
The chances that you picked 2X in the first place are 50/50.
If you change envelopes, the chances of changing to X are 50/50.
If you change envelopes, the chances of changing to 2X are 50/50.

[TheFallen]

Okay after a night's troubled sleep where I couldn't get this damn thing off my mind, I've come to the following conclusions (which of course may be entirely wrong):

Vraith, yes I totally agree that "There IS NO GENERAL rule about whether to switch or not in 2-envelope conditions". However, your point about literally not being able to lose is irrelevant - the question was whether you could do anything to influence your chances of "winning bigger", as you put it. My answer is that no, you cannot.

This would suggest that my Scenario #2 above...

Let's call the amount of money in one envelope F and in the other envelope 2F. This is also a correct mathematical expression of the conditions at this point. You pick an envelope - it MUST BE irrelevant whether you open it or not, since you'll either have chosen F or 2F, and opening the envelope won't tell you which you have chosen. If you have unknowingly chosen F and then swap, you'll end up having gained an extra F in comparison to your original envelope choice in so doing. If you have unknowingly chosen 2F and then swap, you'll end up having lost an F over your original choice in so doing. The same is true in reverse if you don't swap. This scenario would seem to show that there is no difference in benefit at all, whether you swap or don't swap.

...is correct in all aspects. But there's still a problem, more on which in a second.

However firstly, Hashi, I'm sorry, but I've decided that you're talking rubbish when you state that opening the first envelope to discover the amount inside has any bearing on things whatsoever. It can't - you're still every bit as blind as to the amount in the second envelope at that stage. All you know is that it's either double or half the amount that you've discovered in the first envelope - which you already knew before you discovered the first amount. That doesn't affect any equation in any possible way, nor can knowing the first amount (as compared to not knowing the first amount) weight strategy to maximise gain in any possible way. The difference between "knowing" and "not knowing" is I am afraid a complete red herring

Having said all that, I still can't fault the reasoning in Scenario #1:

Let's call the amount of money in the first envelope you choose A. You know that the amount of money in the second envelope is either 2A or A/2. This is a correct mathematical expression of the conditions at this point. Because of this, it'd make sense to swap to the second envelope, since your potential gain is higher. If you have A in your hand (and to my mind, utterly regardless of whether you know what the value of A is), then swapping to an alternate that is either 2A or A/2 would seem advantageous on average... as per Hashi's reasoning. He points out that swapping would on average convey a 25% increased gain as compared to not swapping.

Vraith, you say that I am EITHER
a) Thinking the two are asking the same question/describing the same problem [which they are not]. Where are they not? Where is the inaccuracy?
OR
b) Mis-writing one of the equations. Where?

Fist, I completely agree wth the following statement of yours:

-If the envelope you originally pick has X, the other has 2X.
-If the envelope you originally pick has 2X, the other has X.
-You will never reveal X/2. There's no such thing.
OR
-If the envelope you originally pick has X, the other has X/2.
-If the envelope you originally pick has X/2, the other has X.
-You will never reveal 2X. There's no such thing.

You are exactly right. That's a re-hash of my Scenario #2 above, as is your post immediately preceding this. I accept that this is entirely correct.

However, what I REALLY want to know is this:-

What is wrong (because something must be wrong) with viewing the problem in Scenario #1 terms? So, you open envelope 1 and find $100. AT THAT POINT, surely it is both logical and correct to assume that Envelope 2 MUST contain either $200 or $50. On that basis, Hashi's "you should definitely swap" would seem to apply... except that I know that it's wrong. I just don't know why it's wrong.

Hang on...

Because, no matter how many times you play, and no matter if you keep your envelope or switch, one of those three amounts will never show up. If X is 100, your math in figuring out the average of many trials can include 100 and 200, or 100 and 50. It cannot include 50, 100, and 200, because all three amounts do not exist. You don't sometimes get X/2, and sometimes 2X...

...But it could not have gone either way. It could only be one or the other.

Hmmm. That's the closest thing I've seen so far to explaining why my delineation of Scenario #1 is fallacious (and why Hashi's "always swap" reasoning is wrong). It's not entirely satisfactory, but I'm kind of getting my head around it... has anyone got a simpler/more clear way of highlighting the fallacy?

[Fist and Faith]

The reason Scenario #1 is fallacious is that you're trying to get information that doesn't exist. That is, you're trying to figure out whether it's better to keep your original envelope, or switch. It's not better either way. It's 50/50. No fiddling with things, or looking at things in a different way, will change that. And anything that says one way is better than the other is wrong. It must have been arrived at through illegitimate means. In Scenario #1, it's because you're trying to make the two possibilities of #2 into one possibility. You might be on X, and switch to 2X. Or you might be on 2X, and switch to X. But you're not on X, and might switch to .5X or 2X. If X is 100, 50 and 200 are not both possibilities. No, you don't know if you're originally on X or 2X, so it looks like switching might give you .5X or 2X. But it won't. It only looks that way because you lack too much information. You're either on X and you might switch to 2X, or you're on 2X and you might switch to X. Trying to find an equation that includes .5X, X, and 2X, so that you come up with anything other than 50/50, is wrong.

