User:Martin Hogbin/Two envelopes

Work in progress - getting there

The problem
The first problem is the problem. The WP statement says:

''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.''

Much information is omitted here and it is up to us to decide on what it should be.

We make the natural simplifying assumptions that the player is risk neutral and wants to maximise the average monetary value in the envelope he eventually holds.

A simple setup
In dealing with complex problems the best approach is to try to reduce the problem to its simplest terms. I therefore propose to start by addressing the simplest version of this problem. It is generally assumed that one envelope is taken randomly from some distribution of possible envelopes and that twice the sum in this envelope is put into the other envelope. We start with a trivial distribution of one envelope containing £2. Thus:

There are two sealed envelopes, one containing £2 and the other £4. A person who knows this chooses one uniformly uniformly at random and is then given the opportunity, without opening it, to swap it with the other one.

Argument for switching
The following argument for switching is presented. The problem is to find the error in this line of argument:

1 Denote by A the amount in the selected envelope.

2 The probability that A is the smaller amount is 1/2 and the probability that it is the larger is 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 1/2 2A + 1/2 A/2 =5/4 A

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.8+ Therefore you should swap and continue to do so.

Solution
The various possible resolutions are numbered R1 etc for future reference

Step 1
What is A?

1a A random variable from the sample space (2,4).

1b The unconditional expectation value of the envelope that you hold E(A), in that case, we know its value, it is 3.

1c A constant, an ordinary fixed number. If that is the assumed meaning the paradox is already solved because an ordinary number cannot have two values. The problem is then one of obvious arithmetical equivocation (R1).

1d A constant representing the sum in the original envelope, which might be £2 or £4. No such animal! This is a mangled attempt to define a random variable and again, if this is the assumed meaning, the paradox is resolved right here (R2).

Step 2
2a If A represents any of the values that might be in the envelope, a random variable, then 2 is fine. The probability that A will turn out to be £2 (P(A=2)) is 1/2, as is the probability that it will turn out to be £4 (P(A=2)).

2b If A is the average amount you expect to find in the envelope that you hold (E(A)) then the paradox is resolved here (R3). There are no probabilities connected with this value - it is 3.

Step 3
By this stage, A must be a random variable, or the paradox is already resolved.

Steps 4 and 5
These steps simply restate the obvious that if A turns out to be £2 B will be £4 and vice versa.

Step 6
What does the figure of 1/2 mean here?

6a The average probability that the unchosen envelope will contain twice the sum in the chosen one (P(B=2A) = P(A=2B)) is 1/2. 6b For every possible sum of money that might be in the first envelope, the probability that the other envelope will contain twice as much is 1/2, that is to say P(B=2A|A)= 1/2 or P(B=2A | A=2 ) = P(B=2A| A=4) = 1/2 where B is a random variable representing the sum in the other envelope. This is clearly false and thus, if this is what is being asserted, the paradox is resolved here(R4); statement 6 is wrong, in particular P(B=2A|A=4)=0, P(B=2A|A=2)=1, P(B=A/2|A=4)=1, P(B=A/2|A=2)=0

Step 7
Is a failed attempt to calculate an expectation value because it neglects to take into account the dependency of the probabilities on the value A turns out to have.

If we are intending to calculate the conditional expectation in the second envelope given any value that we might have in our original envelope E(B|A) then we cannot say E(B|A) = 1/2 2A + 1/2 A/2 because the values of probabilities of 1/2 used in this formula are wrong for some values of A, as shown above (R4).

If the 1/2 mentioned in step 6 is only the average probability that the unchosen envelope will contain twice the amount in the chosen envelope then we cannot use it in an expectation formula as if it applies to all possible values thus the paradox is resolved(R5)

If we want to calculate the unconditional expected value in the other envelope we need to do it thus as E(B)= ( 2 * P(A=2) + 4 * P(A=4) ) / 2  which is 3.

In this simple version of the problem there is no way round this error thus the paradox is resolved here if it has not already been resolved.

Extension to more complicated cases
The above argument can be easily extended to more complex scenarios.

A bounded distribution of envelopes
Suppose the first envelope, instead of containing the fixed value 2, is chosen randomly from some bounded distribution. Say the minimum value is 1 and the maximum is 128.

The above argument still applies exactly. To be more specific

Step 6
What does the figure of 1/2 mean here?

6a The average probability that the unchosen envelope will contain twice the sum in the chosen one (P(B=2A) = P(A=2B)) is 1/2. 6b For every possible sum of money that might be in the first envelope, the probability that the other envelope will contain twice as much is 1/2, that is to say P(B=2A|A)= 1/2 or P(B=2A | A=2 ) = P(B=2A| A=4) = 1/2 where B is a random variable representing the sum in the other envelope. This is clearly false and thus, if this is what is being asserted, the paradox is resolved here(R4); statement 6 is wrong, in particular P(B=2A|A=256)=0, P(B=2A|A=1)=1, P(B=A/2|A=256)=1, P(B=A/2|A=1)=1

Step 7
Is a failed attempt to calculate an expectation value because it neglects to take into account the dependency of the probabilities on the value A turns out to have. If we are intending to calculate the conditional expectation in the second envelope given any value that we might have in our original envelope E(B|A) then we cannot say E(B|A) = 1/2 2A + 1/2 A/2 because the values of probabilities of 1/2 used in this formula are wrong for some values of A, as shown above (R4).

