Straight Dope Message Board > Main 2 envelopes problem revisited
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#1
06-20-2012, 11:23 AM
 Saffer Guest Join Date: Feb 2009
2 envelopes problem revisited

The short version for those who don't want to click the link:
I have 2 sealed envelopes, one of which has exactly twice as much money as the other.
I tell you if you pick one of the envelopes you can keep the cash.
After you pick an evelope I offer you the opportunity to take the other one instead.
You reason thusly:
Let X be the value of the envelope you are holding.
The other envelope must have either 2X or X/2 value with 50-50 probability.
The expected value of the other envelope is therefore 50%(2X) + 50%(X/2) = (5/4)X
The expected value of the other envelope is more than the one in your hand, therefore you should switch.
Doing so leaves you in the same position as before so according to the above analysis, you should continue to swap forever.

Im sure there have been many threads asking to explain how to resolve this problem, this is not one of those.

Clearly the expected value (in reality) of the other envelope is the same as the one in your hand.
Let p be the probability that the other envelope contains twice as much money (2X) as the one on your hand. Therefore the probability of the other envelope containing X/2 is (1-p).

The expected value of the other envelope is p(2X) + (1-p)(X/2) which as stated above is equal to the value of the envelope in your hand.
So solve for p:
p(2X) + (1-p)(X/2) = X
2p + (1-p)/2 = 1
p = 1/3

So it turns out that the probability of the other envelope containing twice as much cash as the one in your hand is always 1/3

What is wrong with this reasoning?
#2
06-20-2012, 12:14 PM
 dracoi Guest Join Date: Dec 2008
If one envelope has twice as much money, you should be using 2x and x, or x and x/2. To use 2x and x/2 means that one envelope has four times as much money.
#3
06-20-2012, 12:21 PM
 Darth Panda Guest Join Date: Mar 2010
I would just say that the total amount of money is x, and that the envelope you hold in your hand (let's call it envelope a) has an expected value of:

(1/2 * 1/3 * x) + (1/2 * 2/3 * x) [that is, a 50% chance of having 1/3 of all of the money and a 50% chance of having 2/3 of all of the money]

the envelope that you don't hold (let's call it envelope b) also has an expected value of (1/2 * 1/3 * x) + (1/2 * 2/3 * x)

Therefore, ev(a)=ev(b) and switching accomplishes nothing

Last edited by Darth Panda; 06-20-2012 at 12:24 PM.
#4
06-20-2012, 01:02 PM
 TriPolar Member Join Date: Oct 2007 Location: rhode island Posts: 21,602
Quote:
 Originally Posted by dracoi If one envelope has twice as much money, you should be using 2x and x, or x and x/2. To use 2x and x/2 means that one envelope has four times as much money.
No. If this envelope contains X, the other envelope contains either 2X or X/2.

I don't have the time or patience to read the whole wiki article in detail and then not understand it in the end anyway. But their description of a paradox seems to be a calculation that shows switching is an advantage based on the average of the two amounts, but doesn't see that the same calculation could be used for not switching with the same result. I don't really understand the paradox here.
#5
06-20-2012, 01:16 PM
 Saffer Guest Join Date: Feb 2009
Quote:
 Originally Posted by TriPolar No. If this envelope contains X, the other envelope contains either 2X or X/2. I don't have the time or patience to read the whole wiki article in detail and then not understand it in the end anyway. But their description of a paradox seems to be a calculation that shows switching is an advantage based on the average of the two amounts, but doesn't see that the same calculation could be used for not switching with the same result. I don't really understand the paradox here.
The paradox is that anyone with half a brain can see that the expected value of each envelope is the same as the other, whilst the analysis I demonstrated says otherwise.
#6
06-20-2012, 01:19 PM
 Saffer Guest Join Date: Feb 2009
Quote:
 Originally Posted by Darth Panda I would just say that the total amount of money is x, and that the envelope you hold in your hand (let's call it envelope a) has an expected value of: (1/2 * 1/3 * x) + (1/2 * 2/3 * x) [that is, a 50% chance of having 1/3 of all of the money and a 50% chance of having 2/3 of all of the money] the envelope that you don't hold (let's call it envelope b) also has an expected value of (1/2 * 1/3 * x) + (1/2 * 2/3 * x) Therefore, ev(a)=ev(b) and switching accomplishes nothing
Thats fine but to be fair all you have accomplished is proving something that everyone knows anyway.
The problem is to identify where the reasoning that leads to the paradox goes wrong.

But anyway, as I stated that is not what this thread is about.
#7
06-20-2012, 01:24 PM
 Buck Godot Guest Join Date: Mar 2010
The problem is that there is no information as to how the values in the envelopes are chosen. You are naively assuming that since there is no information all situations are equally likely, but there is no reason for this to be true.

In stat speak, you are attempting to apply Bayes rule, without a fixed prior.
#8
06-20-2012, 01:45 PM
 OldGuy Charter Member Join Date: Dec 2002 Location: Very east of Foggybog, WI Posts: 2,416
Quote:
 Originally Posted by Buck Godot The problem is that there is no information as to how the values in the envelopes are chosen. You are naively assuming that since there is no information all situations are equally likely, but there is no reason for this to be true. In stat speak, you are attempting to apply Bayes rule, without a fixed prior.
In a bit more detail there is no proper prior for which it is equally likely that the other envelope has either twice as much or half as much. This prior would have to have a equal value for all numbers of the form a2n where n runs from minus infinity to plus infinity. But such a prior cannot be normalized to sum to unity.
#9
06-20-2012, 01:51 PM
 Acsenray Charter Member Join Date: Apr 2002 Location: U.S.A. Posts: 24,781
If you say that the second envelope holds either 1/2 A or 2A, there's your problem, because the A in 1/2 A is not equal to the A in 2A. You've got two different As.
#10
06-20-2012, 01:53 PM
 Buck Godot Guest Join Date: Mar 2010
The proper way to analyze it is to assume that there is a fixed probability density f(x) which represents the relative probability that there are x dollars in the smaller envelope.