EDIT: You might say, "But if I switch, I might end up with half of what I have now, or I might end up with twice what I have now. How is that not .5X or 2X?" It's really half of 2X or twice X.

[TheFallen]

Okay. THANK YOU!

I got disheartened when I read your saying "And anything that says one way is better than the other is wrong. It must have been arrived at through illegitimate means. " I knew this already - from the same a posteriori reasoning... I just didn't know WHAT the illegitimate means were.

But then - thank GOD - you highlighted where the fallacy was. And nailed it with that final edit. So, FnF, you've managed to stop my brain leaking out through my ears.

Mind you, I'd feel even more relaxed if Hashi - who's normally pretty good with these sorts of conundrums (conundra? ) would post his agreement with your summation - and in so doing, admit that he originally got this one wrong. Shock horror!

*** SUDDEN LIGHTBULB MOMENT ADDITION ***

Okay I suddenly realised where the mis-assumption was. There is NOTHING wrong with the equation in Statement #1... the amount of money in the 2nd envelope MUST BE either double or half the money in the 1st envelope... BUT THE CRUCIAL POINT IS that the amount in envelope #2 IS NOT A VARIABLE. It's ALREADY BEEN DEFINED at game-start, even though it's unknown to the player so no actions on the part of the player can validly be based upon the possibility of it being variable.

Vraith and FnF, I realise that this is what you've both been trying to say to me with your "it's a pure binary situation" statements, but I get it now. Woohoo!

Here's what I finally realised:-

The ONLY way that Hashi's "you oughtta swap" insistence would have been right would have been for there to be no evelope #2, but for the host to tell the player "Okay, you can either stick with envelope #1 or you can choose to flip a coin. Heads you double your money, tails you lose half. Whaddya wanna do? Stick or flip?" In that situation, of course you flip as per Hashi's reasoning... but that's only because the alternate amount HAS NOT YET BEEN DEFINED. Treating it as a possible variable in the original two envelope scenario is therefore just plain wrong.

Whew! Now I'm happy - whether Hashi posts his agreement or not

[Hashi Lebwohl]

Whew! Now I'm happy - whether Hashi posts his agreement or not

I would hope that you would be happy whether I post any agreement or not under any cicumstances

This thread shows how a relatively simply scenario can become extremely complicated when analyzed sufficiently.

[Vraith]

Vraith and FnF, I realise that this is what you've both been trying to say to me with your "it's a pure binary situation" statements, but I get it now. Woohoo!

Cool...Props where props are due: FnF came up with the necessary clarity I was searching for but couldn't manage for some reason, so the credit is all his, really. [a couple times I was confusing myself trying to come up with it and dragged myself away from my own goal...honestly, I fairly regularly get caught in my own brain loops like you did with this...a couple times you nearly had me convinced the inconsistency was real!]

[wayfriend]

WF, you really need to read through the entire thread! I've already pointed out that the logical endpoint of Hashi's reasoning is an infinite swapping of envelopes (but only presuming that one doesn't know the value of the first envelope - and Hashi states that his "swap" reasoning doesn't apply if this is the case - a thing I find mind-boggling and scarcely credible).

Yes, sorry about that.

But this merits some further discussion.

Logic dictates that you swap. But then logic dictates that you swap again? Even though this brings you back to the original envelope -- which logic had already dictated you should trade away?

It's plain there is a flaw in this thinking.

Here's the flaw.

Consider the second swap. You can only increase your money if the first swap decreased it. And you will only decrease your money if the first swap increased it.

What it means is that these events are not ORTHOGONAL. They aren't independent - the results of events are influenced by the results of earlier events. When events are not orthogonal, calculating probablities is much harder. It's not P(A) times P(B) any longer.

Flipping a coin twice produces orthogonal events. The second flip is independent of the first slip.

Therefore the flaw is treating these as orthogonal events - the flaw is assuming that the result of the second swap is independent of the result of the first swap.

In actually, the result of the second swap is highly dependent on the result of the first swap. There is no possibility to swap and improve your money, then swap again and improve your money again.

Which is fairly obvious in retrospect.

Here's the math:

The odds of first swap increasing your money is: 0.5

The odds of the second swap increasing your money (over the result of the first swap) is (0.5)*(0) + (0.5)*(1.0) = 0.5.

