My Math Forum Solution to the Two Envelope Paradox

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April 20th, 2017, 04:06 PM   #21
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Quote:
 Originally Posted by Yooklid Since this was posted in the number theory section rather than the probability section, my first thought was that perhaps the amount of money in the envelopes must be an integer number of dollars. If that were the case, then there would certainly be some strategies available to maximize your gain. For example if the first envelope contains an odd amount, then obviously you should exchange it for the other envelope.
A more formal statement of the paradox restricts x and y to the set {1, 2, 4, 8, 16, ... } where y = 2x.

In the formal setting, if you see 1 in your envelope you always swap because you know you'll get 2. Otherwise, you are tricked into thinking you have a better shot by swapping by incorporating an impossible outcome. Here is a clearer exposition:

Traditionally, the paradox arises from comparing what appears to be someone’s odds when they are allowed to see what’s in their envelope vs. their odds when they aren’t.

Without loss of generality, let x = 10 and 2x = 20. Assume the person’s envelope to be selected uniformly at random from those two options.

Without seeing what is in the envelope, the person makes a choice that is true to reality. They realize that it doesn’t matter whether or not they trade envelopes because they can’t distinguish. Half the time they get 10 and half the time they get 20 regardless of whether or not they swap. They are ok with it, and they average 15.

If someone sees what is in the envelope, they incorporate an impossible outcome. If the person sees 10 in the envelope, they think they have a 50% chance of swapping for 20 and a 50% chance of swapping for 5, but they don’t. They have a 100% chance of swapping for 20. Alternatively, if they see 20 in the envelope, they think they have a 50% chance of swapping for 40 and a 50% chance of swapping for 10, but they don’t. They have a 100% chance of swapping for 10. The equation that is given to construct the paradoxical sense of needing to swap to improve one's odds…:

$$0.5( \,0.5x) \, + 0.5( \,2x) \, = 1.25x$$

… is just a red herring. It does not properly establish the one and only one true average outcome of $1.5x$.

That said, I see no true paradox here. A sample space is not dependent upon someone’s frame of reference as the paradox suggests.

Last edited by AplanisTophet; April 20th, 2017 at 04:08 PM.

April 20th, 2017, 04:57 PM   #22
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Quote:
 Originally Posted by AplanisTophet the one and only one true average outcome of $1.5x$.
You have two different formulations of the problem with a non-trivial difference in the definition of $x$. Your $\frac32$ is only the "one and only one true average outcome" for one of them.

The calculation you took from the other thread is fine, but there $x$ represents not the minimum "win" but the amount you found in your envelope.

Thus, whatever you do, the average win is $\frac32$ times the minimum "win". But, if you change every time, the average win is $\frac54$ times the amount in your envelope.

The real challenge is to show that these to formulations mean the same thing. I don't suppose it's awfully difficult.

April 20th, 2017, 05:56 PM   #23
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Quote:
 Originally Posted by v8archie You have two different formulations of the problem with a non-trivial difference in the definition of $x$.
In one version of the problem, the person doesn't get to peek in the envelope before deciding whether or not to swap. In the second, the person does get to take a peek.

Does taking a peek really change the probability of what that person will get? I assume you are saying "yes" to that question, so what is this non-trivial difference in the definition of x that you speak of?

Either way, one envelope has x dollars in it and another 2x dollars in it. There are two possibilities, no more and no less, and they average to 1.5x.

The folly is to assume your envelope contains x and therefore there are three possibilities depending on whether you swap or not: 0.5x, x, and 2x, which you assume by thinking you have a 50% chance of swapping for 0.5x, a 100% chance of keeping x if you don't swap, and a 50% chance of swapping for 2x, when in truth you have a 100% chance of swapping for either x or 2x depending on whether your envelope contains 2x or x, respectively.

Quote:
 Originally Posted by v8archie The calculation you took from the other thread is fine, but there $x$ represents not the minimum "win" but the amount you found in your envelope.
See the paper I cited in the OP. (sorry, looks like the link wasn't working)

It states that "the expected value of the other envelope is higher than the given amount in yours, more precisely:"

$E( \,Y|X = x) \, = \begin{cases} \frac{11}{10}x & \text{for } x > 1 \\ 2 & \text{for } x = 1 \end{cases}$

I just gave the usual "folly" of...