Then if you see Y dollars when you open the envelope, your expected value for staying is obviously Y
While your expected value for switching according to Bayes Rule is (Y/2*f(Y/2)+2Y*f(Y)))/(f(Y/2)+f(Y))

So you should switch if (Y/2*f(Y/2)+2Y*f(Y)))/(f(Y/2)+f(Y))>Y or more simply if 2*f(Y) > f(Y/2).

The first version of the analysis made the assumption that there was a flat prior but this leads to an improper distribution, hence the apparent contradiction. That doesn't mean that the assumption that switching envelopes makes no difference is any better, since that would inherently assume that 2*f(Y) = f(Y/2).
#11
06-20-2012, 01:54 PM
 aux203 Guest Join Date: Apr 2005
Quote:
Okay

Quote:
 Clearly the expected value (in reality) of the other envelope is the same as the one in your hand.
Yup

Quote:
 Let p be the probability that the other envelope contains twice as much money (2X) as the one on your hand. Therefore the probability of the other envelope containing X/2 is (1-p).
Nope

When you use 2X, you are assigning the variable X to the value of the money in your envelope if that money is the smaller amount. Let's call this X1 instead. When you use X/2, you are assigning the variable X to the value of the envelope if that money is the larger amount. Let's call this X2 for clarity.

But by definition, X1 is X2/2. They are not the same thing. But you proceed to treat them as if they are. The rest of your calculation errors flow from here.

There are a number of ways to analyze this problem but that's the simplest way to look at it, in my opinion.
#12
06-20-2012, 02:20 PM
 KneadToKnow Voodoo Adult (Slight Return) Charter Member Join Date: Jul 2000 Location: Charlotte, NC, USA Posts: 22,275
Quote:
 Originally Posted by Acsenray You've got two different As.
#13
06-20-2012, 02:34 PM
 Cheesesteak Guest Join Date: Jan 2001
Quote:
 Originally Posted by aux203 When you use 2X, you are assigning the variable X to the value of the money in your envelope if that money is the smaller amount. Let's call this X1 instead. When you use X/2, you are assigning the variable X to the value of the envelope if that money is the larger amount. Let's call this X2 for clarity.
I was thinking this way, then I changed the way I considered the problem.

Instead of X, let's say you opened your envelope, it has \$100. What is in the other envelope? Absent information regarding the total amount of money available, the other envelope is going to have either \$200 or \$50.

If \$100 is X1, then the other envelope has \$200. If \$100 is X2, then the other is \$50.

The expected value of each envelope is the same, but once you open your envelope and find it's value to be \$100, how do you get the value of the other envelope to equal \$100, without changing the probability of \$200 vs \$50? Does that mean it's more likely that \$100 is X2? How does that happen?
#14
06-20-2012, 02:44 PM
 Dr. Love Charter Member Join Date: Mar 2002 Posts: 935
Quote:
 Originally Posted by Cheesesteak The expected value of each envelope is the same, but once you open your envelope and find it's value to be \$100, how do you get the value of the other envelope to equal \$100, without changing the probability of \$200 vs \$50? Does that mean it's more likely that \$100 is X2? How does that happen?
You and the OP are mixing expectation with conditional expectation. Before you open your envelope both must have the same conditional expectation. But after opening and seeing how much money it contains, the expected value of the other envelope changes. Consider a simple example where the envelopes contain \$1, \$2 or \$4. Before you open yours, both envelopes have a conditional expectation of \$7/3 = \$2.33... But when you open:
• If your envelope contains \$1, the conditional expectation of the other is \$2 (because it must contain \$2).
• If yours contains \$4, the conditional expectation of the other is \$2.
• If yours contains \$2, then the conditional expectation of the other is (1/2)\$4 + (1/2)\$1 = \$2.50.
Once you open your envelope you've changed the information available to you about the other envelope.

The OP's reasoning wrongly assumes that if the two envelopes have equal unconditional expectation, they must also have equal conditional expectation.
#15
06-20-2012, 02:46 PM
 Ethilrist Guest Join Date: Nov 2000
Schroedinger wept.
#16
06-20-2012, 02:47 PM
 Terr Guest Join Date: Sep 2011
Quote:
 Originally Posted by Dr. Love You and the OP are mixing expectation with conditional expectation. Before you open your envelope both must have the same conditional expectation. But after opening and seeing how much money it contains, the expected value of the other envelope changes. Consider a simple example where the envelopes contain \$1, \$2 or \$4. Before you open yours, both envelopes have a conditional expectation of \$7/3 = \$2.33... But when you open:If your envelope contains \$1, the conditional expectation of the other is \$2 (because it must contain \$2). If yours contains \$4, the conditional expectation of the other is \$2. If yours contains \$2, then the conditional expectation of the other is (1/2)\$4 + (1/2)\$1 = \$2.50. Once you open your envelope you've changed the information available to you about the other envelope. The OP's reasoning wrongly assumes that if the two envelopes have equal unconditional expectation, they must also have equal conditional expectation.
That's only if you know the totals. If you don't, but only know that the envelopes contain X, 2X, and 4X, when you open an envelope and it contains \$20, you still don't know which one of the three you opened.
#17
06-20-2012, 03:45 PM
 mcgato Guest Join Date: Aug 2010
How did we get three envelopes all of the sudden?