(That is, the odds of the first swap increasing your money times the odds of the second swap increasing your money, plus, the odds of the first swap decreasing your money times the odds of the second swap increasing your money.)

The odds of swapping twice and resulting in an overall increase is (0.5)*(0) + (0.5)(0) = 0.

(That is, the odds of the first swap increasing your money times the odds of the second swap not decreasing it, plus, the odds of the first swap decreasing your money times the odds of the second swap gaining back more than what was decreased.)

It turns out I do remember some of my college probability class.

- - - - - - - -

I will keep working at some other things. Hashi's argument "(.50)(50) + (0.50)(200) = 125" is also flawed, but I need to work out how.

[Hashi Lebwohl]

I will keep working at some other things. Hashi's argument "(.50)(50) + (0.50)(200) = 125" is also flawed, but I need to work out how.

It likely has to do with the fact that I--at random--chose that the envelope you open contains $100, so the other envelope could contain either $50 or $200. I am uncertain that an actual dollar amount was given in the original problem.

[wayfriend]

I will keep working at some other things. Hashi's argument "(.50)(50) + (0.50)(200) = 125" is also flawed, but I need to work out how.

It likely has to do with the fact that I--at random--chose that the envelope you open contains $100, so the other envelope could contain either $50 or $200. I am uncertain that an actual dollar amount was given in the original problem.

All I know is, the odds of having the larger envelop after swapping should be exactly the same as when not swapping. So there should be no advantage to swapping.

The great thing about math is you can calculate things in different ways but the consistency of mathematics is such that you get to the same answer regardless.

So if you have two ways of calculating an answer and the answers are different, one of them must be flawed somewhere.

In this particular case, there is one possible universe where the envelopes are $100 and $200. And there is another possible universe where the envelops are $50 and $100. If you have an envelope containing $100, which universe are you in? I think the flaw is thinking that this has a probability of 50/50 - or any probability at all. It's a very subtle issue, I fear.

[Fist and Faith]

You guys don't agree with me, or don't understand me. In case you don't understand me, let me try it this way. (If you don't agree with me, then you're just wrong. Wrong wrong wrong. Wrong-a-dy wrong, all the day long. [Hey, I just made that one up just now! ])

The X in Scenario #1 is not a mathematical X. It's just an "I wonder how much money is in this envelope" X. Why? Because the mathematical X must represent a single, specific number. If you have X in the first envelope, and switching envelopes might get you .5X or 2X, then X=0. That's the only way the math can work.

But we're talking about a positive amount of money, so neither envelope can have 0 in it. And, if we want to use math to determine the odds of coming out ahead by switching envelopes, we cannot say "The original envelope has X, and the envelope we switch to might have .5X or 2X." If we switch and come out ahead, then we originally had X, and switched to 2X. If we switch and lose, then we originally had 2X, and switched to X.

[TheFallen]

Okay, fortunately, my calm is not shattered...

WF, you are entirely right. Being absolutely no mathematician - gee, really? Whoda thunk? - I didn't know the term "orthogonal" and I'll take it that this is the correct term for what I more prosaically attempted to describe in talking about pre-defined amounts and true variables. Actually, my halting effort to describe why scenario #1 was wrong - as it duly had to be - mirrors your "two universe" statement. As to Hashi's argument being wrong, you've already answered this yourself (as I did). I *think* Hashi's argument can only be right when the amount in envelope #2 is a true variable (as per my coinflip exemplar, where he'd be right to recommend a swap - see below). However, since, as you have completely correctly pointed out, the amount in envelope #2 has been pre-defined at game start, it is thus not a potential variable and therefore there is no benefit to swapping.

Hashi, you're right when you say that I defined no monetary amounts in my initial exposition of the apparent paradox, but I think (but am not at all sure) that this is irrelevant. Or is it? Can the simple act of opening the first envelope (i.e. actually defining X) radically and mathematically alter the state and condition of things?

FnF, I'd like to think I *have* understood you. Well, I must have to a certain extent, since it was your posts that got my tortured brain to a resolution. However, I think your most recent post has gone off on an incorrect tangent:

The X in Scenario #1 is not a mathematical X. It's just an "I wonder how much money is in this envelope" X. Why? Because the mathematical X must represent a single, specific number. If you have X in the first envelope, and switching envelopes might get you .5X or 2X, then X=0. That's the only way the math can work.

It doesn't have to be an "I wonder how much money is in this envelope" X if you open the first envelope. Does the act of opening the envelope (and thus giving X a value) change things? How can it? What happens if you do open the first envelope and find a $100 bill in it? That's then a clearly defined, single, specific number... so therefore we absolutely can say "by the known rules of the game, the envelope we switch to might have .5X or 2X." In fact it must have either 0.5X or 2X. And just because I'll get either get $50 or $200 if I swap, surely doesn't make the $100 I already have in my hand zero, mathematically speaking, does it?

