Return of the two envelope paradox

[TheFallen]

Okay the last time (three years ago?) I tried to have one of you brainiacs explain this to me, you utterly failed...

...which I am of course blaming you lot for.

Anyhow, let's try again and reprise.

There are two sealed envelopes and one contains twice the amount of money than the other. That's all you know.

You are given one of the sealed envelopes.

You are told that you can either keep the envelope you've got... or exchange for the other sealed envelope. Is there anything you can do in order to maximise potential gain?

I say "switch" on the apparently logical basis that if I decide to term the unknown amount I am holding x and I switch, then I have a 50% chance of ending up with x/2 and a 50% chance of ending up with 2x. To me, switching then produces an average result of ending up with 1.25x, so must be a good thing.

But then what happens if, having switched, you're offered the exact same choice again? Don't you end up in a position where you should infinitely switch? That's clearly ludicrous?

And another thing... why does the maths apparently change if - going back to the startpoint - I alter the frame of reference and instead merely decide to term the unknown amount in the OTHER envelope x? If I do that, then in that frame of reference, I've ALREADY GOT either x/2 or 2x, so I SHOULDN'T then switch?

This one seriously twists my melon, man.

Oh and before any smartass starts with some reference to zero sum games and it making no difference, because regardless, you'll come out with some level of gain, that answer is not gonna be accepted... you tried to bamboozle me with that one before. Let's presume that whoever loaded up and sealed the envelopes used your own money to fill one of them... and you don't know how much he used, or if he filled the other envelope with double the amount of his own money or half the amount of his own money.

[wayfriend]

I say "switch" on the apparently logical basis that if I decide to term the unknown amount I am holding x and I switch, then I have a 50% chance of ending up with x/2 and a 50% chance of ending up with 2x. To me, switching then produces an average result of ending up with 1.25x, so must be a good thing.

That's bogus math.

The problem is that you are not using X as a specific amount of dollars. In one instance X represents the smaller amount, and in another case it represents the larger amount. All your math from then on is fraudulent.

Honest math would say let X be the smaller amount, 2X be the larger amount. 50% of the time you gain X by trading, 50% of the time you lose X by trading. The average of all possible results is +/- 0%. Exactly as one would expect.

But I commend you for working so hard to improve your skills at deception. You spend a lot of time at it, as you say.

[Hashi Lebwohl]

Just like last time, envelope A contains x and envelope B contains 2x. You have a .50 chance of being given A and a .50 chance of being given B, so the expected value when you are handed an envelope is

.50(x) + .50(2x) = 1.5x (we don't care what the value of x actually is)

if you were given A and you switch then your expected value is .5(2x) (the probability you were given A * the expected value of B) = x

if you were given B and you switch then your expected value is .5(x) = .5x

Since you started off with 0, I would recommend keeping the first envelope you are given. Either way, you win.

[TheFallen]

Okay, wf and Hashi (and ignoring your cheap shot wf... save it for the Tank if you must)...

I absolutely understand and agree with both your formulations of the problem. I also freely admit that my formulation of the problem must be fallacious...

...but I cannot see where my description is in error.

Okay, say I get my first envelope and open it, revealing a crisp $20 bill. I then know that the other envelope must contain either $10 or $40. It cannot contain anything else. Again on that basis, swapping would seem to be a very sensible policy?

Again, both your descriptions of the scenario make perfect sense and balance out to the expected zero difference - but where's the error in mine (which must be there)? Where's the "bogus"?

And gently please... linguist here, not any sort of mathematician.

[Hashi Lebwohl]

Okay, say I get my first envelope and open it, revealing a crisp $20 bill. I then know that the other envelope must contain either $10 or $40. It cannot contain anything else. Again on that basis, swapping would seem to be a very sensible policy?

Opening the envelope is a slightly different problem than the one originally posed. If you are allowed to open the envelope first and then switch you now have a version of the Monty Hall problem, in which case you definitely want to switch.

In your specific case, you are holding a $20 bill. The expected value of the other envelope is (.5)(10) + (.5)(40) = 25, which is higher than the value you are holding in your hand; therefore, it is in your best interest to switch.

[TheFallen]

Ok - and by the way I fully get the Monty Hall problem - I don't see why opening my envelope changes the mathematical frame of reference in any way? Why is it relevant? What's different about calling the amount in my envelope $20 or X? Why does it matter?

I cannot believe for a second that establishing what's in my envelope makes any difference to the maths.

[wayfriend]

Let me try to show you this way.

Suppose I say to you, flip a coin, if it's heads, you get $50. I go to Hashi and say, flip a coin, if it's heads, you get $100. Both of you get to do this 3 times.

Odds are that Hashi will win more money than you. Because there is more money involved in Hashi's game to begin with. Both of you will win 1.5 times on average, but because there is more money in one game, this skews the net results in favor of Hashi - $75 average vs. $150 average. So: The two scenarios are not equal when you consider the net result.

