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]

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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.

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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]

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[ 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.

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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]

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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"?

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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]

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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.

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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 is an in-depth analysis.

[callydrias]

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

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