The money in 2 evelopes "paradox"

[Ra_]

let the two amounts be A and 2A

I think argument 1 only holds true if A is randomly selected
so that 0<A<inf, where a A can be any positive number where A mod 0.01 = 0. If this is the case there is a greater than 99.99999999999999999999999999999999999999999999999 99999999999999999999999999999999999999999999999999 99999999999999999999999999999999999999999999999999 9999999
percent chance that A is greater than
10000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 00000000000000000000000000000000000000000000000000 0000000
so i dont think it matter which envelope you pick, because your expected value is near infinity.

If A has a finite number of values or a finite range
(unknown to you) then i think your EV is predetermined no
matter your desision.

imagine you give this challenge to a million people using
A=50 and to a million people using A=100. Your EV =
-75,000,000 - 150,000,00 = -225,000,000. assuming people
always try to switch.

Edited to make A only even cents.

[Beermantm]

Not sure on how this qualifies as a paradox although it is intresting.

In the first equation with the envelopes you have a 100% Chance of making money so there is no gamble.

In the second equation you are asked to wager half of the first envelope contents to possibly double that envelope.

You are only +EV to switch because there was no risk to get to the first amount and now you are only risking half meaning you are a gauranteed winner regardless if you switch or not. Lets change the rules just slightly and the paradox goes away. Lets say you can only get paid out in whole dollar amounts (meaning no fractions of a dollar) and say the first envolope you pick is 1 dollar. Now you are asked to risk the full dollar amount because anything under a dollar can not get paid off. The next envolope would be (50 cents)a zero dollar value or 2 dollars. Now your back to 1:1 instead of 3:1 and probably better off just keeping the buck.

The first envelope could also be a penny and then you can not get paid because there is no coin value less than the penny. 1:1

The trick just seems to be that if the first envelope has any value at all other than 0 you can devide it by two for infinity.

The answer is you switch because you can't lose after the first pick and you want to know like wholly hell what is in the second envelope. LOL

[FortunaMaximus]

[ QUOTE ]

Argument 1: It's +EV to switch. You had a 50/50 chance of picking the high or low envelope so there's a 50% chance that the other envelope is the high and a 50% chance it's the low. Therefore, EV of switch = 0.5*(+100) + 0.5*(-50) = +25.

Argument 2: It's EV neutral. If always switching was a +EV strategy then it would be more profitable to choose envelope A first and then switch to B then to just choose envelope B and not switch.

Who's right?

[/ QUOTE ]

Argument 1 makes the assumption the second envelope contains 50. If it contains 200, isn't this equation then [.5(100) + .5(+100) = +100] or [.5(100) + .5(-50) = +25]?

This makes #1 +EV, IMHO.

[Alan3]

[ QUOTE ]

Argument 1 makes the assumption the second envelope contains 50. If it contains 200, isn't this equation then [.5(100) + .5(+100) = +100] or [.5(100) + .5(-50) = +25]?

This makes #1 +EV, IMHO.

[/ QUOTE ]

No, that isn't what argument 1 says.

Argument 1 says there is a 50% chance that you chose the smaller envelope and a 50% chance you chose the larger envelope. So if you switch there is a 50% chance that you gain 100 and a 50% chance that you lose 50.

I made the same mistake you did when I read it at first.

The flaw in the paradox is that there is a 50/50 chance of you having the smaller envelope once you open it. In fact, seeing the amount of money in the envelope gives you money and you have to make a judgement call on if the person offering the envelopes would have offered you .5A/A or A/2A.

At least that is what I got out of the wikipedia explanation.

[Ra_]

For a quick answer on why 1 is incorrect.

Without knowing the distribution of A, its like saying you can have a 1)Red Snapper or 2)whats in the box.

You have to make some assumptions before you can calculate the EV in switching.

[rufus]

Well, let's say we do the following:

Assume that I have infinite money, and I start with a dollar ($1), and flip a coin. Every time that the coin comes up heads, I tripple the amount.

Then I put a check for the amount in one envelope, and a check for three times teh amount in another.

Now, if you pull an envelope with $4 there are the following two possibilities:
I flipped heads once (.5) and you got the big envelope (.5)
or
I flipped heads twice (.25) and you got the small envelope (.5)

So the probability that you're loosing money if you switch is 2/3, and the probability that you gain is 1/3. So the EV of switching is $1*2/3+$9*1/3=3 + 2/3 > $3, right?

