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The two envelopes problem
 Posted: 04 January 2012 08:27 PM [ Ignore ]
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What is it? From the wiki

The problem:

The two envelopes problem, also known as the exchange paradox, is a brain teaser, puzzle or paradox in logic, philosophy, probability and recreational mathematics, of special interest in decision theory and for the Bayesian interpretation of probability theory.

Specifically:

Let us say you are given two indistinguishable envelopes, each of which contains a positive sum of money. One envelope contains twice as much as the other. You may pick one envelope and keep whatever amount it contains. You pick one envelope at random but before you open it you are offered the possibility to take the other envelope instead.

The implications of switching:

The switching argument: Now suppose you reason as follows:

1. I denote by A the amount in my selected envelope.
2. The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.
3. The other envelope may contain either 2A or A/2.
4. If A is the smaller amount the other envelope contains 2A.
5. If A is the larger amount the other envelope contains A/2.
6. Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2.
7. So the expected value of the money in the other envelope is

8. This is greater than A, so I gain on average by swapping.
9. After the switch, I can denote that content by B and reason in exactly the same manner as above.
10. I will conclude that the most rational thing to do is to swap back again.
11. To be rational, I will thus end up swapping envelopes indefinitely.
12. As it seems more rational to open just any envelope than to swap indefinitely, we have a contradiction.

The puzzle: The puzzle is to find the flaw in the very compelling line of reasoning above.

Non-probabilistic variant:

The following plainly logical arguments lead to conflicting conclusions:

1. Let the amount in the envelope chosen by the player be A. By swapping, the player may gain A or lose A/2. So the potential gain is strictly greater than the potential loss.

2. Let the amounts in the envelopes be Y and 2Y. Now by swapping, the player may gain Y or lose Y. So the potential gain is equal to the potential loss.

The conflict between human reasoning, logic and rationality. To switch or not to switch? :lol:

[ Edited: 06 January 2012 08:42 AM by kkwan ]
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 Posted: 04 January 2012 10:02 PM [ Ignore ]   [ # 1 ]
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Nice

:)

Let’s say that one envelope has \$1 and one envelope has \$2. If the player chooses the \$1 envelope, the player will gain \$1 by switching. If the player chooses the \$2 envelope, the player will lose \$1 by switching. So, to say that the player may gain A or lose A/2 isn’t a complete way of stating the problem. Those are two different As. It’s true that \$1+\$1=\$2 and \$2-\$1=\$1, but \$1/=\$2, as the assumption with the “may gain A or lose A/2” implies.

[ Edited: 04 January 2012 10:05 PM by TromboneAndrew ]
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 Posted: 05 January 2012 10:10 AM [ Ignore ]   [ # 2 ]
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TromboneAndrew - 04 January 2012 10:02 PM

Let’s say that one envelope has \$1 and one envelope has \$2. If the player chooses the \$1 envelope, the player will gain \$1 by switching. If the player chooses the \$2 envelope, the player will lose \$1 by switching. So, to say that the player may gain A or lose A/2 isn’t a complete way of stating the problem. Those are two different As. It’s true that \$1+\$1=\$2 and \$2-\$1=\$1, but \$1/=\$2, as the assumption with the “may gain A or lose A/2” implies.

There is no implication that \$1=\$2 because the different values of A refer to two distinct chosen situations.

Let the amount in the envelope chosen by the player be A. By swapping, the player may gain A or lose A/2. So the potential gain is strictly greater than the potential loss.

If the player chose the envelope where A=\$1, by swapping she gained \$1. (Situation 1)

OTOH, if the player chose the envelope where A=\$2, by swapping she lost \$1. (Situation 2)

Hence, in situation 1 she gained A and in situation 2 she lost A/2.

So, the potential gain in situation 1 (100%) is strictly greater than the potential loss in situation 2 (50%).

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 Posted: 05 January 2012 10:24 AM [ Ignore ]   [ # 3 ]
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kkwan - 05 January 2012 10:10 AM

So, the potential gain in situation 1 (100%) is strictly greater than the potential loss in situation 2 (50%).

I think the solution is that the amount of money that can be gained or lossed is what matters.

It’s irrelevent what that is as a percentage of what’s in the envelope.

