Exchange Paradox

[callydrias]

Saw this written up in Nature last month and thought it was interesting. I'm not sure where the hole is. Maybe you can spot it.

There are two envelopes each containing money. One of the envelopes contains twice as much as the other and you can not tell which is which. You choose an envelope at random (you truly have no information on which to base your choice) and are then given the option to exchange if you'd like.

Consider the expectation of exchanging. There is a 50% chance that the other envelope contains twice as much money as you have now. There is also a 50% chance that the other envelope has half as much. The expectation for exchanging then is 0.5(2x-x) + 0.5(x/2-x) = 0.25x. Therefore exchanging envelopes is +EV so you should exchange.

If given the option to exchange again (if you haven't opened any of the envelopes yet), you would exchange over and over, ad infinitum, using the same reasoning each time. Discuss.

[Skipbidder]

[ QUOTE ]
Saw this written up in Nature last month and thought it was interesting. I'm not sure where the hole is. Maybe you can spot it.

There are two envelopes each containing money. One of the envelopes contains twice as much as the other and you can not tell which is which. You choose an envelope at random (you truly have no information on which to base your choice) and are then given the option to exchange if you'd like.

Consider the expectation of exchanging. There is a 50% chance that the other envelope contains twice as much money as you have now. There is also a 50% chance that the other envelope has half as much. The expectation for exchanging then is 0.5(2x-x) + 0.5(x/2-x) = 0.25x. Therefore exchanging envelopes is +EV so you should exchange.

If given the option to exchange again (if you haven't opened any of the envelopes yet), you would exchange over and over, ad infinitum, using the same reasoning each time. Discuss.

[/ QUOTE ]

The amounts in the envelope are determined before you start picking them. They contain X and 2X. You don't know which you've originally picked.
For the exchange, if you originally started with X, you will definitely gain X with the exchange. If you started with 2X, you will definitely lose X with the exchange. Not knowing which envelope you've originally chosen, you have a 50% chance of gaining X and a 50% chance of losing X. Push.

[elitegimp]

[ QUOTE ]
[ QUOTE ]
Saw this written up in Nature last month and thought it was interesting. I'm not sure where the hole is. Maybe you can spot it.

There are two envelopes each containing money. One of the envelopes contains twice as much as the other and you can not tell which is which. You choose an envelope at random (you truly have no information on which to base your choice) and are then given the option to exchange if you'd like.

Consider the expectation of exchanging. There is a 50% chance that the other envelope contains twice as much money as you have now. There is also a 50% chance that the other envelope has half as much. The expectation for exchanging then is 0.5(2x-x) + 0.5(x/2-x) = 0.25x. Therefore exchanging envelopes is +EV so you should exchange.

If given the option to exchange again (if you haven't opened any of the envelopes yet), you would exchange over and over, ad infinitum, using the same reasoning each time. Discuss.

[/ QUOTE ]

The amounts in the envelope are determined before you start picking them. They contain X and 2X. You don't know which you've originally picked.
For the exchange, if you originally started with X, you will definitely gain X with the exchange. If you started with 2X, you will definitely lose X with the exchange. Not knowing which envelope you've originally chosen, you have a 50% chance of gaining X and a 50% chance of losing X. Push.

[/ QUOTE ]

I don't really like this answer, because the values don't _have_ to be predetermined. How is this scenario different from me saying "I'll flip a coin, and heads I double what is in the envelope, tails I halve it"?

[Mano]

The "paradox" is that x is a different number depending on which envelope you hold, and you have to factor that in. One envelope has x in it the other has 2x in it you have 50% chance of gaining x and 50% chance of losing x. The way they have it written with the chance of doubling or halving your money, you can either double x or halve 2x, so EV = .5(2x - x) + .5((2x)/2 - 2x) = 0.

[PokerProdigy]

This does NOT seem correct because the EV should be zero???

[elitegimp]

[ QUOTE ]

I don't really like this answer, because the values don't _have_ to be predetermined. How is this scenario different from me saying "I'll flip a coin, and heads I double what is in the envelope, tails I halve it"?

[/ QUOTE ]

too late to edit, but after reading the last two posts, I see the difference in the two. My bad.

[chris_a]

Of course. The quantity x is itself a random variable.