If we switch and come out ahead, then we originally had X, and switched to 2X. If we switch and lose, then we originally had 2X, and switched to X.

Yes that is absolutely true, but again, all you're doing here is validating Scenario #2, rather than disproving Scenario #1.

Oh no... I'm starting to have a wobble! I think I need a recap again.

Scenario #1.
Let's call the amount of money in the first envelope you choose A. You know that the amount of money in the second envelope is either 2A or A/2. This is a correct mathematical expression of the conditions at this point. Because of this, it'd seem to make sense to swap to the second envelope, since your potential gain is higher. The theory goes that, if you have A in your hand, then swapping to an alternate that is either 2A or A/2 would seem advantageous on average... as per Hashi's argument. He points out that swapping would on average convey a 25% increased gain as compared to not swapping.

Okay, everyone now agrees that the above is fallacious, although some of us are unsure as to where the fallacy lies. I ***know*** that it's just wrong. Wrong wrong wrong. Wrong-a-dy wrong, all the day long (to quote Fisty). I'm just not entirely sure where it's wrong, because it looks right.

Scenario #2.
Let's call the amount of money in one envelope F and in the other envelope 2F. This is also a correct mathematical expression of the conditions at this point. You pick an envelope - it MUST BE irrelevant whether you open it or not, since you'll either have chosen F or 2F, and opening the envelope won't tell you which you have chosen. If you have unknowingly chosen F and then swap, you'll end up having gained an extra F in comparison to your original envelope choice in so doing. If you have unknowingly chosen 2F and then swap, you'll end up having lost an F over your original choice in so doing. The same is true in reverse if you don't swap. This scenario would seem to show that there is no difference in benefit at all, whether you swap or don't swap.

Okay, everyone agrees that the above is completely correct. So much so that it's the prime reason that Scenario #1 is wrong - but this a posteriori reasoning doesn't in itself highlight where the fallacy in Scenario #1 lies.

Let me help - or possibly confuse things more.

Scenario #3.
There's only one envelope which you are given and open. It contains a $100 bill. You're then told that you may either keep that $100 bill, or agree to a coin flip. If you agree to this and call the coin flip correctly, you're given $200. If you call the coin flip wrongly, you're given $50.

Does anyone doubt that, in the above scenario, it's a good idea to go for the coin flip? Because Hashi's argument about a 1.25 return seems to now apply, on the basis of (0.5 * 50) + (0.5 * 200) = 1.25? If so, the fallacy in Scenario #1 lies in the difference between it and this Scenario #3. And the only substantive differences I can see are that:-

A) in Scenario #1, the amount in the second envelope is pre-defined before the game starts, whereas in Scenario #3, the amount that's effectively in the absent second envelope gets defined by the coin flip,

But hang on... what happens if the gamemaster, having stuffed a $100 into envelope #1, decided the amount to be stuffed into envelope #2 prior to game start by a coinflip - which you of course know nothing about? It cannot be the coin flip itself that matters, so it must be the rigid pre-definition of the amounts prior to game start, no? But can such pre-definition really make that huge difference? Are my and WF's "two universe" theories really correct? (I actually hope so - because then it'd fit into what Terry Pratchett calls the"trousers of time" view on the cosmos )

Anyway, here's the other substantive difference.

B) in Scenario #3 you open the first envelope and thus identify the amount.

If this is the important difference, it seems really quantum/Schroedinger's cat though. How can *knowing* the contents of envelope 1 affect things mathematically? Or is this a red herring and it's the the fact that the winnable amount has already been pre-defined that's important?

And what happens if there's a 4th Scenario?

Scenario #4.
There's only one envelope which you are given but DO NOT OPEN. You're then told that you may either keep whatever's in that envelope, or agree to a coin flip. If you agree to this and call the coin flip correctly, you're given twice the amount that turns out to be in your envelope. If you call the coin flip wrongly, you're given half the amount.

Is there any advantage to going for the coin flip in this scenario?

*** ADDED FURTHER LIGHTBULB MOMENT ***

Hang on a second... Hashi, I now think you may be right (but I do not understand why) when you say, provided that you open the first envelope and have identified the amount therein (call it $100), you should swap! Geez... because I think this then becomes exactly the same as ignoring envelopes altogether, taking a $100 bill out of your wallet and betting it on a coin-flip, but one where the odds aren't double or nothing, but double or a half.

Guys... I'm tail-spinning here. And after I thought I'd got it. Help!

[Fist and Faith]

Can't read your post right now, TF, but I wasn't referring to you. You understood and agreed with me. Which is what I like best. My wife does neither, ever.

