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#21 |
FFR Player
Join Date: Jan 2006
Age: 36
Posts: 116
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![]() 1. I know there is X {the amount} in my envelope.
2. The probability that X is the smaller of the two amounts is 1/2, and also it is a 1/2 possibility that it's the larger amount also. 3. The other envelope may contain either 2X or X divided by 2 {1/2X} 4. If X is the smaller amount the other envelope contains 2X, BUT, If X is the larger amount the other envelope contains 1/2X 5. This means, the other envelope contains 2X with the probability 1/2 and 1/2X with probability 1/2 6. So the value of the money in the other envelope is: {1\2} * {2X} + {1\2} * {1\2X} = {5\4}X 7. This is greater than X, so, on average, I gain money by switching 8. After the swap, I can denote that content of the other envelope {Y} and reason in exactly the same way as above 9. I conclude that the most rational thing to do is to swap back again 10. To be rational I will then end up swapping envelopes forever. 11. As it seems more rational to open just any envelope than to swap indefinitely we have a contradiction
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#22 | |
FFR Player
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nevermind im a dumbass. I just created a whole different question and made up an answer for it |
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#23 |
FFR Simfile Author
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![]() Why isn't anyone listening to me?
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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 |
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#24 |
Private College
![]() ![]() ![]() Join Date: Feb 2006
Location: Lol badger
Posts: 536
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![]() Because your solution doesn't cause problems. We're trying to figure out why the other approach fails.
Plus you just stole it off Wikipedia.
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<img src="Bent Lines" /> |
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#25 |
FFR Simfile Author
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![]() I didn't even look at wikipedia. And if my approach is true, how can the other one also be true? Problem solved?
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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 |
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#26 |
FFR Player
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![]() You also stand to lose more than you already have gained, which makes the arguement that you stand to gain more than you lose false.
I would keep the $100 because $200 is an obscure amount compared to an even $50, so I would figure that the gameshow would more likely put $50 in one and $100 in the other. If it turns out I was wrong, then oh well. I still made $100. If you always did things because you stood to gain more than you lose, then you'd always play at the dollar slots and spend more money than you'd make because hey, that dollar can turn into $5000 if you're lucky. It can also net you absolutely nothing. Rule of gambling is that if you are ahead, stick with what you got because chancing a definate gain on a possible gain doesn't pan out more often than not.
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SIG PICTURES: POINTLESSLY TAKING UP BANDWIDTH SINCE THE INCEPTION OF THE INTERNET |
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#27 |
Admiral in the Red Army
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![]() The thing is, Laharl, that you have the same chance to gain or lose.
You have a 50% chance of getting 2x as much. You have a 50% chance of losing half of what you already had. Since it's equal chances, it makes most sense to go with the better option, because it's not like some risky situation.
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#28 |
FFR Player
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![]() Prisoner's Dilemma.
Nash Equilibrium. Game Theory. Econ. Q |
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#29 |
FFR Player
Join Date: Oct 2006
Posts: 14
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![]() There is a solution to this...somehow...well when dealing with paradoxes you have to make up whatever you can think of that could give you a solution.
The game show picks two amounts of $, and one is twice the other (this is where we use are imagination) what if the money they picked was solid gold? then all you would have to do is look at or feel which envelope has more money thats how i slove this paradox. |
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#30 | |
FFR Simfile Author
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I still seriously don't get what's wrong with my solution, though.
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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 |
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#31 |
FFR Player
Join Date: Oct 2006
Posts: 6
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You, sir, win the Intelligent Poster of the Year of 2006 Award. To claim your reward, please pose with this piece of Uranium at groin level.
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#32 | |
FFR Player
Join Date: May 2006
Posts: 224
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![]() ..I really don't understand how this is confusing, or much less, a paradox.
Quote:
Say one envelope had 100, and the other had 50. Two situations: 1. You pick the one with 100 dollars. You immediately think that there are two possibilities. One envelope had 100, the other had 200, OR one had 100, and the other 50. 2. You pick the one with 50 dollars. You immediately think one had 100, and the other 50, OR one had 50, the other had 25. The fact that you don't know the value of the envelopes allows different conclusions without it being a paradox. |
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#33 |
Banned
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