The two envelope paradox... AAARGH!

[TheFallen]

Okay, I'm sure this one's not new, but I'd not heard it before - and it's seriously twisting my melon, man.

So you're invited - free of charge - to play a game of chance. It's very simple... there are two sealed envelopes and you know for a fact that both contain money, but that one contains double the amount that's in the other. Or, equally truthfully stated, that one contains half the amount that's in the other.

So you are told you can pick a sealed envelope and you duly do. You open it to reveal a crisp $100 bill. You're now told that you can either keep that $100, or you can discard it and choose whatever's in the second still-sealed envelope instead. Needless to say, if you do choose to switch envelopes, there's no going back.

Very simple, right? And the resultant question is simple too - what is it best to do do after opening the first envelope - stick or switch?

Okay, here's a very logical statement:

1. Let's call the amount you've revealed in the first envelope X. So, having opened the first envelope, if you switch, you may reveal 2X or you may reveal X/2. That is an inarguable fact. Your potential gain is therefore X, but your potential loss is only X/2. That'd seem to suggest it's obviously better to switch... since you'll have a 50% chance of doubling your winnings and worst case you'll only lose half of them. Those are 2 to 1 odds... pretty good in a 50/50 situation.

With me so far? Good. BUT here's another and equally very logical statement:

2. Let's call the amounts in the two sealed envelopes Y and 2Y. That is also an inarguably correct statement. So, no matter which envelope you first pick, if you then switch, you'll either gain Y or lose Y. On that basis, it's completely indifferent if you switch or not... there's no benefit in either course. Those are even odds, 1 to 1... which you'd expect in a 50/50 situation.

Both of those two statements seem to me to be entirely factual, entirely correctly phrased and both make entire logical sense. But despite this, they define two radically differing recommendations to the self-same question (stick or switch)...

How can this possibly be? Where is the fallacy that I am not spotting?

[Vain]

Your assumption around the X values is too simplistic. It
should be divided into two scenarios:

Scenario 1. Let's call the amount you've revealed in the first envelope X ($100). So, having opened the first envelope, if you switch, you may reveal 2X ($200)

Scenario 2. Let's call the amount you've revealed in the first envelope X ($200). So, having opened the first envelope, if you switch, you may reveal 1/2X ($100)


which is the same as the Y scenario of even odds. Well that's my story and I'm sticking to it

[Hashi Lebwohl]

Find the expectation of the situation, which means finding the sum of [{probability of event A)(gain from event A)]. It doesn't matter whether we say the envelopes contain "k and 2k" or "h and h/2"; the result should be the same.

sum = (0.5)(x) + (0.5)(2x) = 0.5x + x = 1.5x. So if x is $100 then playing the game nets you 1.5x in the long run.

This almost seems like a reduced version of the Monty Haul problem: three doors, the car is behind one door and the other two doors contains goats. You choose door 1 so Monty opens door 3 to reveal a goat. Do you switch your choice to door 2? Strangely, the correct answer is that you switch--check it out at mathworld.wolfram.com to see--so in this case if we follow the same logic the answer would be "switch".

I suppose we could analyze the situation differently. You open the envelope and find $100. The other envelope could contain half, meaning your envelope was the larger one, or it could contain twice, meaning your envelope is the smaller one. The expectation of the other envelope is now (0.5)(50) + (0.5)(200) = 125 so again the conclusion is that you switch envelopes.

Conclusion: switch envelopes.

[TheFallen]

Heh. Vain supports my proposition 2, but Hashi supports my proposition 1. I'm glad to see I'm not alone in my confusion.

Hashi, hang on a second...

sum = (0.5)(x) + (0.5)(2x) = 0.5x + x = 1.5x. So if x is $100 then playing the game nets you 1.5x in the long run.

Okay... that'd apply whether you swapped or didn't. If the amounts in the envelopes were $100 and $200, then randomly picking (and sticking with) an envelope 200 times would get you on average (100 x $100) + (100 x $200) = $3,000. Divide that by the 200 attempts and you get an average "per turn" return of $150, or a factor of 1.5 as you state. However, this is regardless of whether you stick or switch.

I suppose we could analyze the situation differently. You open the envelope and find $100. The other envelope could contain half, meaning your envelope was the larger one, or it could contain twice, meaning your envelope is the smaller one. The expectation of the other envelope is now (0.5)(50) + (0.5)(200) = 125 so again the conclusion is that you switch envelopes.