[Fist and Faith]

FnF, I'd like to think I *have* understood you. Well, I must have to a certain extent, since it was your posts that got my tortured brain to a resolution. However, I think your most recent post has gone off on an incorrect tangent:

The X in Scenario #1 is not a mathematical X. It's just an "I wonder how much money is in this envelope" X. Why? Because the mathematical X must represent a single, specific number. If you have X in the first envelope, and switching envelopes might get you .5X or 2X, then X=0. That's the only way the math can work.

It doesn't have to be an "I wonder how much money is in this envelope" X if you open the first envelope. Does the act of opening the envelope (and thus giving X a value) change things? How can it? What happens if you do open the first envelope and find a $100 bill in it? That's then a clearly defined, single, specific number... so therefore we absolutely can say "by the known rules of the game, the envelope we switch to might have .5X or 2X." In fact it must have either 0.5X or 2X. And just because I'll get either get $50 or $200 if I swap, surely doesn't make the $100 I already have in my hand zero, mathematically speaking, does it?

If we switch and come out ahead, then we originally had X, and switched to 2X. If we switch and lose, then we originally had 2X, and switched to X.

Yes that is absolutely true, but again, all you're doing here is validating Scenario #2, rather than disproving Scenario #1.

But you don't see any .5X in Scenario #2. Because there's no such thing. Not when we're figuring the odds mathematically. Even if we don't know, even if we never learn, the amounts in each envelope, if we accept that there is 100 in one and 200 in the other, then we only ever have X and 2X. When we're mathematically figuring out the odds, there is no .5X. No envelope has 50. And when we try to use .5X to figure out the odds, we come up with something other than 50/50.

Scenario #1 is correct only when you have to decide one time. Yes, there's a 50/50 that you'll end up with half of what's in your hand or twice what's in your hand. But when you start calling it "X", and putting it into equations, X must always equal the same thing. You can't have X=100 in one trial, when you ended up with 2X (200) after switching; and X=200 in another trial, when you ended up with .5X (100) after switching.

[TheFallen]

But you don't see any .5X in Scenario #2. Because there's no such thing. Not when we're figuring the odds mathematically. Even if we don't know, even if we never learn, the amounts in each envelope, if we accept that there is 100 in one and 200 in the other, then we only ever have X and 2X. When we're mathematically figuring out the odds, there is no .5X. No envelope has 50. And when we try to use .5X to figure out the odds, we come up with something other than 50/50.

I KNOW there's no 0.5X in Scenario #2. Scenario #2 is so simply expressed that it has to be true. One can totally correctly state that in the game, there's ONLY EVER X and 2X involved - you're going to pick one of those, so if you swap, you're going to end up with the other one. 50/50 that you'll gain or lose X by swapping, so it is NEITHER advantageous NOR disadvantageous to swap. This is surely a clear fact.

Scenario #1 is correct only when you have to decide one time. Yes, there's a 50/50 that you'll end up with half of what's in your hand or twice what's in your hand. But when you start calling it "X", and putting it into equations, X must always equal the same thing. You can't have X=100 in one trial, when you ended up with 2X (200) after switching; and X=200 in another trial, when you ended up with .5X (100) after switching.

But but but... Scenario #1 CANNOT be right, even for one single occasion. Why? Because Scenario #2 is ALWAYS right. Thus there can NEVER be an advantage in swapping, even on one single occasion, can there? And what's more, if there was an advantage in swapping, even on one single occasion, then there would be an advantage in swapping on every occasion... because every occasion can be quite correctly viewed as its own single occasion. It's just that the actual value of X would differ on every occasion, but so what? This can't be right.

Anyhow, some further questions...

a) In a two envelope game, does it make a difference if you open the first envelope or not? (I can't see why it would).

b) In a single envelope game, where you are offered the chance to either keep the money in that envelope, or call a coin flip to either double the money or halve it, should you call the coin flip if you HAVE opened the envelope in advance of it? (I'd say absolutely yes, as per Hashi's (0.5 * 0.5) + (0.5 * 2) = 1.25 argument, but I'm just checking).

c) In a single envelope game, where you are offered the chance to either keep the money in that envelope, or call a coin flip to either double the money or halve it, should you call the coin flip if you HAVE NOT opened the envelope in advance of it? (I'd still say absolutely yes, again as per Hashi's (0.5 * 0.5) + (0.5 * 2) = 1.25 argument, but I'm just checking).

Re b) and c) above, can we agree upon it making no difference if you open the envelope or not? And can we agree that you should go for the coin flip, either in one or both scenarios?

d) Say there's a single envelope game. You're given the envelope. You're told that you may either keep what's in the envelope, or you can lift a cup to reveal a coin that's already been randomly flipped. You're told that, if that coin is heads, your money is doubled. And that, if it's tails, your money is halved. So... the coin has already been flipped at game start. Should you go for the coin option in this case?