To make it more clear, look at a concrete example, imagine there's $1 in one and $2 in the other. Then the original formulation is equivalent to stating that the EV is:

EV = 0.5*(2-1) + 0.5*(1/2 - 1) = 0.25

But this is incorrect since when you switch from the $2 to the $1 you are losing $1 and not $0.50.

The real random variable is ONLY your initial choice. So then you can formulate it as X being the initial choice. Then there's a 0.5 prob. of having started at $1 and thus the gain from switching is $1 and a 0.5 prob. of having started at $2 and thus the loss from switchin is $1. The EV is then clearly 0.

This isn't so much a paradox as much as just the wrong formula IMO.

[ThinkQuick]

[ QUOTE ]
Of course. The quantity x is itself a random variable.

To make it more clear, look at a concrete example, imagine there's $1 in one and $2 in the other. Then the original formulation is equivalent to stating that the EV is:

EV = 0.5*(2-1) + 0.5*(1/2 - 1) = 0.25

But this is incorrect since when you switch from the $2 to the $1 you are losing $1 and not $0.50.

The real random variable is ONLY your initial choice. So then you can formulate it as X being the initial choice. Then there's a 0.5 prob. of having started at $1 and thus the gain from switching is $1 and a 0.5 prob. of having started at $2 and thus the loss from switchin is $1. The EV is then clearly 0.

This isn't so much a paradox as much as just the wrong formula IMO.

[/ QUOTE ]

agreed.
You cannot substitute x for "N or 2N". The expected result of switching is always 1.5N (although achieving this result is not actually possible in the game)

[callydrias]

If you haven't googled it already, here (http://www.pitt.edu/~jdnorton/papers/Exchange_paradox.pdf) is an in-depth analysis.

[callydrias]

[ QUOTE ]
This does NOT seem correct because the EV should be zero???

[/ QUOTE ]

That's what makes it a paradox /images/graemlins/smile.gif

[Ponts]

ok my turn!
this was a nice one and here is my take:
the two envelopes X and 2X, u have 50% chance at each giving u EV of your envelope 1,5X. changing will give u either a gain of ,5X or loss of ,5X from what u are holding right now (not in real terms of course, but from the perspective of what u know). its a 50/50 chance so
,5*(-,5x) + ,5*(+,5x)= 0

what u really are doing is to pick one of the two again, first time u pick A and now u either i switch (that is picking envelope B) or u dont switch (that is picking envelope A)

I wouldn't switch, like my 1,5X as they are /images/graemlins/wink.gif

/Ponts

[Lexander]

There is some of this topic in "Statistical Inference" (Casella & Berger 2002, p 203). That text lists some other good discussions of the issue.

I agree the problem is in the incorrect determination of the conditional probabilities used to calculate the expectation.

[chris_a]

It's not in the "conditional probabilities", but in the values themselves. The probabilities are trivial.

Simple stated you are looking for E[Gain from switching].
The unknown random variable, Y is the amount in your original envelope.

In technical terms, the "EV" is the expectation of a derived random variable, since Y is the amount in the envelope, but we are really interested in the derived expectation, E[G(Y)] of the gain from switching, G(Y).

E[G(Y)] = G(Y_1)*P(Y_1) + G(Y_2)*P(Y_2)

where Y_1 and Y_2 are the different amounts in the envelopes and P(Y_1) and P(Y_2) are the probabilities of chosing those envelopes originally.

So the problem lies in the statement of G(Y_1) and G(Y_2) in the original problem formulation. G(Y_1) is the gain from switching from Y_1 to Y_2 and G(Y_2) is the gain from switching Y_2 to Y_1.

G(Y_1) is thus Y_2 - Y_1
G(Y_2) is thus Y_1 - Y_2
P(Y_1) = P(Y_2) = 0.5

Thus, the proper EV is:

E[Gain from Switching] = (Y_2-Y-1)*0.5 + (Y_1-Y_2)*0.5 which cancels out for any value of Y_1 and Y_2. So regardless of what the values are, the EV of switching is 0.

[pzhon]

Usually, this paradox is stated so that you get to see how much is in the first envelope.