Your reasoning is again apparently right - your average "per turn" return is now $125, or a factor of 1.25 as you state. However, this is now entirely dependant upon your switching envelopes.

The difference in those two factors would suggest that you're better not to switch, surely? The conundrum remains. Plus you haven't explained where the fallacy is in my original statement 2 - as supported by Vain - namely:

2. Let's call the amounts in the two sealed envelopes Y and 2Y. That is also an inarguably correct statement. So, no matter which envelope you first pick, if you then switch, you'll either gain Y or lose Y. On that basis, it's completely indifferent if you switch or not... there's no benefit in either course. Those are even odds, 1 to 1... which you'd expect in a 50/50 situation.

PS This isn't at all like the car/goats Monty Haul classic problem. That's easily understood by realising that 2 out of 3 times, the host is forced to show you the other goat (because he never reveals what's behind the door you first pick). Only if you're unlucky enough to hit the door with the car behind it on your first pick (1 in 3 odds) does the host have the luxury of being able to reveal to you either goat, or to put it another way, does the host have the luxury of being able to reveal what's behind either of the remaining two closed doors. If you hit a door with a goat behind it (2 in 3 odds), the host is then compelled to reveal the other goat behind one of the remaining two closed doors.

[Hashi Lebwohl]

I said it seems like a Monty Haul-type problem.

My final analysis is the correct one. Fact: you opened an envelope and find $100. The expectation of the other envelope is (.50)(50) + (0.50)(200) = 125, thus switching will, on average, return you more money --> switch envelopes.

Even if the other envelope contains $50, though, you still win because you started the game with $0.

[I'm Murrin]

I suppose it's fairly simple, really. It's a 50/50 shot, but the risk is mitigated by the fact you only lose half in the worst case, so the risk is worth it.

That's for one. Over many tries, the average would settle directly between the minimum and maximum, somewhere around 2/3rds of the most you could have gotten - and it would averae in that area whether you switched or not.

[Vraith]

Even if the other envelope contains $50, though, you still win because you started the game with $0.

That. All appearances of conflict in the earlier posts/equations are misapplications, not a paradox in the math. Plus the fact that everyone who plays, gains.
The simple truth is that switching or not switching, in this case, unlike Monty Haul, has no effect on the odds at all.
Every single person who "plays" will end up with one of two amounts.
Every person has precisely a 50/50 chance to be the low or the high.
That 50/50 chance is not changed in any way by switching or not switching.
The situation is completely binary.
Suppose everything stayed the same except on the second draw you know you will either quadruple your money or owe double it?
[you MIGHT owe 200, you MIGHT gain 400, the calculated return is a whopping DOUBLE]
Whether it is SMART to risk it has changed...because you could actually lose money. But the ODDS are precisely the same, 50/50, for everyone switchers and non-switchers alike.

[TheFallen]

My final analysis is the correct one. Fact: you opened an envelope and find $100. The expectation of the other envelope is (.50)(50) + (0.50)(200) = 125, thus switching will, on average, return you more money --> switch envelopes.

Well, so you say, but I think it's a lot freakier than you at first think...

What happens if you aren't allowed to open the first envelope. That should logically make no difference, right? I mean, how can the act of opening the envelope possibly affect expectation?

So, according to your theory, you should swap, but now you're swapping blind. So you do swap and so open the 2nd envelope and find $50. You have no idea whether the first envelope contained $100 or $25... so where does that leave you and your calculations?

Plus, what happens if - again presuming you're not allowed to open the first envelope - you swap to the 2nd envelope (following your advice here, Hashi). Before you open that 2nd envelope, you're offered another chance to swap... back to the first envelope. Surely, if your logic is correct, you should take that choice and swap back. And then if you're offered a further swap? And another? And again ad infinitum? Doesn't that effectively mean that the logical progression of your advice is that you should eternally swap envelopes and never actually open one?

And while we're about it, if you're correct Hashi, can someone please point out the logical fallacy in this summation of the situation:

Suppose the two amounts of money are X and 2X. Here are the four possible outcomes.

i) You unknowingly pick X and don't swap. In ending up with X, you've lost out on a further amount of X.

ii) You unknowingly pick X and do swap. In ending up with 2X, you've gained a further amount of X.

iii) You unknowingly pick 2X and don't swap. In ending up with 2X, you've avoided losing an amount of X.

iv) You unknowingly pick 2X and do swap. In ending up with X, you've lost an amount of X.

In all cases above, your risk is either to gain X or to lose X. That's a perfectly balanced situation... you'll either end up with X or 2X exactly 50% of the time, completely irrelevant of whether you swap or not.

