Envelope Paradox: A simple solution?

[RainGame53]

To this date I have still not been able to find an easily understood solution to this famous paradox. Now I'm counting on your guys for one. Here is the paradox (copied from a google search) -


Suppose you're on a gameshow where you can choose either of two sealed envelopes, A or B, both containing money. The host doesn't say how much money is in each, but he does let you know that one envelope contains twice as much as the other.

You pick envelope A, open it and see that it contains $100. The host then makes the following offer: you can either keep the $100, or you can trade it for whatever is in envelope B.

You might reason as follows: since one envelope has twice what the other one has, envelope B either has 200 dollars or 50 dollars, with equal probability. If you switch, then, you stand to either win $100 or to lose $50. Since you stand to win more than you stand to lose, you should switch.

But just before you tell the host you would like to switch, another thought might occur to you. If you had picked envelope B, you would have come to exactly the same conclusion. So if the above argument is valid, you should switch no matter which envelope you choose. But that can't be right. What's wrong with your reasoning?

Can any of you can to explain what is wrong with this in simple english and basic math?

[masterhickle]

From how I see it, regardless of which envelope you choose, and not knowing which contains what amount of money, you always stand to make more than you'll lose.

Then again, my thinking may be flawed from being sick and being in bed for the past 15ish hours.

?_?

[skyrunner06]

i think that no matter wich enveloe u choose you will loose money

[MalReynolds]

Originally Posted by skyrunner06

i think that no matter wich enveloe u choose you will loose money

Yes...

Also, you're a disgrace to 19 year olds.

[Squeek]

Wikipedia:

"The two envelopes problem is a puzzle or paradox within the subjectivistic interpretation of probability theory; more specifically within Bayesian decision theory. This is still an open problem among the subjectivists as no consensus has been reached yet."

You're asking for an answer among people who've never seen this problem, when the people who spend years on it haven't solved it yet.

Edit: Summary.

If there's no game show involved and you don't see the values of the money, allowing you to switch as many times as you want, you'll never be able to decide based on logic alone. That is what makes it complicated. You have a higher probability that the other envelope has the higher amount no matter what envelope you choose. Even after you switch, the other envelope (the one you JUST had) has a higher probability of having the higher amount.

There.

[Patashu]

Well, let's say the gameshow picks the monetary values for both before you open either. This is true. We'll call them x and 2x.

If you take x and decide to switch, you'll get x more. If you take 2x and decide to switch, you'll take x less.

Clearly, it doesn't matter if you switch or not.

EDIT: But if you're one of those paranoid or better-safe-then-sorry types, ALWAYS switch. No matter how you attack the problem there's no way to come to the conclusion that staying with your envelope will produce a gain compared to switching.

[jewpinthethird]

Which ever you choose, you will still end up with more money than you originally started with, so what does it matter?

[Reach]

Seems like switching or not switching, looking at it from a probability perspective, doesn't matter. You could gain money or lose money. It's irrelevant.

But then again i'm incredibly tired.



Both arguments are valid because you personally don't know if the other envelope has less or more. However, it's still either less, or more, and this applys to both envelopes naturally because of the unknown variable. Simple. it can't be both. XD


Squeek, explain why switching is more probable. I don't get it. Unlike 3 or 4 envelopes, the chances of you picking the highest or lowest value is the same since there are only 2 envelopes...why would it matter if you switched?

You are no more unlikely to pick the highest envelope than you would switching envelopes.

[talisman]

I'm with jewpin.

[Squeek]

Reach, it's just that switching is better because of reasons already stated. You'd win more than you'd lose.

[Patashu]

Originally Posted by Squeek

Reach, it's just that switching is better because of reasons already stated. You'd win more than you'd lose.

See my post.

[Squeek]

Originally Posted by Wiki

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's the larger 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

{1 \over 2} 2A + {1 \over 2} {A \over 2} = {5 \over 4}A

8. This is greater than A, so I gain on average by swapping
9. After the switch I can denote that content 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

Read.

[Patashu]

Nono, read.

The events occur in this order:

1. The gameshow picks the two monetary values, one double the other (thus, x and 2x)

2. You are handed either x or 2x. You don't know whether you were given x or 2x.

If you have x and swap you get 2x, or a profit of x (compared to not swapping).
If you have 2x and swap you get x, or a deficit of x (compared to not swapping).

If I swapped every time I would get, overall, 1.5x per play.
If I didn't swap every time I would get, overall, 1.5x per play.
If I swap or don't swap at random every time I would get, overall, 1.5x per play.

The numbers aren't just magically teleporting around, they're set in stone before you touch any envelopes. Unless you can predict the future no strategy will ensure more money then any other.

Is this so really so hard to understand?

[lord_carbo]

Originally Posted by Squeek

Wikipedia:

"The two envelopes problem is a puzzle or paradox within the subjectivistic interpretation of probability theory; more specifically within Bayesian decision theory. This is still an open problem among the subjectivists as no consensus has been reached yet."

You're asking for an answer among people who've never seen this problem, when the people who spend years on it haven't solved it yet.

Edit: Summary.

If there's no game show involved and you don't see the values of the money, allowing you to switch as many times as you want, you'll never be able to decide based on logic alone. That is what makes it complicated. You have a higher probability that the other envelope has the higher amount no matter what envelope you choose. Even after you switch, the other envelope (the one you JUST had) has a higher probability of having the higher amount.

There.

http://en.wikipedia.org/wiki/Gamblers_fallacy
http://www.fallacyfiles.org/gamblers.html

It's always going to be a 50:50. This isn't like ap's door problem where you can apply a bit of math and logic to it.

Okay, look at it like this:

You pick an envelope. That envelope will be defined as x. Now, the second envelope is either F(x) = 2x or F(x) = .5x. There's no greater chance of losing money by switching, it's a complete 50:50 because you don't know how to define the second one. It's all psychological.

[Reach]

I thought it was kind of obvious it was going to be 50/50. What squeek is saying makes no sense. You don't gain anything by swaping because the chances of you picking the envelope with the higher amount of money in it the first time is 50:50, therefore making it impossible to some how increase your chances by swamping.

[Kilgamayan]

Squeek is saying that switching is the logical thing to do because you stand to gain more than you would lose. The probability of either event happening is .5, but the amount of money you would gain is greater than the amount of money you would lose, so theoretically you want to switch.

[Squeek]

Why is it that Kilga is always the one to understand.

I am well aware that this is a 50/50 chance, but you'd always switch because you'd stand to gain more than you'd lose.

Besides, $100 from a game show sucks these days. Especially when you have game shows giving away millions.

[kentbball]

You should always switch.

Lets say there is a gameshow (bad gameshow but oh well) and the rules are as follows:

There are three amounts: 50$, 100$, and 200$.

Each round they give you the 100$ to start out with and they have another envolope that either contains 200$ or 50$.

So you start out with the 100$ envolope and your reasoning is that you will always gain more than you lose so you switch and get 200$ (WOOHOO). next round you switch and get 50$ (BOO). you than have 250$. So if you hadnt of switched at all you wouldve only had 200$. 200$ vs 250$ i know what i cwould like more.

probability says that it will always go 200-50-200-50-200-50 etc. It obviously wont but it will even out to that in the end. That means every two turns you will get an average of 250$ instead of 200$ if you decide not to switch.

No freakin clue how it works out to that but it just does.

[FFR4EVA_00]

Easy, the problem is that you created the illusion that your thinking is wrong.
Your brain is working correctly.
Stop wasting time or you won't get anything at all.

[kentbball]

huh?
who are you talking to? :S