It is remarkable how many smart people are stumped by this old puzzle. It was discussed by Mike Caro here (http://www.poker1.com/newsmanager/templates/mculib_articles.asp?articleid=94&zoneid=6), but he got the answer wrong (http://www.poker1.com/newsmanager/templates/mculib_articles.asp?articleid=95&zoneid=6).

You have a 50% chance of being given the smaller envelope. However, once you have seen the amount x, the conditional probability that the other envelope has x/2 is not necessarily 50%. That you see x gives you some information about whether you have been given the smaller amount.

For example, lets suppose the envelopes have a 50% chance to contain ($2,$4) and a 50% chance to contain ($4,$8). If you see $2, the probability the other envelope is larger is 1. If you see $4, the probability the other envelope is larger is 1/2. If you see $8, the probability the other envelope is larger is 0.

The computation 1/2(x/2 + 2x) is wrong because it assumes the conditional probabilities, once you know x, are 1/2. Even if you don't know the distribution, you can't assume the unknown probabilities are 1/2.

It is quite counterintuitive to some people, but there are strategies that are better than accepting the envelope (which is equivalent to switching 100% of the time). You can create such a strategy that works even if you don't know the distribution. Suppose you decide to switch whenever you see less than $100, and stay whenever you see more than $100. If the amount of money in the lesser envelope is less than $50, you will make the right choice half of the time, so you will break even. If the amount of money in the lesser envelope is greater than $100, you will make the right choice half of the time, so you will break even. However, if the amount of money in the lesser envelope is between $50 and $100, you will make the right choice 100% of the time, so you gain over the strategy of never switching.

You can adjust this strategy by randomizing $100 to have positive probability on every interval on the positive real line. This modified stategy is better than always staying with the first envelope, no matter how the amounts in the envelopes are chosen.

Another paradox (mentioned elsewhere) is that there are distributions in which no matter what amount you see, you should want to switch. In these distributions, your expected amount of money is infinite, and you are always infinitely disappointed with the amount you see. You will always want to switch. As an example, let the smaller envelope contain 2^n (n=1,2, ...) with probability (1/9)/(0.9^n). No matter what you see, you would be willing to pay to switch. Nevertheless, the strategies of always switching and of never switching have the same expected value, which is infinite in this case.

[chris_a]

Ah, I misunderstood the problem statement from the original poster. Thanks. The point of this article makes sense now.

[mostsmooth]

im not good with math so the posts above may be stating this already, but you dont lose X or gain X, you lose 1/2x or gain X, no?

[PairTheBoard]

This is the heart of the problem - the distribution from which the envelope amounts are chosen. Even advanced probabilty students miss the underlying bogus assumption that all envelope amounts are equally likely. ie. that the envelope amounts were chosen from a uniform distribution over the positive real numbers. Of course there is no such probabilty distribution.

You could just say, well the amounts are chosen, who cares how. Just let them be fixed at the amounts X and 2X. They are fixed but we don't know what they are. If the experiment is repeated the amounts will remain X and 2X. In other words, assume the envelope amounts come from an unknown single point distribution (Delta distribution?). If the experiment is repeated the SAME UNKNOWN single point distribution will be used. Now Skip's answer is absolutely correct. You open an envelope. Half the time you will be looking at X and will gain X if you switch. Half the time you will be looking at 2X and you will lose X if you switch.

Here's the real brain twister. Under the above assumptions, When you open the envelope and look at an amount there is NOT a 50-50 chance the other envelope is half or double. There IS EITHER a 100% chance the other envelope is double OR a 0% chance the other double DEPENDING on whether you are looking at X or 2X.

P(Other Env is double | you look at X) = 100%
P(Other Env is double | you look at 2X) = 0%
P(Other Env is half | you look at X) = 0%
P(Other Env is half | you look at 2X) = 100%

Using these conditional probabilities and the 50-50 chance of seeing X or 2X when you open your envelope you get the same result that Skip gave above.

PairTheBoard

[PairTheBoard]

Actually I've never liked that Brain Twister part. Given the Fixed X,2X envelopes. You open one and see an amount A. You have no idea if A is X or 2X. There is a 50-50 chance of doubling or halving with a switch. If you double then A=X. If you half then A=2X. Either way you gain or lose X.

Notice when you double you double the Small Amount and when you Halve you Halve the Large Amount.

PairTheBoard