Stephen

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 Posted: 05 January 2012 10:52 AM [ Ignore ]   [ # 4 ]
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I think the problem lies in steps 4-6:

The switching argument: Now suppose you reason as follows:

1. I denote by A the amount in my selected envelope.
2. The probability that A is the smaller amount is 1/2, and that it is the larger amount is also 1/2.
3. The other envelope may contain either 2A or A/2.
4. If A is the smaller amount the other envelope contains 2A.
5. If A is the larger amount the other envelope contains A/2.
6. Thus the other envelope contains 2A with probability 1/2 and A/2 with probability 1/2.
7. So the expected value of the money in the other envelope is

8. This is greater than A, so I gain on average by swapping.
9. After the switch, I can denote that content by B and reason in exactly the same manner as above.
10. I will conclude that the most rational thing to do is to swap back again.
11. To be rational, I will thus end up swapping envelopes indefinitely.
12. As it seems more rational to open just any envelope than to swap indefinitely, we have a contradiction.

The puzzle: The puzzle is to find the flaw in the very compelling line of reasoning above.

Steps 4 and 5 are conditionals. The other envelope contains 2A if A is the smaller amount, and A/2 if A is the larger amount. Step 6 combines these two equations into one, which is invalid because you have not yet resolved whether A is the larger or smaller amount.

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 Posted: 05 January 2012 10:57 AM [ Ignore ]   [ # 5 ]
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domokato - 05 January 2012 10:52 AM

Steps 4 and 5 are conditionals. The other envelope contains 2A if A is the smaller amount, and A/2 if A is the larger amount. Step 6 combines these two equations into one, which is invalid because you have not yet resolved whether A is the larger or smaller amount.

That’s basically what I tried to say, but more clearly.

:)

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 Posted: 06 January 2012 08:41 AM [ Ignore ]   [ # 6 ]
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domokato - 05 January 2012 10:52 AM

Steps 4 and 5 are conditionals. The other envelope contains 2A if A is the smaller amount, and A/2 if A is the larger amount. Step 6 combines these two equations into one, which is invalid because you have not yet resolved whether A is the larger or smaller amount.

Steps 4 and 5 are not conditionals.

They are only statements of what is in the other envelope if A is smaller or larger.

And they are not equations.

Step 6 does not combine steps 4 and 5 into one at all. It just expresses them as simple probabilities.

Obviously, with only 2 envelopes, the probability of each is 1/2.

So, steps 4, 5, and 6 are valid.

How about step 7 (which is an equation)? From the same wiki:

Proposed resolution

This follows from the recognition that the symbol A in step 7 is effectively used to denote two different quantities, committing the fallacy of equivocation. This error is brought into relief if we denote by X the smaller amount, making 2X the larger amount, then reconsider what happens in steps 4 and 5:

4. If A=X then the other envelope contains 2A (or 2X).

5. If A=2X then the other envelope contains A/2 (or X).

Each of these steps treats A as a random variable, assigning a different value to it in each possible case. However, step 7 continues to use A as if it is a fixed variable, still equal in every case. That is, in step 7, 2A is supposed to represent the amount in the envelope if A=2X, while A/2 is supposed to represent the value if A=X. However, we cannot continue using the same symbol A in one equation under these two incompatible assumptions. To do so is equivalent to assuming A=X=2X; which, for nonzero A, implies 1=2.

However, there is a second variant:

The solution above doesn’t explain what’s wrong if the player is allowed to open the first envelope before being offered the option to switch. In this case, A is indeed a fixed variable. Hence, the proposed solution in the first case breaks down and another explanation is needed.

Then, there is a third variant:

Though probability theory can defuse the second variant of the paradox, it turns out that examples can still easily be found of proper probability distributions, such that the expected value of the amount in the second envelope given that in the first does exceed the amount in the first, whatever it might be.

Proposed resolution

Thus if we switch for the second envelope because its conditional expected value is larger than what actually is in the first, whatever that might be, we are exchanging an unknown amount of money whose expectation value is infinite for another unknown amount of money with the same distribution and the same infinite expected value.

So, we now have a situation where infinity is invoked.  :cheese:

There is much more exotic stuff to read in the wiki which will really bend your mind.  ;-)

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 Posted: 06 January 2012 09:23 AM [ Ignore ]   [ # 7 ]
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StephenLawrence - 05 January 2012 10:24 AM
kkwan - 05 January 2012 10:10 AM

So, the potential gain in situation 1 (100%) is strictly greater than the potential loss in situation 2 (50%).

I think the solution is that the amount of money that can be gained or lossed is what matters.

It’s irrelevent what that is as a percentage of what’s in the envelope.

It is relevant because if the potential gain is greater than the potential loss, that is the guide to switch.