If the 1/2 mentioned in step 6 is only the average probability that the unchosen envelope will contain twice the amount in the chosen envelope then we cannot use it in an unconditional expectation formula as if it applies to all possible values thus the paradox is resolved(R5)

If we want to calculate the unconditional expected value in the other envelope we need to do it thus as E(B)= (1 * P(A=2) + 2 * P(A=2) + ... 256 * P(A=256) ) / 9 .

In this version of the problem there is no way round this error thus the paradox is resolved here if it has not already been resolved.

An unbounded distribution of envelopes
Suppose the first envelope is chosen randomly from some unbounded distribution. If this distribution is 1, 2, 4, 8, 16... with no upper bound then there is is a immediate problem in that this distribution constitutes an improper prior. This is considered to be a resolution in itself by some (R6).

However it is possible to contrive distributions that constitute an proper prior and for which there is still a rationale to swap. All these distributions, however, have an infinite expectation in both envelopes if neither is opened.

Step 6
6b For every possible sum of money that might be in the first envelope, the probability that the other envelope will contain twice as much is sufficient to make swapping advantageous, that is to say ( P(B=2A|A) * 2A + +(B=A/2|A *A/2 ) > A where B is a random variable representing the sum in the other envelope. This statement is now true (except for the unimportant case A=1 where there is even more reason to swap)

Step 7
We now can calculate the conditional expectation in the second envelope given any value that we might have in our original envelope E(B|A) but this in infinite, as is the expectation in our original envelope.

Step 8
If you do not look in your envelope then the expectation in both envelopes is infinite, thus there is no gain in swapping (R7).

It is slightly more complicated, however, if you do look in your envelope. Although your expectation is infinite you will almost certainly be disappointed to find a finite sum in your envelope. Whatever this sum is you can calculate that your expectation in the other envelope is greater than the sum in your envelope E(B|A) > B. (This argument can be used to justify swapping before you look).

There is still a problem with this step, however. It is claimed that on average you will gain. On average, the sum in your envelope is still infinite so there is nothing to be gained, on average, by swapping (R8).

This situation may be regarded a paradoxical by some but such things are common in calculations involving infinities and, in my opinion, these 'paradoxes' are no more remarkable that the fact that 1/0 = 2/0.

Step 9
I might be noted that this step also fails if you look in your original envelope. If you do swap, the expectation in your unopened envelope will be more than the value you have just looked at in your original envelope, thus there is no reason to swap back. (R9)

Summary of resolutions
If you open your second envelope it will become obvious what to do.

Result for the coin tossing version
Any resolution of the simple paradox needs to also show why you  should swap  once in Falk's coin toss version (the same as Nalebuff's Ali-Baba version). Falk does not do this in her paper she only shows why relentless swapping is not justified. Below is my simplified coin tossing version.

Simple Setup
There is a sealed envelope containing £2 which is given to the player. A fair coin is tossed and, upon the fall of the coin, either £1 or £4 is placed in another envelope, which is then sealed. The same argument is presented for switching.

Solution
In 1, what is A? It could be random variable from the trivial sample space (2) but why bother?. It can just be a simple number, 2, in fact.

2-6 Now present no problem

7 Is a perfectly correct calculation, using a number rather than a random variable.

False resolutions
There are some proposed resolutions that are a paradoxical as the original problem. In some ways, these claimed resolutions, mirror the original paradox.

if we denote the lower of the two amounts by C we can write the expected value calculation as


 * $${1 \over 2} C + {1 \over 2} 2C = {3 \over 2}C$$

Here C is a constant throughout the calculation and we learn that 1.5C is the average expected value in either of the envelopes. So according to this new calculation there is no contradiction between the decisions to keep or to swap, and hence no need to swap indefinitely.

Problem 1 - What is C?
1a A random variable? it is described as a constant.

1b The unconditional expectation value of the envelope that you hold E(A)? Not that.

1c A constant, an ordinary fixed number. In that case we cannot use it to describe the sum in envelope taken from some distribution of possible values, that would require a random variable. If C is a constant then there can only be two possible known envelope values (as in the simple setup above). In that case we do not need a symbol we might just as well use a number, say '2'.

1d A constant representing the sum in the original envelope. It cannot be this as it is always the smaller sum.

Problem 2
The most common way to explain the paradox is to observe that A isn't a constant in the expected value calculation, step 7 above. In the first term A is the smaller amount while in the second term A is the larger amount. To mix different instances of a variable or parameter in the same formula like this shouldn't be legitimate, so step 7 is thus the proposed cause of the paradox.

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