I think that Ascenray is the correct one that there are two As.
#18
06-20-2012, 03:55 PM
 Ethilrist Guest Join Date: Nov 2000
Quote:
 Originally Posted by Dr. Love You and the OP are mixing expectation with conditional expectation. Before you open your envelope both must have the same conditional expectation. But after opening and seeing how much money it contains, the expected value of the other envelope changes. Consider a simple example where the envelopes contain \$1, \$2 or \$4. Before you open yours, both envelopes have a conditional expectation of \$7/3 = \$2.33... But when you open:If your envelope contains \$1, the conditional expectation of the other is \$2 (because it must contain \$2). If yours contains \$4, the conditional expectation of the other is \$2. If yours contains \$2, then the conditional expectation of the other is (1/2)\$4 + (1/2)\$1 = \$2.50. Once you open your envelope you've changed the information available to you about the other envelope. The OP's reasoning wrongly assumes that if the two envelopes have equal unconditional expectation, they must also have equal conditional expectation.
Does this change if you have no idea what the amounts are (as implied in the OP)? Upon finding \$4, you don't know whether the other envelope has \$2 or \$8.

There's also nothing in the OP indicating that you make your choice AFTER opening your envelope. Does it change anything if you have a sealed envelope in your hand?
#19
06-20-2012, 04:05 PM
 Dr. Love Charter Member Join Date: Mar 2002 Posts: 935
To clarify my example, there are only two envelopes. The possible outcomes are
• Your envelope contains \$1; the other contains \$2
• Yours: \$2; other: \$1
• Yours:\$2; other: \$4
• Yours \$4; other :\$2

If you really have no idea what amounts could be in the two envelopes then it's meaningless to talk about the expected value of the envelopes, because there is no probability distribution that would model such a situation (as OldGuy pointed out).
#20
06-20-2012, 04:45 PM
 Colophon Guest Join Date: Sep 2002
Quote:
 Originally Posted by Acsenray If you say that the second envelope holds either 1/2 A or 2A, there's your problem, because the A in 1/2 A is not equal to the A in 2A. You've got two different As.
This is the key. You can't have the same variable standing for two different amounts.

E.g. let's say the envelopes contain \$500 and \$1000.

If the second envelope contains A/2, then A must equal 1000. If the second envelope contains 2A, then A must equal 500.

What you need to do is say, the two envelopes hold A and 2A dollars respectively, where A = 500, in this case.

The two cases are that you are holding A in your hand, or you are holding 2A in your hand. The probabilities of each of those happening are 0.5, agreed? (Barring any psychic ability on your part.)

Therefore, with p(0.5), the value of the second envelope is 2A (because you are holding A), and with p(0.5), the value of the second envelope is A (because you are holding 2A).

Expected value of envelope in your hand = (0.5 x A) + (0.5 x 2A) = 1.5A
Expected value of second envelope = (0.5 x 2A) + (0.5 x A) = 1.5A

So the expected value of both envelopes is the same, and is the mean of the higher and the lower values (in this example \$750), which is, er, as you would expect, right?

Now, I didn't study maths beyond school, so maybe I'm missing some subtlety here, but I am really not seeing that there is any paradox here at all, other than using one letter to stand for two different values and expecting an equation to work.

Can anyone explain to me in layman's terms why this so-called problem is worthy of its own Wikipedia entry, etc?

Last edited by Colophon; 06-20-2012 at 04:47 PM.
#21
06-20-2012, 05:12 PM
 Chronos Charter Member Join Date: Jan 2000 Location: The Land of Cleves Posts: 50,877
Once you've opened one of the envelopes, then you need more information to make any meaningful judgment. If, for instance, you know that the person offering you the envelopes has a maximum available budget of less than \$300, and your envelope has \$100, then you should always keep your envelope, because you know that it must be the larger of the two.
#22
06-20-2012, 05:20 PM
 Cheesesteak Guest Join Date: Jan 2001
Quote:
 Originally Posted by Colophon Expected value of envelope in your hand = (0.5 x A) + (0.5 x 2A) = 1.5A Expected value of second envelope = (0.5 x 2A) + (0.5 x A) = 1.5A [/i] So the expected value of both envelopes is the same, and is the mean of the higher and the lower values (in this example \$750), which is, er, as you would expect, right? Can anyone explain to me in layman's terms why this so-called problem is worthy of its own Wikipedia entry, etc?
What happens when you open the envelope, and see that it contains \$100? Is \$100 A, 2A or 1.5A?
#23
06-20-2012, 05:43 PM
 Indistinguishable Guest Join Date: Apr 2007
One way to "resolve" the paradox is to say "Hey, let's suppose that there are actually only finitely many possible monetary values" or "Let's suppose that the values aren't all equiprobable". But these aren't really confronting the core issue.

The paradox is essentially just the phenomenon of conditionally convergent series. It's the same as the paradox that 1 - 1 + 1 - 1 + 1 - 1 + ... can be made to look positive, negative, or zero, depending on how you re-arrange and group the terms.

To make this clear, let's frame the problem in a different way:

Suppose we place an equal weight at every point (X, Y) where X and Y are adjacent powers of 2 [thinking of X as representing the value of your envelope and Y as representing the value of the other envelope]. Like so:

Code:
```YOUR ENVELOPE

| 1 | 1 | 1 |   |   |   |   |
| - | - | - | 1 | 2 | 4 | 8 |
| 8 | 4 | 2 |   |   |   |   |
+---+---+---+---+---+---+---+

+---+---+---+---+---+---+---+ +--
|   |   |   |   |   |   |   | |
|   |   |   |   |   | @ |   | | 8  T
|   |   |   |   |   |   |   | |    H
+---+---+---+---+---+---+---+ +--  E
|   |   |   |   |   |   |   | |
|   |   |   |   | @ |   | @ | | 4  O
|   |   |   |   |   |   |   | |    T
+---+---+---+---+---+---+---+ +--  H
|   |   |   |   |   |   |   | |    E
|   |   |   | @ |   | @ |   | | 2  R
|   |   |   |   |   |   |   | |
+---+---+---+---+---+---+---+ +--  E
|   |   |   |   |   |   |   | |    N
|   |   | @ |   | @ |   |   | | 1  V
|   |   |   |   |   |   |   | |    E
+---+---+---+---+---+---+---+ +--  L
|   |   |   |   |   |   |   | | 1  O
|   | @ |   | @ |   |   |   | | -  P
|   |   |   |   |   |   |   | | 2  E
+---+---+---+---+---+---+---+ +--
|   |   |   |   |   |   |   | | 1
| @ |   | @ |   |   |   |   | | -
|   |   |   |   |   |   |   | | 4
+---+---+---+---+---+---+---+ +--
|   |   |   |   |   |   |   | | 1
|   | @ |   |   |   |   |   | | -
|   |   |   |   |   |   |   | | 8
+---+---+---+---+---+---+---+ +--```
(I've drawn the axes on log-scale for convenience)

Where is the center of mass of this system?

One argument says: Well, this system is symmetric under reflection across the line Y = X, so the center of mass must be on that line.

Another argument says: For each particular column, the center of mass is on the line Y = 1.25X, so the overall center of mass must also be on that line.

Another argument says: For each particular row, the center of mass is on the line Y = 0.8X, so the overall center of mass must also be on that line.

But these are three different lines!

Yes, troubling... But no more so than the ambiguity in the question "Where is the center of the integers?". One might say "0", because the integers are symmetric around 0... but the integers are just as symmetric around 17, or -1/2.

We can see the connection to conditionally convergent series even more clearly if we look at the profit from switching envelopes:

Code:
```YOUR ENVELOPE

| 1 | 1 | 1 |   |   |   |   |
| - | - | - | 1 | 2 | 4 | 8 |
| 8 | 4 | 2 |   |   |   |   |
+---+---+---+---+---+---+---+

+---+---+---+---+---+---+---+ +--
|   |   |   |   |   |   |   | |
|   |   |   |   |   | 4 |   | | 8  T
|   |   |   |   |   |   |   | |    H
+---+---+---+---+---+---+---+ +--  E
|   |   |   |   |   |   |   | |
|   |   |   |   | 2 |   |-4 | | 4  O
|   |   |   |   |   |   |   | |    T
+---+---+---+---+---+---+---+ +--  H
|   |   |   |   |   |   |   | |    E
|   |   |   | 1 |   |-2 |   | | 2  R
|   |   |   |   |   |   |   | |
+---+---+---+---+---+---+---+ +--  E
|   |   | 1 |   |   |   |   | |    N
|   |   | - |   |-1 |   |   | | 1  V
|   |   | 2 |   |   |   |   | |    E
+---+---+---+---+---+---+---+ +--  L
|   | 1 |   |-1 |   |   |   | | 1  O
|   | - |   | - |   |   |   | | -  P
|   | 4 |   | 2 |   |   |   | | 2  E
+---+---+---+---+---+---+---+ +--
| 1 |   |-1 |   |   |   |   | | 1
| - |   | - |   |   |   |   | | -
| 8 |   | 4 |   |   |   |   | | 4
+---+---+---+---+---+---+---+ +--
|   |-1 |   |   |   |   |   | | 1
|   | - |   |   |   |   |   | | -
|   | 8 |   |   |   |   |   | | 8
+---+---+---+---+---+---+---+ +--```
Naturally, along the top left line, where the other envelope has more money, you get positive profits, while along the bottom right line, where the other envelope has less money, you get corresponding negative profits.

Now, the question is, on average, is it good, neutral, or bad to switch envelopes? This is essentially the question as to whether the sum of the profits over all cases is positive, zero, or negative. And... there is an ambiguity here. It's not clear what the value, or even the sign, of the sum of all the boxes comes out to. For example, consider the following diagram:

Code:
```+---+---+---+---+---+---+---+
|   |   |   |   |   |   |   |
|   |   |   |   |   | 4 |   | ...
|   |   |   |   |   |   |   |
+---+---+---+---+---+---+---+
|   |   |   |   |   |   |   |
|   |   |   |   | 2 | + |-4 | = -2
|   |   |   |   |   |   |   |
+---+---+---+---+---+---+---+
|   |   |   |   |   |   |   |
|   |   |   | 1 | + |-2 |   | = -1
|   |   |   |   |   |   |   |
+---+---+---+---+---+---+---+
|   |   | 1 |   |   |   |   |
|   |   | - | + |-1 |   |   | = -1/2
|   |   | 2 |   |   |   |   |
+---+---+---+---+---+---+---+
|   | 1 |   |-1 |   |   |   |
|   | - | + | - |   |   |   | = -1/4
|   | 4 |   | 2 |   |   |   |
+---+---+---+---+---+---+---+
| 1 |   |-1 |   |   |   |   |
| - | + | - |   |   |   |   | = -1/8
| 8 |   | 4 |   |   |   |   |
+---+---+---+---+---+---+---+
|   |-1 |   |   |   |   |   |
|   | - |   |   |   |   |   | ...
|   | 8 |   |   |   |   |   |
+---+---+---+---+---+---+---+
=   =   =   =   =
... 1/8 1/4 1/2  1   2  ...```
This shows that if you add the numbers down each column first, you get a bunch of positive values (which add up to an infinitely large total). On the other hand, if you add the numbers across each row first, you get correspondingly negative values (which add up to an infinitely negative total). And, of course, if you add the numbers along each diagonal pair first [in the \ direction], you'll just get a bunch of zeros (which add up to a total of zero).

So there is this ambiguity as to what the sum of the numbers in that square comes out to, and that is precisely where all the ambiguity as to whether one should switch comes from. When it looks like switching envelopes is a great idea, this is because one is essentially adding down the columns to get a positive total for the square. When it looks like switching envelopes is a terrible idea, this is because one is essentially adding across the rows to get a negative total for the square. And when it looks like switching envelopes shouldn't make any difference, this is because one is essentially noting the symmetry and adding diagonally to get a total of zero. Three different ways of totalling that square, which give three different kinds of results. And there's nothing more to it than looking at that square and observing, yup, that can happen.

Does that help illustrate anything?

(Note: This is largely a repost from here)

Last edited by Indistinguishable; 06-20-2012 at 05:47 PM.
#24
06-20-2012, 05:54 PM
 Colophon Guest Join Date: Sep 2002
Quote:
 Originally Posted by Cheesesteak What happens when you open the envelope, and see that it contains \$100? Is \$100 A, 2A or 1.5A?
You know that with p(0.5), A = \$50, and with p(0.5), A = \$100.

In other words, the expected value of A is \$75.

If A = \$50, then the expected value was \$75. If you stick, you are \$25 up on the EV, if you switch, you are \$25 down.

If A = \$100, then the EV was \$150. If you stick, you are \$50 down on the EV, if you switch, you are \$50 up.

Regardless of which case is true, you have an equal chance of winning or losing.

I see now what the "paradox" is - the average EV would seem to be more than \$100, suggesting you should always switch.

But that only works once you know what the amount is and there is no hidden variable. It's like saying, "Here's \$100. Would you like to swap that for an even chance of \$50 or \$200?" Well of course you would - the upside is bigger than the downside.

You can't extend that to say "you should always keep switching", because clearly you can't. Either you switch and you win, or you switch and you lose.
#25
06-20-2012, 06:05 PM
 Dr. Love Charter Member Join Date: Mar 2002 Posts: 935
The paradox comes from the apparent difference in outcomes from the player's perspective vs. the game maker's perspective. Those arguing that A (or X) stands for two different things are seeing this exclusively from the game maker's perspective, i.e., someone who know exactly what the two amounts are, but perhaps not which envelope contains which amount.

The player OTOH has no idea at all about what the two amounts are. Imagine you go into the home of an eccentric infinataire (who has an infinite amount of money). He hands you an envelope with some money in it, and holds up a second envelope. He says,

"I put some amount of money in the envelope you're holding. Then I flipped a coin. If the coin turned up heads, I put double that amount of money in this envelope. If it turned up tails, I put half in this envelope. Would you like to trade envelopes with me?"

Suppose you look into the envelope before making your decision, and you found \$100. Then there is a 50-50 shot that the other envelope contains \$50 or \$200. Then the expected value of the other envelope is (1/2)*50 + (1/2)*200 = \$125, which is more than your \$100, so you switch.

Suppose instead you found \$450 in your envelope. In this case the other envelope either contains \$225 or \$900, with expected value \$562.50. So again, you would want to switch.

In fact, no matter what your envelope contains, the other envelope has an expected value of 25% more. You'll switch no matter what your envelope contains, so why even bother opening it?

So the infinitaire leaves and comes back with a third envelope, and tells you the same thing about the coin flip, so this envelope either contains half or double the amount you currently have, and offers to switch with you. By the same analysis, you would switch without even looking in your envelope, right?

Let's change things a bit. Same scenario, but instead of giving you a long explanation about coin flips, the infinitaire tersely says, "This envelope either has half or double the amount in your envelope, with equal probability." Surely the same analysis applies, and you would switch without even opening your own envelope? And when he goes back to get a third envelope and gives the same terse explanation, you would switch again without opening the envelope?

Finally, suppose that before the Infinitaire gives you an envelope, he thoroughly mixes the first two up so that even he doesn't know which is which. So he hands you one and tells you the other contains either half or double that amount, with equal probability. This statement is still true, yes? So you switch without looking in your own envelope, and he leaves, but this time, unbeknown to you, he comes back with the same envelope you just gave him. He says "his envelope either has half or double the amount in your envelope, with equal probability." Neither you nor he knows which one holds more money, so it's 50-50 that he has the larger envelope. So, you do the same analysis in your head, conclude that the envelope he holds has an expected value 25% higher than the one you hold, and decide that you should switch envelopes.

Thus the paradox. It seems that the player would be made better off by continually switching envelopes, but we know that it cannot be so.
#26
06-20-2012, 06:30 PM
 Darth Panda Guest Join Date: Mar 2010
The thing is once you open one envelope, and discovered the amount of money inside, it probably makes sense to trade for the other one.

For example, if you open the envelope and it holds \$100 and someone gives the option of trading it, you should take it.