So, in both situations, one should switch because one does not know whether the other envelope has more or less money

OTOH, can your solution guide you to whether any money can be gained or lost at all, i.e. to switch or not to switch?

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 Posted: 06 January 2012 01:03 PM [ Ignore ]   [ # 8 ]
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kkwan - 06 January 2012 08:41 AM
domokato - 05 January 2012 10:52 AM

Steps 4 and 5 are conditionals. The other envelope contains 2A if A is the smaller amount, and A/2 if A is the larger amount. Step 6 combines these two equations into one, which is invalid because you have not yet resolved whether A is the larger or smaller amount.

Steps 4 and 5 are not conditionals.

They are only statements of what is in the other envelope if A is smaller or larger.

(emphasis mine)

Isn’t that what a conditional is?

And they are not equations.

Yes, sorry for inaccurate terminology, but aren’t you being a bit nitpicky? I think my solution is essentially the same as the one you quoted (and Andrew’s :)).

Then, there is a third variant:

Though probability theory can defuse the second variant of the paradox, it turns out that examples can still easily be found of proper probability distributions, such that the expected value of the amount in the second envelope given that in the first does exceed the amount in the first, whatever it might be.

Proposed resolution

Thus if we switch for the second envelope because its conditional expected value is larger than what actually is in the first, whatever that might be, we are exchanging an unknown amount of money whose expectation value is infinite for another unknown amount of money with the same distribution and the same infinite expected value.

Hm, I don’t understand why the expectation value is infinite. Also, this solution doesn’t point out the flaw in the original argument (of the original variant) - it only posits a correct resolution. So then what’s the problem with the first proposed resolution?

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 Posted: 06 January 2012 09:16 PM [ Ignore ]   [ # 9 ]
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domokato - 06 January 2012 01:03 PM
kkwan - 06 January 2012 08:41 AM
domokato - 05 January 2012 10:52 AM

Steps 4 and 5 are conditionals. The other envelope contains 2A if A is the smaller amount, and A/2 if A is the larger amount. Step 6 combines these two equations into one, which is invalid because you have not yet resolved whether A is the larger or smaller amount.

Steps 4 and 5 are not conditionals.

They are only statements of what is in the other envelope if A is smaller or larger.

(emphasis mine)

Isn’t that what a conditional is?

From the wiki on indicative conditional

In natural languages, an indicative conditional is the logical operation given by statements of the form “If A then B”. Unlike the material conditional, an indicative conditional does not have a stipulated definition. The philosophical literature on this operation is broad, and no clear consensus has been reached.

Because an indicative conditional does not have a stipulated definition, one can consider steps 4 and 5 as statements of the logical operation (given A is larger or smaller).

Yes, sorry for inaccurate terminology, but aren’t you being a bit nitpicky? I think my solution is essentially the same as the one you quoted (and Andrew’s :)).

No, your solution only considered steps 4,5, and 6 which are valid whereas the proposed resolution found inconsistencies with step 7.

Hm, I don’t understand why the expectation value is infinite. Also, this solution doesn’t point out the flaw in the original argument (of the original variant) - it only posits a correct resolution. So then what’s the problem with the first proposed resolution?

The expected value becomes infinite because of the term “whatever it might be” in the third variant:

Though probability theory can defuse the second variant of the paradox, it turns out that examples can still easily be found of proper probability distributions, such that the expected value of the amount in the second envelope given that in the first does exceed the amount in the first, whatever it might be.

It purports to show the absurdity of comparing unknown infinite values, the flaw in the third variant. Since the original and all the variants are related, the flaw in the third variant implies the original paradox is similarly flawed because it is also absurd to switch ad infinitum in step 11, notwithstanding the logic of steps 1 to 10.

Whenever infinity is invoked in any argument, it is flawed as infinity cannot be defined at all. :lol:

If any philosopher had been asked for a definition of infinity, he might have produced some unintelligible rigmarole, but he would certainly not have been able to give a definition that had any meaning at all.

Bertrand Russell

The first proposed resolution collapses with the second variant:

The solution above doesn’t explain what’s wrong if the player is allowed to open the first envelope before being offered the option to switch. In this case, A is indeed a fixed variable. Hence, the proposed solution in the first case breaks down and another explanation is needed.

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 Posted: 06 January 2012 11:51 PM [ Ignore ]   [ # 10 ]
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kkwan - 06 January 2012 09:23 AM
StephenLawrence - 05 January 2012 10:24 AM
kkwan - 05 January 2012 10:10 AM

So, the potential gain in situation 1 (100%) is strictly greater than the potential loss in situation 2 (50%).