Now, in the envelope game, if you have $50, you think that the other envelope can be $25 or $100. However, there are actually two different possible universes: one with envelopes $25 and $50, the other with envelopes $50 and $100. In the second case, there is more money in the game to begin with. So: the odds are the same, but the net result is skewed toward the second game - switching envelopes in the second game gains you more than you lose switching envelopes in the first game.

Also.

Suppose you reframe your original outline of the problem to make X the amount of money in the other envelope. Thus, you may have either X/2 or 2X when you begin.

If you DON'T switch, on average you will end up with 1.25X dollars.

Huh? Do nothing, make money?!?!?!

This tells you that you have an absurdity in your thinking.

[TheFallen]

I think this problem is more complex than it seems. We're sort of getting into the realms of quantum, the multiverse and Schroedinger's cat. Uncertainty is certainly (heh) involved.

WF, again I understand that there is a clear absurdity in how I've framed the problem... I also got to that bizarre 1.25x figure. But I still don't see where it lies - for all that your alternate framings completely stack up. I don't dispute a thing you're saying - but that's not telling me where my problem framing is incorrect.

And surely Hashi is wrong when he says that opening my envelope changes things?

[wayfriend]

You went wrong when you averaged the two possible outcomes.

If you create an expression using X several times, the value X has to represent the same thing every time.

Side trip: Suppose I say "Let X be the amount I spend at the restaurant tonite". And then I say, "Let X be the amount I spend on my phone plan". Now, both of these may actually be $50. But this doesn't mean that I can consider these things to be defined the same. The two things are not the same BY DEFINITION. They are only the same BY HAPPENSTANCE. Yes, this is a contingent vs. necessary thing.

Returning: In your case, it SOUNDS like they are the same - "X is the amount of money I have". However, they aren't really the same.

Because there actually two different hands being dealt: The {$25,$50} hand. And the {$50,$100} hand.

In once case, X is "the amount of money I have if I have the larger of $25 and $50". And the other is "the amount of money I have if I have the smaller of $50 and $100".

It seems like they are the same because they are both $50. Equality is contingent. But one X lives in a completely different game than the other X. There is no scenario where they both exist at the same time, because the two games are mutually exclusive of each other. So equality is not necessary.

The X's are not the same. So when you add them together the results are undetermined.

[Hashi Lebwohl]

Let me try to show you this way.

Suppose I say to you, flip a coin, if it's heads, you get $50. I go to Hashi and say, flip a coin, if it's heads, you get $100. Both of you get to do this 3 times.

Odds are that Hashi will win more money than you. Because there is more money involved in Hashi's game to begin with. Both of you will win 1.5 times on average, but because there is more money in one game, this skews the net results in favor of Hashi - $75 average vs. $150 average. So: The two scenarios are not equal when you consider the net result.

Now, in the envelope game, if you have $50, you think that the other envelope can be $25 or $100. However, there are actually two different possible universes: one with envelopes $25 and $50, the other with envelopes $50 and $100. In the second case, there is more money in the game to begin with. So: the odds are the same, but the net result is skewed toward the second game - switching envelopes in the second game gains you more than you lose switching envelopes in the first game.

Wayfriend nailed it perfectly.

And surely Hashi is wrong when he says that opening my envelope changes things?

No, I am not. Opening the envelope is like opening the box holding the cat--looking at the money inside changes what happens.

[TheFallen]

You must be wrong.

To use WF's frame of reference, knowing that your envelope contains $20 does not in any way tell you which of the two possible "universes" you're in.

[Hashi Lebwohl]

You must be wrong.

Check my signature.

Your envelope contains x, which means the other envelope could contain either .5x or 2x. The expected value of that other envelope is .5(.5x) + .5(2x) = 1.25x.

[TheFallen]

..and now you're arguing in favour of my original (and clearly absurd) premise. That is exactly what I originally said - and know to be absurd - but am still unclear exactly where it's wrong.

WF, tell Hashi he's wrong - as he must be if you're right.in your explanation. This is not Schroedinger's cat. The mere act of observing the $20 in your own envelope does not change the problem mathematically speaking, or alter the frame of reference, nor does it define which of the two possible universes/games you're in.

Hashi you surely cannot agree with WF's summary in a case where you have not opened your first envelope, but then disagree with him if you have opened your first envelope?

ADDED EDIT

In fact, Hashi, what you're effectively also saying is that, if instead of opening my envelope, the OTHER envelope is opened to reveal $20, then in that case I should definitely NOT swap. That's surely nonsense?

[I'm Murrin]

I cannot believe looking in the envelope changes things. The Monty Hall problem requires certain knowledge that the odds have changed - which is gained by removing one of the "wrong" choices. Looking in the envelope reveals the scale of the options but does not affect the odds at all - it's still a straight 50:50 where changing makes no difference.

[TheFallen]

I'm sure you're correct Murrin. See my added edit above.

[Hashi Lebwohl]

In fact, Hashi, what you're effectively also saying is that, if instead of opening my envelope, the OTHER envelope is opened to reveal $20, then in that case I should definitely NOT swap. That's surely nonsense?