Think about. There two possibilities, that the other envelope contains \$50 or that it contains \$200.

It's like being offered to play a game with a 50% chance to win \$200 and a 50% to win \$50, with a play price of \$100. That's clearly a good game.

Another way to look at it is even odds to lose \$50 (going from \$100 to \$50), or to gain \$100 (going from \$100 to \$200). Again, this is a good deal.

But you may not want to make the trade in very extreme circumstances, as the relationship between utility and money can become strange at the extremes and the ultimate goal is to maximize utility, not dollars and so you might need to incorporate a risk adjusted point of view. For example, if you opened the envelope and it contained 2 million dollars, you might not want to risk losing 1 million dollars for the chance of winning another \$2 million even though the game a has a positive expected value. It would all depend on the curve of utility to dollar ratio.

But the point is that once you have a piece of information, the paradox goes away. Trading for the other envelope makes sense. In the absence of that information, the expected value is equal. It's also different in that once you open one envelope you have a constrained view of the options and you know that once you switch, you now have a new game. In that when you have an unopened envelope in your hand and the \$100 as the other option, you know that if you trade you have a 50% chance to gain \$50 and 50% to lose \$100 - so the proposed trade is non-paradoxical in that you would have no inclination to trade back.

Last edited by Darth Panda; 06-20-2012 at 06:31 PM.
#27
06-20-2012, 07:57 PM
 Colophon Guest Join Date: Sep 2002
Quote:
 Originally Posted by Dr. Love Finally, suppose that before the Infinitaire gives you an envelope, he thoroughly mixes the first two up so that even he doesn't know which is which. So he hands you one and tells you the other contains either half or double that amount, with equal probability. This statement is still true, yes? So you switch without looking in your own envelope, and he leaves, but this time, unbeknown to you, he comes back with the same envelope you just gave him. He says "his envelope either has half or double the amount in your envelope, with equal probability." Neither you nor he knows which one holds more money, so it's 50-50 that he has the larger envelope. So, you do the same analysis in your head, conclude that the envelope he holds has an expected value 25% higher than the one you hold, and decide that you should switch envelopes. Thus the paradox. It seems that the player would be made better off by continually switching envelopes, but we know that it cannot be so.
That's not really a paradox - that's playing with incomplete knowledge of the game. If the man is truly an infinitaire, and can keep preparing fresh envelopes, each of them checked against the previous envelope and either doubled or halved with equal probability, then the more times you play the better off you are likely to be. That's surely a fact, not a paradox. At each step your expected win increases by 25%, but that's OK because the man has infinite wealth and can keep stuffing new envelopes with more cash.

The paradox only arises if he switches back to your original envelope without your knowledge, in which case all bets are off. And that's effectively what happens if there are only ever two envelopes to start off with - if you keep switching, you're just getting the same envelope "switcherooed" on you.

Last edited by Colophon; 06-20-2012 at 07:59 PM.
#28
06-20-2012, 09:06 PM
 Chronos Charter Member Join Date: Jan 2000 Location: The Land of Cleves Posts: 50,877
Quote:
 The thing is once you open one envelope, and discovered the amount of money inside, it probably makes sense to trade for the other one.
If you know that seeing the amount is going to make you want to switch, then why not just say right at the start "I'm going to open the envelope to look, and then switch"? And if you're going to do that anyway, then what's the point of looking in the envelope?
#29
06-20-2012, 09:29 PM
 Little Nemo Charter Member Join Date: Dec 1999 Location: Western New York Posts: 52,637
Logically, you can look at the situation as a whole and it's clear that you should just pick an envelope at random and take its contents. You have no information on which to make a choice that will be more informed than a random guess.

But when you try to prove this answer is correct, it falls apart. No matter what amount of money there is in your envelope, by switching you will either gain an amount equal to what you have or lose an amount equal to half of what you have. So it appears that the possible gain is twice as large as the possible loss and the possibilities of a gain or loss are equal so you logically should take the risk of switching.

The paradox is that you shouldn't get this result. You "know" that there's no reason why one envelope is "better" than the other. So when you examine it in detail, you should confirm this intuitive expectation. But somehow you don't and one envelope somehow appears to be a better choice than the other.
#30
06-20-2012, 09:36 PM
 Indistinguishable Guest Join Date: Apr 2007
This is why I prefer the geometric rephrasing in terms of center of mass... so we don't get stuck on the epistemological bits of the presentation, which are irrelevant to the underlying phenomenon of the paradox.

Last edited by Indistinguishable; 06-20-2012 at 09:36 PM.
#31
06-20-2012, 09:45 PM
 Darth Panda Guest Join Date: Mar 2010
Quote:
 Originally Posted by Chronos If you know that seeing the amount is going to make you want to switch, then why not just say right at the start "I'm going to open the envelope to look, and then switch"? And if you're going to do that anyway, then what's the point of looking in the envelope?
To close the loop. Once you look in one envelope, you know all of the possible outcomes.
#32
06-20-2012, 09:46 PM
 kaylasdad99 Charter Member Join Date: Sep 1999 Location: Anaheim, CA Posts: 19,451
Do you get to keep the money if you chose the one with LESS cash?
#33
06-21-2012, 01:54 AM
 Little Nemo Charter Member Join Date: Dec 1999 Location: Western New York Posts: 52,637
Quote:
 Originally Posted by kaylasdad99 Do you get to keep the money if you chose the one with LESS cash?
Yes, if you switch you get the contents of the second envelope regardless of whether it's twice the amount or half the amount in the first.
#34
06-21-2012, 01:58 AM
 Little Nemo Charter Member Join Date: Dec 1999 Location: Western New York Posts: 52,637
If you didn't get to keep the lesser amount the paradox wouldn't exist. If that was the case then you have an envelope with a known quantity X and you'd have the option of switching to a second envelope which had an effective quantity of either 2X or 0 (the X/2 amount wouldn't matter if you don't get to keep it). Under that situation, the expected value of the second envelope would be equal to X and there'd be no reason to prefer either envelope over the other. The proof would match the intuitive answer.
#35
06-21-2012, 02:19 AM
 Indistinguishable Guest Join Date: Apr 2007
