Monty Hall-esque question

[dtbog]

Say you have two unmarked envelopes. You're on a new game show, and the host has put money in the envelopes in the following manner: the value of one envelope is twice the value of the other envelope. You have no idea whatsoever about the expected magnitude of the prizes.

You open one of the envelopes, and you see a check for $100. Now, the host offers you the chance to switch to the other envelope. (Again, no game theoretical assumptions about whether or not it is more likely that the game show would have a $200 prize or a $50 prize.) Do you switch?

Simply stepping through a calculation, it seems like you should: the expected value of the unseen envelope is (50+200)/2 or $125.

Now, what if you are presented with the same conundrum, but this time the host didn't show you what was in the first envelope. Should you still switch?

It seems like you should; the EV of the other envelope should always be higher through conventional calculation. However, this clearly makes no sense!

Is there a way to explain this seemingly paradoxical fact?

[PairTheBoard]

Switching envelopes amounts to betting half the amount in the First Envelope at 2-1 odds that the second envelope contains the larger amount. That would be a good bet if you could bet any amount you like, at 2-1 odds, that the second Envelope will be larger. But you're Not allowed to bet any amount you like. You are required to bet an amount that is being determined by the outcome. If the outcome is a win you are being required to bet a small amount. If the outcome is a loss you are required to bet twice as much.

Here is a comparable proposition. Using a standard deck of 52 cards I will let you draw a random card face down and allow you to bet that it will be a red card. I will pay you 2-1 odds if you win. Sounds like a good deal right? But there's a catch. I have written dollar amounts on the back of all the cards and when you draw the card you are required to bet the amount written on the back of the card. It's a one shot deal. Do you make the bet? If you do you will be getting the worst of it because I've writtn $10 on the back of all the red cards and $50 on the back of all the black cards.

The card game would be fair if I'd written $10 on all the red cards and $20 on all the black cards. If it's a 2 card deck, 1-red 1-black, that's exactly the 2 Envelope Situation.

This is the best explanation you'll ever see for what's going on here. Credit for it goes to me,

PairTheBoard

[AaronBrown]

You will find extensive discussion of this if you Search on this forum, with no consensus. You can also use Google to search for "Necktie paradox," "Wallet Game" or "Envelope paradox."

[elindauer]

Hi PairTheBoard,

I read your explanation, and I thought it was interesting. But I had to think about it some more. So I thought... and I thought... and then I understood... but wait, no... so some more thinking...

While I was thinking, I read some of the other descriptions resolving the paradox. They were long. They involved lots of rigorous math. I basically had to take the author's word on many points. I wondered to myself, could the answer really be as simple as what PairTheBoard has written?

And after a while, I came to really understand what you were saying, and the beauty of it dawned on me like the first light of a glorious new day. Wow. "This guy is really smart", I thought to myself, as I stood and marveled at your creation.

But then!...

It suddenly occured to me that this really elegant description of why the game must be fair is in fact only proving something that everyone already knows! Hell, I can prove the game is fair in one word... "symmetry". QED!

The real "problem" is not to find an explanation of why switching offers no gain... the problem is to find an explanation of why the argument that it does gain something IS WRONG. Everyone, I think, agrees that there is no advantage to switching. The problem is trying to explain why the seemingly sound logic for switching is incorrect.


So, kudos for coming up with a really fantastic description of the two envelopes problem and why there is no advantage to switching. However, to really resolve the paradox, you do have to debunk the other logic (hence the paradox), which means you have to do the math demonstrating that there is no uniform probability distribution with integral of 1 and EV that's not infinite, etc etc.

-eric

PS. Although I don't think your explanation resolves the paradox, I do still admire the elegance of your explanation and that "this guy is really smart" feeling lives on. Well done.

[PairTheBoard]

[ QUOTE ]
It suddenly occured to me that this really elegant description of why the game must be fair is in fact only proving something that everyone already knows! Hell, I can prove the game is fair in one word... "symmetry". QED!

The real "problem" is not to find an explanation of why switching offers no gain... the problem is to find an explanation of why the argument that it does gain something IS WRONG. Everyone, I think, agrees that there is no advantage to switching. The problem is trying to explain why the seemingly sound logic for switching is incorrect.