Those are 1:1 odds. Ergo, it makes no difference if you swap or not.

Now the above is, I admit, entirely counter-intuitive - but I cannot see the flaw in it.

Even if the other envelope contains $50, though, you still win because you started the game with $0.

Yes, of course you're going to end up better off, no matter what. But that wasn't the question - the question was, what, if anything, should you do to maximise your chances of achieving the greatest gain?

*** ADDED EDIT ***
I now notice that Vraith's weighed in on this one and is supporting the "it makes no difference if you swap or not" theory, which is reassuring. But I still don't entirely see where/why Hashi's wrong? (And I suspect he doesn't either).

[wayfriend]

2. Let's call the amounts in the two sealed envelopes Y and 2Y. That is also an inarguably correct statement. So, no matter which envelope you first pick, if you then switch, you'll either gain Y or lose Y. On that basis, it's completely indifferent if you switch or not... there's no benefit in either course. Those are even odds, 1 to 1... which you'd expect in a 50/50 situation.

The fallacy here is the assessment that Y is the same in both cases. This is a bad assessment. The value of Y is different in both cases. You might gain $100, but you would lose $50.

Or ... to be more precise ... if you don't know whether you have Y or 2Y, the value of Y is undetermined. If you have $100 in your hand, you don't know whether Y is $100 or Y is $50. The result of switching is based on which of the two possible values Y is. If you gain, Y was $100. If you lose, Y was $50.

You do NOT gain or lose the same amount.

But wait, that's only 1 of the fallacies.

Consider this. If you conclude that theory 1 is correct, and it's to your benefit to switch, you'd still be wrong.

You picked an envelope randomly, it's 50/50 that you got the big one. If you now switch envelopes, you still have a 50/50 chance that you have the big one.

So nowhere did you improve your odds by switching.

The falacy of your first theory is evident, but a bit harder to explain. But I have to get back on that one.

---

If you think this puzzle was interesting, you should check out the Let's Make a Deal Paradox.

• Suppose you're on a game show, and you're given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what's behind the doors, opens another door, say No. 3, which has a goat. He then says to you, "Do you want to pick door No. 2?" Is it to your advantage to switch your choice?

Edit: doh! :sheepish: Never mind.

[TheFallen]

So WF, I'm unsure if you're supporting the "it makes no difference" theory (as espoused by me, Vain and Vraith), or the "you should swap" theory (as espoused by Hashi and Murrin). Which is it, in your opinion?

I cannot for the life of me see a fallacy in stating that, at game start, one of the envelopes contains Y and one of the envelopes contains 2Y. Though I obviously do agree that the value of Y only becomes apparent once the contents of BOTH envelopes are known. But that's surely irrelevant - I don't believe there's any fallacy at all in my original statement #2.

Because I'm so confident in original statement #2 being correct, I know there MUST be a fallacy in original statement #1... but I cannot for the life of me see it. And it's driving me nuts.

As to car and goats, scroll up a little

PS This isn't at all like the car/goats Monty Haul classic problem. That's easily understood by realising that 2 out of 3 times, the host is forced to show you the other goat (because he never reveals what's behind the door you first pick). Only if you're unlucky enough to hit the door with the car behind it on your first pick (1 in 3 odds) does the host have the luxury of being able to reveal to you either goat, or to put it another way, does the host have the luxury of being able to reveal what's behind either of the remaining two closed doors. If you hit a door with a goat behind it (2 in 3 odds), the host is then compelled to reveal the other goat behind one of the remaining two closed doors.

I have to admit that I find the answer to the Monty Haul classic far FAR easier to understand than I do the apparent answer (it makes no difference if you swap or not) to this seemingly way more simple two envelope problem.

[Hashi Lebwohl]

What happens if you aren't allowed to open the first envelope. That should logically make no difference, right? I mean, how can the act of opening the envelope possibly affect expectation?

This is a completely different situation altogether. If you are not allowed to open the envelope then the expectation becomes (0.5)[k + 2k] = 1.5k. Since the envelope is blind, two situations develop:
1) you have the envelope with k and if you switch then you will receive 2k
2) you have the envelope with 2k and if you switch then you will receive k

We will also presume that the envelopes are opaque enough that you cannot see the contents of the interior and they have been weighted and padded to the point where it is impossible to tell how many bills might be in either envelope.

In this case I would recommend keeping the envelope you are given or the first one you choose. Only once the discarded envelope is opened and its contents revealed are you in a position where you wind up gaining or losing k. Still, the end result is that in the worst-case scenario you walk away with k given that you started with 0.