The paradox would still exist; it just wouldn't be relevant to your decision-making. You could still ask yourself "Do I expect, on average, the other envelope to have more money than this one?", instead of "Do I expect, on average, the other envelope to yield more money for me to keep than this one?", whether or not you expected to do anything with that information.

Like I said, though, I think all the decision-theoretic framing just obfuscates what's going on.

Last edited by Indistinguishable; 06-21-2012 at 02:21 AM.
#36
06-21-2012, 03:52 AM
 Saffer Guest Join Date: Feb 2009
Quote:
 Nope When you use 2X, you are assigning the variable X to the value of the money in your envelope if that money is the smaller amount. Let's call this X1 instead. When you use X/2, you are assigning the variable X to the value of the envelope if that money is the larger amount. Let's call this X2 for clarity. But by definition, X1 is X2/2. They are not the same thing. But you proceed to treat them as if they are. The rest of your calculation errors flow from here.
As others have pointed out, if you open the envelope and it contains 100, you know the other contains either 50 or 200. If you open the envelope and see it contains X, the other contains either 2X or X/2. So why bother opening the envelope? Please don't offer me any Schroedinger's cat type explanations either

It seems to me that the problem is not how 'X' is used in the equation, but rather the use of 50-50 probability. The only way for it to be 50-50 is for the guy bankrolling this to have an infinite amount of money.
So what I did was set the expected value of swapping to be equal to X, and then solve for p. I ended up with p(2X) = 1/3.
This seems to be just about as suspect as gaining from swapping indefinitely, when we assume that p = 1/2.
In other words, what I did was start with the fact that the expected values are the same and then work backwards from there.
#37
06-21-2012, 03:59 AM
 Little Nemo Charter Member Join Date: Dec 1999 Location: Western New York Posts: 52,637
Quote:
 Originally Posted by Saffer It seems to me that the problem is not how 'X' is used in the equation, but rather the use of 50-50 probability. The only way for it to be 50-50 is for the guy bankrolling this to have an infinite amount of money.
Why? It's a one-time deal. The only reason an infinite amount of money would be required would be if one envelope contained infinity/3 and the other contained 2infinity/3. That would fit the criteria but so would any lesser sum that is evenly divisible by six.
#38
06-21-2012, 04:16 AM
 Saffer Guest Join Date: Feb 2009
Quote:
 Originally Posted by Little Nemo Why? It's a one-time deal. The only reason an infinite amount of money would be required would be if one envelope contained infinity/3 and the other contained 2infinity/3. That would fit the criteria but so would any lesser sum that is evenly divisible by six.
Lets say the bankroller had 500 moneys to make the game and he tells you so. (in other words the max he can lose it 500)
If you open the envelope and it contains 400. You know for sure that the other contains 200 because is can't possibly contain more than the banker can afford to lose.
If you open the envelope and see that it contains 250, then it is possible that the other can contain 500.

Now consider that you don't know what the bankers limit is, but you know that it is finite. Imagine a function that randomly picks a number from (but not including) 0 to the limit. He puts that in envelope 1. He then flips a coin to determine whether envelope 2 will contain twice or half envelope 1. However, if envelope 1 contains more than half the limit, there is no point in flipping the coin.
The limit itself is now a random variable but because it is finite, the chances of the other envelope containing the lower amount seem to be higher. The only way for it to be 50-50 is if the limit is infinite.

Last edited by Saffer; 06-21-2012 at 04:17 AM.
#39
06-21-2012, 04:34 AM
 Little Nemo Charter Member Join Date: Dec 1999 Location: Western New York Posts: 52,637
Quote:
 Originally Posted by Saffer Lets say the bankroller had 500 moneys to make the game and he tells you so. (in other words the max he can lose it 500) If you open the envelope and it contains 400. You know for sure that the other contains 200 because is can't possibly contain more than the banker can afford to lose. If you open the envelope and see that it contains 250, then it is possible that the other can contain 500. Now consider that you don't know what the bankers limit is, but you know that it is finite. Imagine a function that randomly picks a number from (but not including) 0 to the limit. He puts that in envelope 1. He then flips a coin to determine whether envelope 2 will contain twice or half envelope 1. However, if envelope 1 contains more than half the limit, there is no point in flipping the coin. The limit itself is now a random variable but because it is finite, the chances of the other envelope containing the lower amount seem to be higher. The only way for it to be 50-50 is if the limit is infinite.
But in both of these examples, you're adding new information in order to make the problem easier to solve.
#40
06-21-2012, 04:39 AM
 Saffer Guest Join Date: Feb 2009
Quote:
 Originally Posted by Little Nemo But in both of these examples, you're adding new information in order to make the problem easier to solve.
Thats true but in the real world no-one has an infinite amount of money so we can assume that information.
#41
06-21-2012, 03:34 PM
 nivlac Charter Member Join Date: May 2001 Location: Golden State Posts: 2,356
Quote:
 Originally Posted by Saffer ... It seems to me that the problem is not how 'X' is used in the equation, but rather the use of 50-50 probability. The only way for it to be 50-50 is for the guy bankrolling this to have an infinite amount of money. So what I did was set the expected value of swapping to be equal to X, and then solve for p. I ended up with p(2X) = 1/3. This seems to be just about as suspect as gaining from swapping indefinitely, when we assume that p = 1/2. In other words, what I did was start with the fact that the expected values are the same and then work backwards from there.
Here's your problem (partly pointed out by others already): You cannot have 2X and (X/2) in the left-hand side of your equation. The amount in the chosen envelop (call it Env1) cannot contain both the smaller and the larger amounts in the envelopes. Here's a correct way of finding the p that you want. Let the two amounts be X and 2X. Choose one of the envelopes (call it Env1) at random. Then the expected value in Env1 is clearly 1.5X. The expected value in the other envelope, Env2, is p(X) + (1-p)(2X) or p(2X)+(1-p)(X) (choose either one, does not matter). Set either one equal to 1.5X and you'll arrive at p=1/2, as expected.
#42
06-21-2012, 03:57 PM
 Dr. Love Charter Member Join Date: Mar 2002 Posts: 935
I can't make any sense of this argument that X is being used in two different ways. It's not. In Saffer's posts, X does not mean the minimum or maximum amount, X is the amount in your envelope, regardless of whether that is more or less than the amount in the other envelope.
#43
06-21-2012, 03:59 PM