$0.5( \,0.5x) \, + 0.5( \,2x) \, = 1.25x$

... in the OP as taken from the stackexchange question I linked to, which just goes to show that there are all kinds of crazy opinions out there for this.

Quote:
 Originally Posted by v8archie Thus, whatever you do, the average win is $\frac32$ times the minimum "win". But, if you change every time, the average win is $\frac54$ times the amount in your envelope.
How can the average win be $\frac{5}{4}$ when you always swap envelopes vs. $\frac{3}{2}$ "whatever you do." I suggest you improve your exposition. If you mean that the minimum win is $x$, then the average win is $1.5x$ "no matter what you do." There is no way to get an average win of $\frac{5}{4}$ "no matter what you do" I do declare.

Quote:
 Originally Posted by v8archie The real challenge is to show that these to formulations mean the same thing. I don't suppose it's awfully difficult.
If it's not awfully difficult, then by all means, show that the two formulations (whatever you mean by that) mean the same thing. This "paradox" (if it is one) has been around for a while, so good luck.

April 20th, 2017, 06:16 PM   #24
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Quote:
 Originally Posted by AplanisTophet Does taking a peek really change the probability of what that person will get?
No
Quote:
 Originally Posted by AplanisTophet Either way, one envelope has x dollars in it and another 2x dollars in it.
No. That is one formulation. The other is that your envelope has $x$ dollars and the other has either $\frac12x$ or it has $2x$. The value of $x$ is not always the same in the two formulations.

Quote:
 Originally Posted by AplanisTophet The folly is to assume your envelope contains x
No. The folly is in misinterpreting the results of the calculation.

Quote:
 Originally Posted by AplanisTophet It states that "the expected value of the other envelope is higher than the given amount in yours
Precisely.

Given that others have written papers on it, I see no reason to work it all out here.

April 20th, 2017, 06:57 PM   #25
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Quote:
 Originally Posted by v8archie No No. That is one formulation. The other is that your envelope has $x$ dollars and the other has either $\frac12x$ or it has $2x$. The value of $x$ is not always the same in the two formulations.
So what if there are three envelopes containing 0.5x, x, and 2x? I hand you the one containing x and say that you can keep your x or try to swap. You'll then be justified in swapping.

That's not the same problem as the two envelopes in reality though, is it? In the two envelopes version, either 0.5x or 2x does not exist!! Incorporating a non-existent solution is no way to set up your sample space. IMHO. Final answer.

April 21st, 2017, 05:44 PM   #26
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Quote:
 Originally Posted by Maschke The grass is always greener on the other side. AKA "Don't it always seem to go that we don't know what we've got till it's gone."
Don't dispair. You'll like this write up I think Maschke. As you should know by now, I always just take a minute or two to warm up.

...

Su, Francis, et. al. have a short description of the paradox here: https://www.math.hmc.edu/funfacts/ff...0001.6-8.shtml

I used that link because it concisely sets forth the paradox both in the basic setting but also given the version where the two envelopes contain $( \,\$2^k, \$2^{k+1}) \,$ with probability $( \,\frac{\frac{2}{3}^k}{3}) \,$ for each integer $k \geq 0$.

Where the paradox is formulated by considering one person’s odds when choosing to swap an envelope, my question is whether the paradox might be resolved by considering the paradox from both swapper’s perspectives instead of just one (i.e. for one person to swap, there must be another person for the original to swap with).

From a single person’s perspective, the paradoxical odds are traditionally given by the equation:
$$0.5( \,0.5x) \, + 0.5( \,2x) \, = 1.25x$$
To incorporate a two-person perspective, the equation would be one person’s odds for gain less “their opponent's” odds for gain:
$$[ \,0.5( \,0.5x) \, + 0.5( \,2x) \,] \, - [ \,0.5( \,0.5x) \, + 0.5( \,2x) \,] \, = 0$$
The result is that neither person improves their odds by swapping. Paradox resolved.

Comments, suggestions, agree, disagree… I’m just fishing here. Thank you?

