Envelope Problem

[gronbog]

Ok. We're in a similar situation as the the Monty Hall problem on Norm's forum. A problem very similar to a well known, intriguing and often mishandled problem has been presented. A slight change to the premise of the well known problem, which corresponds to the most commonly made error when solving that problem has been specified and was missed by those who are already familiar with that problem. Me included. Mea culpa.

Yes, if you get to open the envelope, then you should switch with an EV of 5/4 of the amount you are holding. [Edit: this is incorrect. See my next post below]

Just for the sake of completeness and harmony, Meistro, do you agree that, if the envelopes were indistinguishable and unopened (the original well known problem) that there is no value in switching?

 

[London Colin]

Yes, if you get to open the envelope, then you should switch with an EV of 5/4 of the amount you are holding.

Just for the sake of completeness and harmony, Meistro, do you agree that, if the envelopes were indistinguishable and unopened (the original well known problem) that there is no value in switching?

Are you sure? That's not my understanding.

 

[gronbog]

Which statement, or is it both?

 

[London Colin]

Indistinguishable means not labelled so a to reveal any asymmetry in the way they were populated.

Opening one of them does not tell us anything useful. We knew it contained an amount, X, and that the other contains either 2X or X/2. Now we know that X=100. How does that help?

 

[London Colin]

Which statement, or is it both?

The first!
(See the two papers I cited, plus my comment above.)

And consider the two-player, symmetrical version, where each sees what's in their own envelope, but not in the other player's. It can't be in both their interests to switch.

 

[Meistro]

Opening one of them does not tell us anything useful. We knew it contained an amount, X, and that the other contains either 2X or X/2. Now we know that X=100. How does that help?

because the expected value of switching is $125 and $125 is more than $100

 

[London Colin]

because the expected value of switching is $125 and $125 is more than $100

And if it contained some other amount? $1, $5, $20, $200,000,000. ... ?

What figure would tell you something other than that you believe yourself to have a 1.25X EV from switching, where X is the initially unknown amount in the envelope?

 

[Meistro]

None, no matter the amount in the envelope, so long as the other has 50% chance of 2x and 50% chance of 1/2 you will have a 5/4 return by switching.

 

[London Colin]

None, no matter the amount in the envelope, so long as the other has 50% chance of 2x and 50% chance of 1/2 you will have a 5/4 return by switching.

Hence, as I said, opening the envelope tells you nothing useful. Based on that assessment, you already knew you wanted to switch. There's really no point in opening it at all.

Conversely, if the switching EV before you open the envelope is zero (which it is). Logic (and all that I have read) surely dictates that it remains zero. Knowing the amount doesn't change anything. I think (hope) that Gronbog has just got a little confused when making the above statement. (If not, then I have got very confused indeed!)

 

[Meistro]

Which brings us back to our roulette game. Are you now trying to argue that the roulette game is not plus ev? Perhaps you could demonstrate how the expectation of switching is $100?

 

[sagefr0g]

just me maybe:
the unopened envelope is indistinguishable other than you know it has an even chance of containing half as much or twice as much as the open envelope. so half the time good news, half the time bad news if you were to switch. same is true if you just stick with the open envelope when comparing the prospects of what may or may not be in the unopened envelope.
but again, it's all really good news because the game is a free roll.

 

[London Colin]

Which brings us back to our roulette game. Are you now trying to argue that the roulette game is not plus ev?

I don't know what you feel has brought us back to your roulette game. But the answer to your question is the same as the last time you asked it - No.

Perhaps you could demonstrate how the expectation of switching is $100?

Not easily. I suspect Gronbog has gone off to read those two papers I keep mentioning; the ones you are resolutely refusing to look at. They each run to a few pages and this is the issue they address.

Gronbog is, I think, more of a mathematician than I'll ever be. Hopefully he'll be able to summarize the main points that can be taken away from these, in a way that I struggle to do.

 

[Joe Mama]

So, its positive EV always to switch. Suppose you gave one envelope to person A and the other to person B, then it would be positive EV for Person A to switch with person B and also positive EV for person B to switch with person A? CRAZY!!

 

[gronbog]

Ok, I have re-read the articles at the links that Colin provided. The first link very clearly points out the flaw in the formula that the EV of switching would be

1/2(2A) + 1/2(A/2) = 5/4(A).

The flaw is that the first term is correct only when A is the smaller amount and the second term is correct only when A is the larger amount. That is, two different values of A are assumed within the same formula.

The correct approach is to realize that the envelopes contain the amounts x and 2x and that, if you are holding the envelope with the smaller amount, you gain x by switching (you go from x to 2x) and if you are holding the envelope with the larger amount, you lose x by switching (you go from 2x to x). So the EV of switching is (1/2)x - (1/2)x = 0.

Furthermore, the EV of each envelope is actually (x + 2x) / 2 = (3/2)x.

Colin is correct when he said that opening one envelope changes nothing. Replacing A by 100 in the flawed formula above changes nothing. So I stand corrected again (thanks Colin). Even if you open your envelope and find 100, there is no value in switching.

The article provides several other solutions, but the above is the simplest one.

 

[Meistro]

so gronbog, you also believe that my hypothetical roulette game is not +EV?

 

[gronbog]

No, I believe that it is +EV but not the same as the envelope problem, with or without looking into the envelope and finding 100.

 

[Meistro]

perhaps you could explain how the roulette game is any different?

In both cases you have a 50% return of $200 and a 50% return of $50.

They are the exact same.

 

[Joe Mama]

So if you gave one envelope to one person, and the other to another person, and they continually swapped envelopes with each other they would both become billionaires realizing a 25% average gain for each on each swap. Ridiculous!

 

[Meistro]

So do you claim that the roulette game is not +EV Joe Mama?

 

[Joe Mama]

So do you claim that the roulette game is not +EV Joe Mama?

The roulette game is +EV. If you played the roulette game under your proposed rules over and over you would get rich.
If you swapped the envelopes over and over you would end up where you started. The roulette spins are independent events, the envelope swaps are not.

Look at the envelope problem with 2 people, and two envelopes -- there is a an equivalent winner and loser on each swap, it cannot be positive EV for both persons on average. EV for each will work out to zero. In the roulette game there is always positive EV for the player and negative for the wheel owner.