Now the game seems closer to Schrodinger's Cat rather than Monty Haul--you have both k and 2k in your envelope until the other envelope is opened. I still don't see any dilemma at all--there is no real hazard to any choice unless you get to open the envelope before choosing whether to switch.

[I'm Murrin]

I didn't say you should, I said you might as well.

The reason this is nothing like Monty Hall is that no new information is received after making your initial choice. (Other than, "hey, this is an amount of money I'd like to keep.")

[TheFallen]

What happens if you aren't allowed to open the first envelope. That should logically make no difference, right? I mean, how can the act of opening the envelope possibly affect expectation?

This is a completely different situation altogether. If you are not allowed to open the envelope then the expectation becomes (0.5)[k + 2k] = 1.5k.

Okay... but WHY is it different? Whether you know the amount of money in your first chosen envelope or not, there's still always only twice or half that amount of money in the other envelope. Why the Hell would knowing what was in the first envelope before deciding to swap or not change things in any way whatsoever?

On that basis, you'd now seem to be arguing that it makes no difference if you swap or not. I don't see how "knowing" makes a substantive difference in any way. It's not as if we're talking quantum here... though I agree that your Schroedinger's cat analogy seems to fit somewhat

My dilemma here is that I'm convinced that it makes no difference if you swap or not - as per my extended 4-stage working-out above. The problem that remains for me is that I cannot see the error in this statement that conflicts utterly with my "no difference" theory:

Say you initially pick an envelope containing amount H. The other envelope must then contain either 2H or H/2. It would then make sense to swap, because the potential gain is twice as large as the potential loss.

This is just a rehash of my original statement #1, which is also your original premise, Hashi. I KNOW that this must be wrong... but where?

[Vraith]

The ODDS don't depend on anything at all except number.
You've got 2, so you have a 50/50 chance. Period.
Whether you switch or not does not solely depend on the 50/50...
it also does not CHANGE the 50/50 in any way.
Whether you should switch, or not, doesn't depend on a statistical rule like Monty Haul...where switching increases your chances of winning.
It depends on the specific numerical values of the precise case, and whether you know them or not.
There is no general solution for the 2 envelope problem, only particular ones.
Hashi's switching solution only works because of the 1/2 vs. 2x potential, and because there is a 100% chance to win...only the magnitude of the win differs. There is zero chance of losing. There are many specific numbers that make switching a good bet. There are just as many that make it a bad bet.

[wayfriend]

So WF, I'm unsure if you're supporting the "it makes no difference" theory (as espoused by me, Vain and Vraith), or the "you should swap" theory (as espoused by Hashi and Murrin). Which is it, in your opinion?

It makes no difference. The odds are 50/50 of having the larger envelope if you don't swap, and 50/50 if you do swap, as I mentioned above. So I say it makes no difference.

I cannot for the life of me see a fallacy in stating that, at game start, one of the envelopes contains Y and one of the envelopes contains 2Y.

That's not the fallacy. The fallacy is thinking that when you gain Y by switching and when you lose Y by switching, the same value of Y is used, and so it appears that they are equal amounts. You know that this is wrong because you either gain $100 or you lose $50 ... it's not the same amount. You cannot possibly be correct when you deduce that $100=$50.

Look at it this way: you only gain Y by switching when Y is the whole amount of the first envelope, and you only lose Y by switching when Y is half the amount in the first envelope. Clearly Y cannot be both the whole amount and also half the amount of the first envelope.

Because I'm so confident in original statement #2 being correct, I know there MUST be a fallacy in original statement #1... but I cannot for the life of me see it. And it's driving me nuts.

Here's the fallacy of number one. Suppose you switch envelopes. Now ask, should you switch again? The same reasoning says switching again improves the odds of having more money again ... and your right back with your original envelope. In fact, you can keep switching and keep improving your odds for inifinite!

So that's obviously wrong. I have a hard time pointing out why. I will think on how to explain it. But if swapping improves your odds from 50% to 50%, it's not improving your odds.

[Hashi Lebwohl]

Okay... but WHY is it different? Whether you know the amount of money in your first chosen envelope or not, there's still always only twice or half that amount of money in the other envelope. Why the Hell would knowing what was in the first envelope before deciding to swap or not change things in any way whatsoever?

Let's take a step back and start over.