This last d) question is interesting, especially in comparison to b) and c) above. I'd say that in d) it MUST MAKE NO DIFFERENCE if you go for the coin option or not. If you decide to reveal the coin, it becomes exactly like the random X or 2X of Scenario #2. All the revealed coin does is define whether you had X or 2X in your envelope, even though there's no other envelope. If the revealed coin is heads, you originally had X, if the revealed coin is tails, you originally had 2X... on that specific single occasion. Oh, and again I don't think it can make any difference in d) if you open your envelope or not before deciding to go for the coin reveal.

But compare d) to b) and c) for a second. If you're right to go for the coin flip in either or both of b) and c), but it's irrelevant to go for the coin flip in d), then that must mean that it's either the fact that the coin flip has already happened or the fact that you don't get to call the coin flip that is the significant fact...

This gets weirder - and ever more like Schroedinger's cat.

WF, what are your current thoughts on this?

[Fist and Faith]

I am fully aware that I'm saying things wrong, and even saying at least one wrong thing. I'm having a tough time getting this out correctly.

If we're going to look at it mathematically, we have to use Scenario 2. But if we want to just look at it casually, Scenerio 1 is ok. Two envelopes, one has twice as much as the other. What are the chances switching will give us less? 50/50. What's the chances switching will give us more? 50/50. And yes, we can do it repeatedly that way. I was wrong to say only once for Scenario 1. (Not entirely sure now what I was thinking when I said that. ) Try it any number of times, and the odds are 50/50 each time. And overall, the odds are 50/50.

We just can't use Scenario 1 mathematically, ie, in equations, because of the issue with X needing to represent the same thing each time. Can't use .5X and 2X in equations. The proof of this is that trying to do so lead to something other than 50/50.

[TheFallen]

Hehehehe... this is a head rush for sure.

I am fully aware that I'm saying things wrong, and even saying at least one wrong thing. I'm having a tough time getting this out correctly.

If we're going to look at it mathematically, we have to use Scenario 2.

I'd agree with that fully, if only because Scenario #2 is invariably mathematically correct... which means that Scenario #1, in offering a different recommendation, must invariably be wrong.

But if we want to just look at it casually, Scenerio 1 is ok. Two envelopes, one has twice as much as the other. What are the chances switching will give us less? 50/50. What's the chances switching will give us more? 50/50. And yes, we can do it repeatedly that way. I was wrong to say only once for Scenario 1. (Not entirely sure now what I was thinking when I said that. ) Try it any number of times, and the odds are 50/50 each time. And overall, the odds are 50/50.

Yes, but you're not addressing the key point - that, after swapping, when you get less, you're seemingly getting LESS "less" and when you get more, you're seemingly getting MORE "more". The question isn't about whether you get less or more if you swap, because of course you do - one or the other. It's about whether it's advantageous to swap - because there are only ever two possible outcomes: worst case, where you can only lose half of what's in the first envelope, and best case where you can gain double what's in the first envelope. That'd seem to clearly suggest that it is indeed advantageous to swap - as per Hashi.

I'm now coming down to the fallacy in Scenario #1 possibly being one of misassumed repeatability. Does it instead need to be considered as a completely discrete "one-off" every single time? If so, then averaging wouldn't come into play and things might therefore be entirely random... I dunno.

[Zarathustra]

My head is hurting after a page and a half of this. I'll have to read the rest later, but I think that FnF has the best insight here.

The problem is a labeling issue, I believe.

You have two unknown quantities (two envelopes), so you should be using two variables instead of one. In the original wording of the OP, there’s an issue with your frame of reference that's hidden by using only one variable per scenario. For scenario 1, you're labeling the uncertainty in the other envelope (y) in terms of the uncertainty of your envelope (x). Thus for scenario 1: y (other envelope) = 2x (2 times your envelope) or ½x (half your envelope). Thus x (your envelope) could be either 1/2y or 2y.

The second scenario makes a similar mistake of creating unnecessary confusion, by labeling the uncertainty for both envelopes in terms of single variable: 1st envelope = y and 2nd envelope = 2y. If you pick one envelope, it’s true that it’s either y or 2y. But we're labeling the situation in terms of 1st envelope, when we could just as easily label in terms of the 2nd. Call 1st envelope x and 2nd envelope z, and thus x = y and z = 2y. Now if we define the uncertainty in terms of 2nd envelope, z (instead of the 1st envelope, as TF did) your choices between x and z can be phrased as "2x or 1/2z" which is the same description as Scenario 1. It's also the same as saying "2y or y," which is the description for Scenario 2. But this choice of reference frame, and the possibility of switching reference frames is hidden.