[/ QUOTE ]

I'm not giving the symmetry argument and I believe I am giving an explanation for why the EV calculation is incorrect. If you buy the insight that switching is equivalent to betting half your envelope amount at 2-1 odds that it's the smaller envelope then you should ask, why isn't that a +EV bet? Looked at that way the incorrect EV calc works exactly the same. If it's a $10 envelope, you are betting $5 at 2-1 that it's the smaller envelope. Why isn't the EV on that bet 2(5)(.5) - (5)(.5) = +$2.50 ? It's the same EV you get looking at it as switching envelopes.

The answer I'm giving is I believe an important principle. If you could bet any amount of your choosing on these terms, the EV calc would be ok based on the given envelopes. But you are Not allowed to bet any amount of your choosing. You are required to bet an amount dictated by the already determined outcome.

You can see this easily with the deck of red black cards. You can bet your card is Red and get paid 2-1 odds. If you could bet any amount you wanted you would indeed have +$2.50 EV on every $5 you bet. But in my scenario you are not allowed to bet whatever you want. If it's a Red Card you must bet $10 and if it's a black card you must bet $50.

Being required to bet an amount that is being dictated by the outcome changes the EV calculation. That's the principle. It has nothing inherently to do with symmetry.


This is I believe the simple principle most people are missing when they incorrectly calculate the EV. If you want to go into the more esoteric observations about possible - and impossible - prior distributions for the amounts in the envelopes then pzhon is of course correct on all his points. The most obvious being that no uniform distribution exists on the positive real numbers. But you don't have to go into prior distributions to see the principle I gave above.

PairTheBoard

[mykey1961]

[ QUOTE ]
This is the best explanation you'll ever see for what's going on here. Credit for it goes to me,

PairTheBoard

[/ QUOTE ]

I sure hope not. All you've done is restate ev = (50+200)/2 as getting 2:1 on 1/2 of your envelope.

Both make assumptions that aren't stated in the "paradox".

Both compare unknown apples to imaginary oranges.

2:1 is true under certain conditions.

(50+200)/2 is true under certain conditions.

Both ignore other conditions.

Neither compares the EV of the 1st decision with the EV of the 2nd decision.

[PairTheBoard]

[ QUOTE ]
[ QUOTE ]
This is the best explanation you'll ever see for what's going on here. Credit for it goes to me,

PairTheBoard

[/ QUOTE ]

I sure hope not. All you've done is restate ev = (50+200)/2 as getting 2:1 on 1/2 of your envelope.



[/ QUOTE ]

That's not what I said. When you open the envelope, see $10, and are offered a switch, you are being offered a gamble. The $10 is now yours. It's your money. If you switch and you have the smaller envelope you will win $10. If you switch and don't have the smaller envelope you will lose $5. That's the gamble. What I said was, that gamble is equivalent to betting half the amount in your envelope at 2-1 odds on the proposition that you have the smaller envelope.

But that's only the first step. To look at it in those terms. Given those terms it appears to be a +EV bet. The question remains, why isn't it?

Is it because seeing the $10 means there is no longer a 50% chance that it's the smaller envelope? Well, yes and no. pzhon's first point says yes, and of course he's correct from the perspective of possible prior distributions on the amounts put in the envelopes. In order for all possbile envelope amounts to have equal chance of being the smaller, the envelope amounts would have to be drawn from a uniform distribution on the positive real numbers. There is no such distribution, as elindauer points out.

Masquerade's point also says Yes from the perspective that the envelope amounts are fixed and any repetition of the experiment will use the same envelope amounts. In that case, seeing the $10 determines the outcome. There's no longer a 50% chance that it's the smaller. Either it is or it isn't. Masquerade's point is also correct and agrees with pzhon's points if you consider the prior distribution for the smaller envelope amount to be a point mass distribution.

How can the answer be "Yes and No" if the above points are correct. Well, it depends on what you mean by probabilty, and it's exactly on this "Yes and No" that Aaron basis his opinion that the problem really is not so well understood.

The sense in which the answer is No, seeing the $10 does not change the probalility of 50% that it's the smaller envelope is from a gambler's perspective. Prior to opening the envelope, anybody here would be delighted to bet $100 that it's the smaller envelope at 2-1 odds. We would see it as 50-50 proposition with favorable odds despite Masquerade's perspective that the envelope has already been chosen and so the 50% probabilty has been nulified.