I think the solution is that the amount of money that can be gained or lossed is what matters.

It’s irrelevent what that is as a percentage of what’s in the envelope.

It is relevant because if the potential gain is greater than the potential loss, that is the guide to switch.

Looks like I’ve missed something?

The potential gain must be the same as the potential loss because the potential gain is the difference between what is in the two envelopes, same as the potential loss is.

It’s only if we do something that we have no reason to do, which is take that as a percentage of what’s in the envelope that there appears to be a puzzle.

OTOH, can your solution guide you to whether any money can be gained or lost at all, i.e. to switch or not to switch?

Yeah you can gain the same amount as you can lose so there is no point in switching.

As I say maybe I’ve missed something?

Stephen

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 Posted: 07 January 2012 08:56 PM [ Ignore ]   [ # 11 ]
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StephenLawrence - 06 January 2012 11:51 PM

Yeah you can gain the same amount as you can lose so there is no point in switching.

As I say maybe I’ve missed something?

It is relative. There are 3 possible situations.

1. Gain…..+1 (100%)

2. Even…....0 (don’t switch)

3. Loss…...-1 (50%)

So, as +1 is better than 0 or -1 and generally:

Let the amount in the envelope chosen by the player be A. By swapping, the player may gain A or lose A/2. So the potential gain is strictly greater than the potential loss.

Also, in steps 7 and 8:

7. So the expected value of the money in the other envelope is

8. This is greater than A, so I gain on average by swapping.

Thus, one should switch.

[ Edited: 07 January 2012 09:10 PM by kkwan ]
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 Posted: 07 January 2012 09:30 PM [ Ignore ]   [ # 12 ]
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As long as I don’t have to buy the envelopes and have no investment, I gain either way. If A has \$1, I gain 1\$. If B has \$2, I gain \$2. I can’t lose anything either way, there is no penalty involved. Can’t look a gift horse in the mouth…. :)

[ Edited: 07 January 2012 09:34 PM by Write4U ]
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 Posted: 08 January 2012 02:34 AM [ Ignore ]   [ # 13 ]
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kkwan - 07 January 2012 08:56 PM
StephenLawrence - 06 January 2012 11:51 PM

Yeah you can gain the same amount as you can lose so there is no point in switching.

As I say maybe I’ve missed something?

It is relative. There are 3 possible situations.

1. Gain…..+1 (100%)

2. Even…....0 (don’t switch)

3. Loss…...-1 (50%)

So, as +1 is better than 0 or -1 and generally:

Ok, I haven’t missed anything the potential gain or loss as a percentage of the envelope is irrelevent.

It’s just a mistake to take that into account.

I don’t know if there is a better way to say that, but that’s the answer.

We’ve had a similar example come up before, I’ll see if I can find it and post it.

Stephen

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 Posted: 08 January 2012 02:55 AM [ Ignore ]   [ # 14 ]
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kkwan - 07 January 2012 08:56 PM
StephenLawrence - 06 January 2012 11:51 PM

Yeah you can gain the same amount as you can lose so there is no point in switching.

As I say maybe I’ve missed something?

It is relative.

Another way to make the point.

Yes it is relative but relative to what?

The relative relationship we are interested in is the amount that can be gained relative to the amount that can be lost and these are equal.

The trick is done by irrationally making the potential loss and gain relative to two different figures (the amounts in the envelopes) depending upon which envelope you have.

I think I’m saying the same thing as domokato.

Stephen

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 Posted: 08 January 2012 09:27 AM [ Ignore ]   [ # 15 ]
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StephenLawrence - 08 January 2012 02:55 AM

Yes it is relative but relative to what?

The relative relationship we are interested in is the amount that can be gained relative to the amount that can be lost and these are equal.

The trick is done by irrationally making the potential loss and gain relative to two different figures (the amounts in the envelopes) depending upon which envelope you have.

I think I’m saying the same thing as domokato.

It is relative as follows:

If one does not switch, there is no gain or loss at all, but one loses the opportunity to double the money.

OTOH, if one switches, there are two possible situations:

1. One gets double the money i.e 100% gain

2. One gets half the money i.e. 50% loss

Now, consider it as a calculated gamble to switch.

The odds are in your favor on average as the expected value is more than A.

Also, the potential gain is more than the potential loss.

It is double or half, not double or nothing.

There is no trick.

So, why should one not switch?

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