In this problem you should never swap. In a Monty Hall problem you should always swap.

Unless the other envelope has more money, in which case you should always swap.

[Avatar]

Words dissemble
Words be quick
Words resemble walking sticks
Plant them they will grow
Watch them waver so
I'll always be a word man
Better then a bird man

--A

[TheFallen]

Right... the penny's finally dropped. And mainly courtesy of WF and his "two universes" model. Many thanks for this.

Hashi, you were precisely no help at all

Here's the thing - and actually, WF I'm pretty sure that a valid equation *can* be written for it.

There's an expression which those of you more able at maths will know - it's something like "probability space"... the set of all possibles.

So, looking at the envelope initially in front of me, there are actually FOUR possible "envelopes" that could be, namely:-

Containing X (in universe A where the total value of the two envelopes = 3X)

Containing 2X (in universe A where the total value of the two envelopes = 3X)

Containing 2X (in universe B where the total value of the two envelopes = 6X)

Containing 4X (in universe B where the total value of the two envelopes = 6X)

The total of the possibilities of both universes is thus 9X. It doesn't actually matter that some of these possibilities cannot co-exist. (You can't have envelopes that both contain 2X, or one that contains X and the other 4X).

On the above basis, the probable value of my initial envelope is therefore (0.25 * X) + (0.25 * 2X) + (0.25 * 2X) + (0.25 * 4X), which equals 2.25X, which is of course correct, being a quarter of 9X. And of course the exact same probable value is true for the envelope opposite me.

My fallacy - and WF was onto this (in fact I'm sure he fully understood it - but just didn't explain it well enough to a non-mathematical audience) - was to conflate the two possible occurrences of 2X into just one. I was effectively therefore stating that there were only three possible envelopes (X, 2X and 4X)... and that was just plain wrong, because there's four possible envelopes (X, 2X, 2X and 4X).

In my defence though, the undoubtedly true statement that you KNOW FOR A FACT that the other envelope holds either twice what you have or half what you have remains compellingly misleading and persuasive.

And Hashi? Opening one envelope... ANY envelope... changes precisely nothing. That particular act of measurement or observation doesn't collapse the wave form or whatever in the least... this isn't Schroedinger's cat... you could still quite happily be in either universe. Indeterminacy still reigns.

So Hashi, get that sig quote changed - because you are just plain wrong here re opening one of the envelopes making any difference at all. It doesn't. Period.

[wayfriend]

So, looking at the envelope initially in front of me, there are actually FOUR possible "envelopes" that could be, namely:-

Containing X (in universe A where the total value of the two envelopes = 3X)

Containing 2X (in universe A where the total value of the two envelopes = 3X)

Containing 2X (in universe B where the total value of the two envelopes = 6X)

Containing 4X (in universe B where the total value of the two envelopes = 6X)

The total of the possibilities of both universes is thus 9X. It doesn't actually matter that some of these possibilities cannot co-exist. (You can't have envelopes that both contain 2X, or one that contains X and the other 4X).

I was going to go down that road. But I tried to stay simple.

First of all - you make a mistake when you say the odds of all four are equal at 25%. Because the odds of being in the 3X game vs the 6X game are completely undefined in your problem!

But let's say someone we don't know flips a coin and decides it. Problem solved.

Watch what happens now! (I refactor things just a little bit so that the X we use now matches the X we used originally.)

The four possibilities are:
+ You have 0.5X in a game with 1.5X
+ You have 1X in a game of 1.5X
+ You have 1X in a game of 3X
+ You have 2X in a game of 3X

If you do the math on this, you will see that, on average, you start with 1.125X, and if you switch, you will have 1.125X.

NO ADVANTAGE with switching or not switching. As you expect.

However, you have eliminated two possibilities in your original question. You presume you have X, and the other envelop is either 2X or 0.5X.

So what you leave us with is:
+ You have 1X in a game of 1.5X
+ You have 1X in a game of 3X

If you do the math on this, on average you start with X, and the other envelop is 1.25X. As you pointed out originally.

Notice that the total in the envelopes is the same as before - 2.25X. But the results are now skewed towards switching. Your starting value is a little smaller. The other envelop is a little bigger.

What happened is, because you (unknowingly) removed two possibilities, you left us with two possibilities that no longer balance out. You removed possibilities that leaned towards starting with more - you took out the best case for the starting envelope, which is 2X. Leaving us with possibilities that lean toward starting with less.

Presto.

Now, what didn't happen is a quantum collapse. We didn't change things by seeing what was in the first envelope.

What did happen is very similar, but less spectacular. We inserted hidden assumptions that ruled out certain possibilities. They didn't disappear -- we chose to ignore them.

[Hashi Lebwohl]

So Hashi, get that sig quote changed - because you are just plain wrong here re opening one of the envelopes making any difference at all. It doesn't. Period.

You are welcome to think so but I disagree with your assessment. My degree in mathematics disagrees with your assessment, as well.