Scenario 1: you choose the first envelope, open it, and take a look at the value inside which we will simply call m. At this point, the expectation of the value in the other envelope is (0.5)(m/2) + (0.5)(2m) = 1.25m. Since the expectation is higher than your current amount you should probably switch envelopes.

Scenario 2: you choose the first envelope but do not open it; you are, however, given the option to switch envelopes. Given that you do not know the value in your envelope the expectation of *both* envelopes is, at this time, (0.5)(m) + (0.5)(2m) = 1.5m...but we could just as easily say (0.5)(m) + (0.5)(m/2) = 0.75 m and, in fact, both scenarios are equally possible, thus the final expectation is (0.5)[1.5m + .75m] = 1.125m. In this scenario it ultimately doesn't matter whether you switch or not--you haven't opened the first envelope and so you have nothing to lose. This scenario has no optimum outcome and is dependent entirely upon the subjective whims of the player.

Scenario 1: I would recommend switching envelopes.

Scenario 2: no opinion because either outcome is acceptable.

I simply don't see any fallacy or dilemma here, though. *shrug*

[TheFallen]

I really don't think I'm being dense here. I'm going to try this one more time, so bear with me as I also recap.

First off, no matter what happens, you're going to end up with some money from a zero start... that's a given (and a guaranteed "win").

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.

However, according to Hashi, the above is only true if you know the value of A, i.e. if you have opened the first envelope... REALLY??? Why is the relevance or accuracy of the above equation in the least dependant upon knowing the value of A? How can it be? Whether you know the value of A or not, the second envelope still always contains 2A or A/2, surely? And you have no idea which.

Let's deal with A as an unknown and run 100 goes with no swapping. Unsurprisingly, A will exactly be your average return
(A*100)/100 = A.

Let's deal with A as an unknown again and run another 100 goes, but this time, swapping on every occasion. This time, 1.25A will be your average return
(((A/2)*50)+(2A*50))/100 = (25A +100A)/100 = 1.25A.

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).

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.

Those two scenarios are mutually exclusive, are they not? And yet the two simple mathematical equations describing the parameters look to me like they are both entirely correct. There simply has to be a fallacy - and I reckon it's in Scenario #1 somewhere.

[I'm Murrin]

You have exactly as much chance of losing money as gaining it, in any of these scenarios. The question is: Given what you have in hand, do you personally feel like it is worth gambling to increase it?

[Fist and Faith]

Okay, I'm sure this one's not new, but I'd not heard it before - and it's seriously twisting my melon, man.

So you're invited - free of charge - to play a game of chance. It's very simple... there are two sealed envelopes and you know for a fact that both contain money, but that one contains double the amount that's in the other. Or, equally truthfully stated, that one contains half the amount that's in the other.

So you are told you can pick a sealed envelope and you duly do. You open it to reveal a crisp $100 bill. You're now told that you can either keep that $100, or you can discard it and choose whatever's in the second still-sealed envelope instead. Needless to say, if you do choose to switch envelopes, there's no going back.

Very simple, right? And the resultant question is simple too - what is it best to do do after opening the first envelope - stick or switch?

Okay, here's a very logical statement:

1. Let's call the amount you've revealed in the first envelope X. So, having opened the first envelope, if you switch, you may reveal 2X or you may reveal X/2. That is an inarguable fact.

Yes and No. This is a weird thing, so I'm going to say the same thing a few different ways, hoping one or the other works best.

Yes, it is a fact that you don't know if you have the smaller or larger amount, so, yes, you view it in these terms.

But that's as far as you can go with these labels. You cannot then use X/2, X, and 2X in your equations. 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.

-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 cannot combine those two when doing the math.

If you don't switch, let's say you take 100 home. If they don't show you what was in the other envelope, you don't know if it was 50 or 200. But it could not have gone either way. It could only be one or the other. If they show you 50 was in the other envelope, you know it could not have been 200. If they show you 200, you know it could not have been 50. Your decision is still the same difficulty, because you don't know whether there's more or less in the second envelope. But you can't use both more and less in the math.

[Vraith]

But you can't use both more and less in the math.

Sure you can.

The problem is that between scenario 1 and 2, TF is EITHER:
Thinking the two are asking the same question/describing the same problem [which they are not]
OR
He is mis-writing one of the equations.
HOWEVER: what I said above is true [once you sort out the conflict...which is just mistatements/misapplication] There IS NO GENERAL rule about whether to switch or not in 2-envelope conditions. Whether you should switch or not is entirely dependent on the particulars. [with a/2 and 2a, you literally cannot lose...only win or win bigger. If it's A or -A, things change.