Something like that.

[Vraith]

My head is hurting after a page and a half of this. I'll have to read the rest later, but I think that FnF has the best insight here.

He absolutely explained it best. [at least so far].

Scenario 1 describes what SEEMS to be an odds/best outcome for switching.
But it does NOT in fact describe that. Statistical/probable calculations ONLY work when outcomes are ACTUALLY possible. That is NOT the case.
If there are 2 envelopes, and X is assigned to represent one "value," then the "second draw" is NOT described by Hashi's equation. Hashi's equation only describes one SIDE of the equation...because the second draw depends completely on the first draw.
As Hashi writes it, there are actually 3 "variables" implied, [x, 2x, x/2]. But only 2 of them exist in fact.
If you want to do it H's way, the full description requires 2 equations.
Keeping equation...[.5[50] + .5[200]]=125
Switching equation [.5[50] +.5[200]=125
Two things equal to the same thing are equal to each other.
Keeping=switching.

Everything else is an artifact of us "believing" that 200 is "better" than [the nonexistent] "average" of 125, is "better" than 50, and "believing" that someone COULD end up with 200, others with 50, others with 100...when in fact...as FnF already said...ONE of those will NEVER happen.

[TheFallen]

The problem is a labeling issue, I believe.

You have two unknown quantities (two envelopes), so you should be using two variables instead of one.

Okaaay...
First off, yes there are two variables, sure - but there are only two very definite pre-ordained relationship possibilities between those two variables. They're not entirely random.

In the original wording of the OP, there’s an issue with your frame of reference that's hidden by using only one variable per scenario. For scenario 1, you're labeling the uncertainty in the other envelope (y) in terms of the uncertainty of your envelope (x). Thus for scenario 1: y (other envelope) = 2x (2 times your envelope) or ½x (half your envelope). Thus x (your envelope) could be either 1/2y or 2y.

You are absolutely right to point out that in Scenario #1, the uncertainties in the envelopes are labelled AFTER the first envelope has been picked. In the OP, I labelled the amount in the first and PICKED envelope as x and thus defined the two possible amounts in the UNPICKED second envelope as either 2x or 0.5x

Okay, what you've done is AFTER picking, is to define things in terms of the UNPICKED second envelope. You've labelled the amount in the second and UNPICKED envelope as y and thus (rightly) defined the two possible amounts in the PICKED first envelope as either 0.5y or 2y.

So... let's run with this ball. Let's say my first picked envelope happens to contain 0.5y... if I swap to the second unpicked envelope, I'll get an envelope containing y, so my gain over my first picked envelope is 0.5y. Or let's say my first picked envelope happens to contain 2 y... if I swap to the second unpicked envelope, I'll get an envelope containing y, so my loss over my original envelope is y.

What you've done here by turning things around, Z is apparently manage to prove that it's DISadvantageous to swap... ??? I thought the general opinion was that there was no advantage or disadvantage to swapping? But hang on... I'm going to combine your thoughts with V's in a moment.

But first...

The second scenario makes a similar mistake of creating unnecessary confusion, by labeling the uncertainty for both envelopes in terms of single variable: 1st envelope = y and 2nd envelope = 2y. If you pick one envelope, it’s true that it’s either y or 2y. But we're labeling the situation in terms of 1st envelope, when we could just as easily label in terms of the 2nd. Call 1st envelope x and 2nd envelope z, and thus x = y and z = 2y. Now if we define the uncertainty in terms of 2nd envelope, z (instead of the 1st envelope, as TF did) your choices between x and z can be phrased as "2x or 1/2z" which is the same description as Scenario 1. It's also the same as saying "2y or y," which is the description for Scenario 2. But this choice of reference frame, and the possibility of switching reference frames is hidden. Something like that.

I think you've got this one wrong. The CRUCIAL difference in the Scenario #2 way of viewing things is that the amounts in the envelopes get labelled BEFORE picking happens. They are correctly labelled y and 2y. I then pick one at random. I must therefore have picked either y or 2y. If I've picked y and swap, I make a gain of y over my original envelope. If I've picked 2y and swap, I make a loss of y over my original envelope. Ergo, no benefit or disadvantage in swapping... I'm randomly going to end up with either y or 2y. That just feels like the proper answer and I'm sure it's absolutely true... but it doesn't explain where the Scenario #1 view is fallacious.


Okay back to a Zar+Vraith fusion...