In practice, there is, of course, an amount of money so large that I can't afford to pay it out - so, if you get a check for slightly more than that amount, the EV of switching is negative.

[TomCollins]

The problem is the question itself is a paradox. It is impossible to set up a situation where each number has an equal chance of being the upper number and lower number.

It's been discussed on here before and there are a lot of proofs that show this.

[AaronBrown]

I wrote this article on the paradox. People who insist on "resolving" the paradox are missing the point. It's easy to add assumptions that make one side or the other seem stronger, but the fact is, both arguments have strength and lead to opposite conclusions. The point is to think about why that's true, and thereby deepen your understanding of probability.

For people who think it's clearly irrational to switch, I offer the following variant. To keep it simple, assume one pound sterling is worth US $2. I'm an American, and I bet $1,000 against 500 pounds sterling with my English friend that the US dollar will get weaker (that is, it will take more than $2 to buy 1 pound) in one year. Assume the odds are 50/50 of this happening.

If I lose, I send my friend $1,000. If I win, he sends me 500 pounds, which will be worth more than $1,000. So clearly this bet is +EV to me. If he loses, he sends me 500 pounds. If he wins, he gets $1,000, which are worth more than 500 pounds. So it's clearly +EV to him as well.

If that's not enough for you, I can go to the foreign exchange option markets and get $60 in cash today for agreeing to exchange the 500 pounds if I win for $1,000, with no obligation if I lose the bet. So I get $60 today to take a 50/50 $1,000 bet; and my friend can get 30 pounds to take the other side of the bet.

[SamIAm]

[ QUOTE ]
People who insist on "resolving" the paradox are missing the point.

[/ QUOTE ]I guess. It's understandable when people say "But, how is X being chosen from the set of all positive reals?". That's not "adding assumptions"; that's clarifying the statement.

[ QUOTE ]
For people who think it's clearly irrational to switch, I offer the following variant. To keep it simple, assume one pound sterling is worth US $2. I'm an American, and I bet $1,000 against 500 pounds sterling with my English friend that the US dollar will get weaker (that is, it will take more than $2 to buy 1 pound) in one year. Assume the odds are 50/50 of this happening.

If I lose, I send my friend $1,000. If I win, he sends me 500 pounds, which will be worth more than $1,000. So clearly this bet is +EV to me. If he loses, he sends me 500 pounds. If he wins, he gets $1,000, which are worth more than 500 pounds. So it's clearly +EV to him as well.

[/ QUOTE ]
This is awesome. I'd never seen this phrasing before, and it's very well done. I assume I can "resolve" the paradox by looking at their financial statements as a whole (and not just the money changing hands). They're just hedging their bets in the financial market, right? If the deal is "If I lose $10, give me $5. If you lose $10, I'll give you $5." then obviously there's no paradox.
-Sam

[AaronBrown]

[ QUOTE ]
It's understandable when people say "But, how is X being chosen from the set of all positive reals?". That's not "adding assumptions"; that's clarifying the statement.

[/ QUOTE ]
I disagree. The original question mentions nothing about how X is chosen. It sets up a realistic situation and asks you to make a choice. To say "I can't make the choice without more information" resolves nothing. To put it in the context of a model, by assuming there's some method for choosing X and reasoning from there, is adding assumptions. Whichever answer you get, you could choose different assumptions and get the other answer.

Real problems, like which envelope to take, do not have to conform to simple mathematical models. To insist that they do is a form of blindness. The logic, "I can't figure out a mathematical solution to this problem without more assumptions, so I'll make more assumptions to get a well defined answer, so that answer is proven mathematically correct," is obviously circular.

I think the paradox makes clear that the concept of expected value, especially as a criterion for decision-making, is more subtle than commonly understood. While the two envelope problem doesn't come up a lot in practice, it's related to the idea of shrinkage, which is important and controversial in practical statistics.

Yes, the two-currency version of the problem uses financial market transactions (foreign currency options) to monetize the extra expected value. It proves that in at least some cases, two people can each have positive expected value from opposite sides of a zero-sum bet. That means the argument that switching envelopes cannot be positive EV, because then both people would have positive EV from a zero-sum swap, is not airtight.