Furthermore, once the envelope has been opened and we see it has $10 in it, I think most people here would still be willing to bet the $100 at 2-1 odds that it's the smaller envelope. We might look at the $10 and consider elindauer's objection and pzhon's first point. But those points don't really effect the EV for our $100 bet as long as we would have been allowed the same $100 bet if we had chosen the other envelope instead. As far as our $100 bet is concerned the chances are still 50% that it's the smaller envelope. And our $100 bet has +EV of $50.

So why doesn't our $5 bet - equivalent to switching - also have +EV of $2.50? Is it really because of elindauer, pzhon, and Masquerade's points? Well, ok, if you try to use a probabilty model to calculate the EV you will run into their points. But if you simply compare the $100 bet with the $5 bet I think the reason is transparent. The $100 bet has +EV because you are betting an amount of your choosing. The $5 bet is not +EV because it is a betting amount that has been determined by the outcome of the bet. A gambler doesn't need to know any fancy probabilty theory to know he's getting suckered there. Flip a coin and bet on heads, I'll pay you 2-1. Great says the gambler. Except you have to bet $10 if it's heads and $50 if it's tails. Not going for that says the gambler. How about $10 for heads and $20 for tails? Why should I says the gambler. Nothing in that for me. What if I hide the same proposition in two envelopes? The gambler is still indifferent.

If you think about it mykey, I'm in complete agreement with your point.

PairTheBoard

[mykey1961]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
This is the best explanation you'll ever see for what's going on here. Credit for it goes to me,

PairTheBoard

[/ QUOTE ]

I sure hope not. All you've done is restate ev = (50+200)/2 as getting 2:1 on 1/2 of your envelope.



[/ QUOTE ]

That's not what I said. When you open the envelope, see $100(ed), and are offered a switch, you are being offered a gamble. The $100(ed) is now yours. It's your money. If you switch and you have the smaller envelope you will win $100(ed). If you switch and don't have the smaller envelope you will lose $5. That's the gamble. What I said was, that gamble is equivalent to betting half the amount in your envelope at 2-1 odds on the proposition that you have the smaller envelope.

But that's only the first step. To look at it in those terms. Given those terms it appears to be a +EV bet. The question remains, why isn't it?

[/ QUOTE ]

It could be +EV to switch, it could be -EV to switch, and it could be EV = 0 to switch. We don't have enough information to determine whether it is, or isn't +EV in this case.


What exactly do we know from the "Paradox"?

We know that one envelope containes $100. We know that the other envelope contains either $50 or $200.

What we don't know is how likely the other envelope is to contain $50, or $200.

If we assume both amounts are equally likely:

Yes it is +EV to switch.

The EV of switching would be 50*(1/2)+200*(1/2) = $125


If the other envelope has a >= 2/3 chance of $50:

EV of switching <= 50*(2/3) + 200*(1/3)

EV <= 100

Bottom line is we don't have enough information to make a solid decision.

[ QUOTE ]

Is it because seeing the $100(ed) means there is no longer a 50% chance that it's the smaller envelope? Well, yes and no. pzhon's first point says yes, and of course he's correct from the perspective of possible prior distributions on the amounts put in the envelopes. In order for all possbile envelope amounts to have equal chance of being the smaller, the envelope amounts would have to be drawn from a uniform distribution on the positive real numbers. There is no such distribution, as elindauer points out.


[/ QUOTE ]

We have no need for "all possbile envelope amounts to have equal chance of being the smaller". We wouldn't always have picked the $100 envelope. All we know is that we did this time.

[PairTheBoard]

[ QUOTE ]
[ QUOTE ]

Is it because seeing the $100(ed) means there is no longer a 50% chance that it's the smaller envelope? Well, yes and no. pzhon's first point says yes, and of course he's correct from the perspective of possible prior distributions on the amounts put in the envelopes. In order for all possbile envelope amounts to have equal chance of being the smaller, the envelope amounts would have to be drawn from a uniform distribution on the positive real numbers. There is no such distribution, as elindauer points out.


[/ QUOTE ]

We have no need for "all possbile envelope amounts to have equal chance of being the smaller". We wouldn't always have picked the $100 envelope. All we know is that we did this time.

[/ QUOTE ]

The point is that people incorrectly conclude that the probabilty that it's the smaller envelope is 50% because all envelope amounts are equally likely and so $50 and $200 must be equally likely for the other envelope. elindauer's point that there is no probabilty distribution for envelope amounts such that all envelope amounts are equally likely speaks to this flawed assumption by people who base their incorrect EV calculation on it.

PairTheBoard

[PairTheBoard]

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