Envelope Paradox: A simple solution?

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  • RainGame53
    FFR Player
    • May 2005
    • 61

    #1

    Envelope Paradox: A simple solution?

    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?
    Last edited by RainGame53; 10-26-2006, 02:18 PM.
  • masterhickle
    FFR Player
    • Apr 2005
    • 557

    #2
    Re: Envelope Paradox: A simple solution?

    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.

    ?_?

    Comment

    • skyrunner06
      FFR Player
      • Jan 2006
      • 154

      #3
      Re: Envelope Paradox: A simple solution?

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

      Comment

      • MalReynolds
        CHOCK FULL O' NUTRIENTS
        • Sep 2003
        • 6571

        #4
        Re: Envelope Paradox: A simple solution?

        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.
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        Now in Paperback!

        Comment

        • Squeek
          let it snow~
          • Jan 2004
          • 14444

          #5
          Re: Envelope Paradox: A simple solution?

          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.
          Last edited by Squeek; 10-27-2006, 06:09 PM.

          Comment

          • Patashu
            FFR Simfile Author
            FFR Simfile Author
            • Apr 2006
            • 8609

            #6
            Re: Envelope Paradox: A simple solution?

            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.
            Last edited by Patashu; 10-27-2006, 06:16 PM.
            Patashu makes Chiptunes in Famitracker:
            http://soundcloud.com/patashu/8bit-progressive-metal-fading-world
            http://img.photobucket.com/albums/v216/Mechadragon/smallpackbanner.png
            Best non-AAAs: ERx8 v2 (14-1-0-4), Hajnal (3-0-0-0), RunnyMorning (8-0-0-4), Xeno-Flow (1-0-0-3), Blue Rose (35-2-0-20), Ketsarku (14-0-0-0), Silence (1-0-0-0), Lolo (14-1-0-1)
            http://i231.photobucket.com/albums/ee301/xiaoven/solorulzsig.png

            Comment

            • jewpinthethird
              (The Fat's Sabobah)
              FFR Music Producer
              • Nov 2002
              • 11711

              #7
              Re: Envelope Paradox: A simple solution?

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

              Comment

              • Reach
                FFR Simfile Author
                FFR Simfile Author
                • Jun 2003
                • 7471

                #8
                Re: Envelope Paradox: A simple solution?

                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.
                Last edited by Reach; 10-27-2006, 09:03 PM.

                Comment

                • talisman
                  Resident Penguin
                  FFR Simfile Author
                  • May 2003
                  • 4598

                  #9
                  Re: Envelope Paradox: A simple solution?

                  I'm with jewpin.

                  Comment

                  • Squeek
                    let it snow~
                    • Jan 2004
                    • 14444

                    #10
                    Re: Envelope Paradox: A simple solution?

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

                    Comment

                    • Patashu
                      FFR Simfile Author
                      FFR Simfile Author
                      • Apr 2006
                      • 8609

                      #11
                      Re: Envelope Paradox: A simple solution?

                      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.
                      Patashu makes Chiptunes in Famitracker:
                      http://soundcloud.com/patashu/8bit-progressive-metal-fading-world
                      http://img.photobucket.com/albums/v216/Mechadragon/smallpackbanner.png
                      Best non-AAAs: ERx8 v2 (14-1-0-4), Hajnal (3-0-0-0), RunnyMorning (8-0-0-4), Xeno-Flow (1-0-0-3), Blue Rose (35-2-0-20), Ketsarku (14-0-0-0), Silence (1-0-0-0), Lolo (14-1-0-1)
                      http://i231.photobucket.com/albums/ee301/xiaoven/solorulzsig.png

                      Comment

                      • Squeek
                        let it snow~
                        • Jan 2004
                        • 14444

                        #12
                        Re: Envelope Paradox: A simple solution?

                        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.

                        Comment

                        • Patashu
                          FFR Simfile Author
                          FFR Simfile Author
                          • Apr 2006
                          • 8609

                          #13
                          Re: Envelope Paradox: A simple solution?

                          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?
                          Patashu makes Chiptunes in Famitracker:
                          http://soundcloud.com/patashu/8bit-progressive-metal-fading-world
                          http://img.photobucket.com/albums/v216/Mechadragon/smallpackbanner.png
                          Best non-AAAs: ERx8 v2 (14-1-0-4), Hajnal (3-0-0-0), RunnyMorning (8-0-0-4), Xeno-Flow (1-0-0-3), Blue Rose (35-2-0-20), Ketsarku (14-0-0-0), Silence (1-0-0-0), Lolo (14-1-0-1)
                          http://i231.photobucket.com/albums/ee301/xiaoven/solorulzsig.png

                          Comment

                          • lord_carbo
                            FFR Player
                            • Dec 2004
                            • 6222

                            #14
                            Re: Envelope Paradox: A simple solution?

                            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.

                            gambler's fallacy, Monte Carlo fallacy, gambling fallacies, statistical fallacies, probability, statistics, critical thinking


                            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.
                            Last edited by lord_carbo; 10-28-2006, 05:54 AM.
                            last.fm

                            Comment

                            • Reach
                              FFR Simfile Author
                              FFR Simfile Author
                              • Jun 2003
                              • 7471

                              #15
                              Re: Envelope Paradox: A simple solution?

                              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.

                              Comment

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