 April 21st, 2017, 07:39 PM #27 Senior Member   Joined: Jun 2014 From: USA Posts: 282 Thanks: 15 Regarding the above post: In being concerned only with what you make, you must factor into account what you lose as well because what your opponent stands to make comes at your expense. Where you stand to make just as much as you stand to lose to your opponent, there is no incentive to switch. That is what the equations say to me.
 April 22nd, 2017, 05:25 AM #28 Senior Member   Joined: Jun 2014 From: USA Posts: 282 Thanks: 15 We can also break it down into the four possible scenarios when swapping where, when ambiguously starting with $x$, we have 25% of the time you’ll lose $0.5x$ to your opponent, 25% of the time you’ll gain $0.5x$ from your opponent, 25% of the time you’ll lose $x$ to your opponent, and 25% of the time you’ll gain $x$ from your opponent. The same is true for your opponent. $$x - 0.25( \,0.5x) \, + 0.25( \,0.5x) \, - 0.25( \,x) \, + 0.25( \,x) \, = x - 0.25( \,0.5x) \, + 0.25( \,0.5x) \, - 0.25( \,x) \, + 0.25( \,x) \,$$ $$x = x$$ In addition to what is noticed in posts 26 and 27, the above shows that there is no incentive to swap. The same general logic works for all consistent versions of the paradox that prey on the ambiguity of the variable $x$ (see the link in post 26 for more insight into that statement). I assert that the paradox is resolved.
April 30th, 2017, 12:56 PM   #29
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Quote:
 Originally Posted by AplanisTophet Statement of the Paradox: Two envelopes exist, one containing $x$ dollars in cash and the other containing $2x$ dollars in cash. You are given one envelope and are told that you may either keep it or swap it for the other envelope. You want to maximize your cash. You have no idea how much cash is in the envelope you are given or how much is in the envelope that you may swap for. The paradox is the supposed result of the probability when swapping: 0.5(0.5x) + 0.5(2x) = 1.25x The above suggests that you should always swap envelopes, but clearly this is absurd.
The underlined parts are incorrect.

The numbers you used, 1/2 in each case, are based on the Principle of Indifference. It says that when you have N random outcomes that occur under functionally identical circumstances, they each should have a probability of 1/N. So when you flip a coin, each side lands face-up under circumstances that are functionally equivalent to the other, and has probability 1/2. When you roll a six-sided die, each side has a 1/6 chance. But when you ask a 3-year old girl what if her favorite color is, from among {red, orange, yellow, green, blue, purple, pink}, you'd be foolish to say each has a 1/7 chance. They don't occur under equivalent circumstances. Almost cerytainly, neither do the possibilities that the other envelope was filled with half, or twice, what you have.

To see this better, consider two variations of your problem:
1. Two pairs of envelopes are prepared, one pair with (\$5,\$10), and one pair with (\$10,\$20). I pick one pair at random, and give it to you.
2. Ten pairs of envelopes are prepared, nine pairs with (\$5,\$10), and one pair with (\$10,\$20). I pick one pair at random, and give it to you.
Both variations satisfy the conditions of your problem, with either x=\$5 or x=\$10.

I haven't looked at your later approach closely, but it a similar one applied to my problems should say that the expected value of every sealed envelope in the first problem is \$11.25, and in the second problem is \$8.25. So why switch?

But now suppose you open your envelope, and see \$10. In the first problem, there is a 50% probability that the other envelope contains \$5, and another 50% probability that it contains \$20. So the expected value for switching is indeed 1.25*x=\$12.50, just like the paradoxical solution you showed said. But note that these probabilities are 50% because we know how the envelopes were prepared, not because there was a 50% chance that you would pick the higher, or lower, envelope of your pair.

This becomes clearer in the second problem: now there is a 90% chance that the other envelope contains \$5, and only a 10% chance that it contains \$20. So the expected value of the envelope you would switch to is only \\$6.50.

The point is, when you try to treat your envelope as having a certain amount of money, even if that amount is called X, then the probability you have the lower amount is not the same as the probability of having the lower amount when you ignore the amount. The latter is always 50%; the former depends on the probabilities that the pair contained (X/2,X) or (X,2X).

There are envelope-filling strategies where it is almost certain that switching will provide the expectation of a gain[1] for most values you could have; but the rare losses will be such large amounts, that the expected gain is zero. There are even strategies where the first switch is guaranteed to be an expected gain, but they require an infinite amount of money.

I don't know why many mathematicians don't understand this error, and so feel the need to write papers on it.

+++++

[1] Fill N pairs with (X,2X), (2X, 4X), ..., and (2^N*X,2^(N+1)*X) There is a (N-1)/N chance that the original "paradoxical" solution is correct.

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