Scenario 1 describes what SEEMS to be an odds/best outcome for switching.
But it does NOT in fact describe that. Statistical/probable calculations ONLY work when outcomes are ACTUALLY possible. That is NOT the case.
If there are 2 envelopes, and X is assigned to represent one "value," then the "second draw" is NOT described by Hashi's equation. Hashi's equation only describes one SIDE of the equation...because the second draw depends completely on the first draw.
As Hashi writes it, there are actually 3 "variables" implied, [x, 2x, x/2]. But only 2 of them exist in fact.
If you want to do it H's way, the full description requires 2 equations.
Keeping equation...[.5[50] + .5[200]]=125
Switching equation [.5[50] +.5[200]=125
Two things equal to the same thing are equal to each other.
Keeping=switching.

Okay, I think I see. V, what you've done is to ALWAYS define things in terms of the envelope you don't end up with. I could reconstrue your two possible equations as follows:-

Keeping Strategy... Envelope 1 = (0.5(0.5 * Env 2)) + (0.5(2 * Env 2)) = 1.25 Env 2.

Switching Strategy... Envelope 2 = (0.5(0.5 * Env 1)) + (0.5(2 * Env 1)) = 1.25 Env 1.

As you say, that shows both choices to be equal, so no benefit, no disadvantage. That's GOOD!

But... heh. How about these related questions?

A. Say I give you $100. I then say to you that you can either keep that $100, or you can go for a coin flip, where I'll flip a coin and you call it. If you call it correctly, I'll give you $200 instead of your original $100. If you call it incorrectly, I'll give you $50 instead of your original $100.

It's obvious what to do for the best, right? Go for the coin flip?

B. Say I give you a sealed envelope with an unknown amount of money in it. I then say to you that you can either keep that unknown amount of money, or you can go for a coin flip, where I'll flip a coin and you call it. If you call it correctly, I'll give you double that original unknown amount of money instead. If you call it incorrectly, I'll give you half that original unknown amount of money instead.

It'd seem obvious what to do for the best, right? Go for the coin flip again?

C. Say I give you $100. I then say to you that you can either keep that $100, or you can reveal an already randomly flipped coin under a cup. If that coin shows heads, I'll give you $200 instead of your original $100. If that coin shows tails, I'll give you $50 instead of your original $100.

What should you do here? Or does it make no difference?

D. Say I give you a sealed envelope with an unknown amount of money in it. I then say to you that you can either keep that unknown amount of money, or you can reveal an already randomly flipped coin under a cup. If that coin shows heads, I'll give you double that original unknown amount of money instead. If that coin shows tails, I'll give you half that original unknown amount of money instead.

What should you do here? Or does it make no difference?

I'm interested if there is any difference between A and B above - i.e. is it significant that the amount in the envelope is unknown in these conditions.

I'm interested if there is any difference between C and D above - i.e. again is it significant that the amount in the envelope is unknown in these conditions.

I'm especially interested if there is any difference between A and C above (or B and D) - i.e. is it significant that the coin has already been flipped?

[Hashi Lebwohl]

A. Say I give you $100. I then say to you that you can either keep that $100, or you can go for a coin flip, where I'll flip a coin and you call it. If you call it correctly, I'll give you $200 instead of your original $100. If you call it incorrectly, I'll give you $50 instead of your original $100.

The expectation here would be (0.5)[50 + 200] = 125 and so technically it would in your best interest to accept the results of the coin toss *but* I would recommend just keeping the $100.

B. Say I give you a sealed envelope with an unknown amount of money in it. I then say to you that you can either keep that unknown amount of money, or you can go for a coin flip, where I'll flip a coin and you call it. If you call it correctly, I'll give you double that original unknown amount of money instead. If you call it incorrectly, I'll give you half that original unknown amount of money instead.

Again, if the amount in the envelope is x the expectation will wind up being 1.25x and this would point to "accept the flip"; however, once again my advice would be "a bird in the hand is worth two in the bush" and thus keep the envelope with the unknown amount of money in it.

C. Say I give you $100. I then say to you that you can either keep that $100, or you can reveal an already randomly flipped coin under a cup. If that coin shows heads, I'll give you $200 instead of your original $100. If that coin shows tails, I'll give you $50 instead of your original $100.

Schrodinger. The coin is neither heads nor tails until the cup is lifted so the real results still equal an expectation of 125. This scenario is no different than scenario A, with the exception that the coin flip happened first.

D. Say I give you a sealed envelope with an unknown amount of money in it. I then say to you that you can either keep that unknown amount of money, or you can reveal an already randomly flipped coin under a cup. If that coin shows heads, I'll give you double that original unknown amount of money instead. If that coin shows tails, I'll give you half that original unknown amount of money instead.

Still Schrodinger like C and my advice would be "keep the envelope and just take the money in it".

In all cases, keep the original amount whether you know how much it is or not. Since you started the game with $0 you will always win.