 Little Nemo Charter Member Join Date: Dec 1999 Location: Western New York Posts: 52,637
Quote:
 Originally Posted by Saffer Thats true but in the real world no-one has an infinite amount of money so we can assume that information.
Yes, but you're not just talking about some theoretical infinite limit. You're creating a situation with a finite limit that's low enough it affects the outcome of the situation. There's a big gap between infinity and 500.
#44
06-21-2012, 04:02 PM
 Little Nemo Charter Member Join Date: Dec 1999 Location: Western New York Posts: 52,637
Quote:
 Originally Posted by nivlac Here's your problem (partly pointed out by others already): You cannot have 2X and (X/2) in the left-hand side of your equation.
Sure you can. If you set X as the amount in the right hand envelope (and there's no reason you can't do this) then the amount in the left hand envelope in 2X or X/2.
#45
06-21-2012, 04:03 PM
 Dr. Love Charter Member Join Date: Mar 2002 Posts: 935
There's an equally big gap between infinity and 100 trillion. Any finite limit is going to resolve the paradox in exactly the same way.
#46
06-21-2012, 04:25 PM
 Little Nemo Charter Member Join Date: Dec 1999 Location: Western New York Posts: 52,637
Quote:
 Originally Posted by Dr. Love There's an equally big gap between infinity and 100 trillion. Any finite limit is going to resolve the paradox in exactly the same way.
But 500 was a factor if you open the first envelope and find \$400 in it. You now know that other envelope can't contain \$800 so it must contain \$200. (And this is an example Saffer gave.) You don't have to raise the finite limit to infinity or 100 trillion - just raise it to 1000 and you've eliminated that out. Or just don't tell the person making the choice what your finite limit is - he wouldn't know there was a limit of 500 unless he was told. Then he would open the envelope containing \$400 and have no reason to think it wasn't possible for the other envelope to contain \$800.

As I said, you may be justified in saying there is an upper limit at some point. But fixing that upper limit, telling the player what the upper limit is, and having the amount in the first envelope be more than half of the upper limit are all new factors that didn't exist in the original paradox. If you add these factors in, you're working on a different question.
#47
06-21-2012, 04:35 PM
 Indistinguishable Guest Join Date: Apr 2007
The paradox is this: Conditioned on any particular value for your envelope, the expected profit from switching is positive, which leads us to say that unconditionally, the expected profit from switching is positive. Conditioned on any particular value for the other envelope, the expected profit from switching is negative, which leads us to say that unconditionally, the expected profit from switching is negative. And conditioned on any particular unordered pair of monetary values for the two envelopes, the expected profit from switching is zero, which leads us to say that unconditionally, the expected profit from switching is zero.

And the explanation is this: Yup. That's right. The sign of the expected profit (equivalently, the sign of the total profit over all cases) changes based on how you calculate it; it is given by a conditionally convergent summation. If a summation has infinitely many positive and negative terms, it is indeed possible for its value to change based on how you group the calculation, as you have just shown with this example.

It's the same phenomenon as that 1 - 1 + 1 - 1 + 1 - 1 + ... = (1 - 1) + (1 - 1) + (1 - 1) + ... = 0 + 0 + 0 + ... = 0, while also 1 - 1 + 1 - 1 + 1 - 1 + ... = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + ... = -1 + 0 + 0 + 0 + ... = -1, and also 1 - 1 + 1 - 1 + 1 - 1 + ... = 1 - (1 - 1 + 1 - 1 + ...), and thus must equal 1/2.

Any finite truncation "resolves" the paradox because a finite summation cannot give two different values based on how you group it. But this is essentially to ignore the actual underlying phenomenon (after all, there is more than one way to truncate; we can specify a lowest and highest value altogether (giving an expected profit of zero by switching), a lowest and highest value for your envelope alone (giving a positive expected profit by switching), or a lowest and highest value for the other envelope alone (giving a negative expected profit by switching)).

Last edited by Indistinguishable; 06-21-2012 at 04:39 PM.
#48
06-21-2012, 04:42 PM
 Indistinguishable Guest Join Date: Apr 2007
Quote:
 Originally Posted by Indistinguishable It's the same phenomenon as that 1 - 1 + 1 - 1 + 1 - 1 + ... = (1 - 1) + (1 - 1) + (1 - 1) + ... = 0 + 0 + 0 + ... = 0, while also 1 - 1 + 1 - 1 + 1 - 1 + ... = 1 + (-1 + 1) + (-1 + 1) + (-1 + 1) + ... = 1 + 0 + 0 + 0 + ... = 1, and also 1 - 1 + 1 - 1 + 1 - 1 + ... = 1 - (1 - 1 + 1 - 1 + ...), and thus must equal 1/2.
Slight negation sign typo corrected in the bold part
#49
06-21-2012, 04:53 PM
 septimus Guest Join Date: Dec 2009
I think most of the answers are misleading and/or overly complex and/or miss the point of OP's question.

To demonstrate this, consider a simplified problem: You know the amounts in the two envelopes are \$1 and \$2 exactly, but you haven't opened either envelope.

Quote:
 Originally Posted by Saffer Let X be the value of the envelope you are holding. The other envelope must have either 2X or X/2 value with 50-50 probability. The expected value of the other envelope is p(2X) + (1-p)(X/2)... What is wrong with this reasoning?
Note that the "paradox" continues, despite that there's only a single case, no infinities, no ambiguity about distributions. You know each envelope has \$1.50 on average, yet still get the paradox. E(other) = 5/4 * E(this).

The fallacy is simple, though I disremember the proper technical terminology. You assume
p(Y=2X) = 1/2 independent of X
but this is not the case.
#50
06-21-2012, 05:05 PM
 Indistinguishable Guest Join Date: Apr 2007
Right, if you know the amount in the two envelopes are \$1 and \$2, you don't have p(Y = 2X) = 1/2 independent of X. This is a (very) finite truncation where you've set a lower and higher bound for the monetary values possible.

But the whole point of the paradox, as I see it, is to consider what would happen if you DID have p(Y = 2X) = p(X = 2Y) = 1/2 independently of X. And in that case, all the "paradoxical reasoning" is flawless; you just happen to have a conditionally convergent summation.

[Here's another simplified problem: You know the value in the other envelope is \$1, but you haven't opened either envelope.

In this case, you don't have p(Y = 2X) = 1/2 independent of X either, but the average profit from switching is negative.

Here's another simplified problem: You know the value in your envelope is \$1, but you haven't opened either envelope.

In this case, you don't have P(Y = 2X) = 1/2 independent of X either, but the average profit from switching is positive.

Picking finite truncations at random doesn't resolve the problem, since you can truncate it in different ways, and it's not about the finite truncations anyway.]

Last edited by Indistinguishable; 06-21-2012 at 05:09 PM.

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