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Dear Long Live Yorke


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#41 hblask

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Posted 15 April 2008 - 11:58 AM

View PostLongLiveYorke, on Tuesday, April 15th, 2008, 2:40 PM, said:

X is whatever value we see in the envelope we open. Half the time it's the larger value and by switching we get .5X, and half the time it's the smaller envelope and by switching we get 2X. So, X is a fixed number and doesn't depend on which envelope we chose first. If I keep it, my expected value is certainly X, because that's how much I'll get every time. If I change, I get either double X or half X, which has a higher expected value. So, why don't I switch? Clearly switching is the wrong answer
I think the problem is that, since you have an envelope in your hand, you can't say that it's either 2X or .5X, because those two X's are not the same -- they each refer to a different situation, so using the same X for both and then mixing them is where the error comes in. The situation you are describing is for this situation:I give you an envelope with 1 dollar in it. There are two other envelopes, one with 0.5 dollars and one with 2 dollars. You can pick either of those two or keep your own envelope. What should you do? Then, your formula that leads to always switching gets the correct answer, because on average you end up with 1.25.
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#42 LongLiveYorke

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Posted 15 April 2008 - 12:01 PM

View Posthblask, on Tuesday, April 15th, 2008, 3:58 PM, said:

I think the problem is that, since you have an envelope in your hand, you can't say that it's either 2X or .5X, because those two X's are not the same -- they each refer to a different situation, so using the same X for both and then mixing them is where the error comes in.
But X is always what we find in the envelope, that's how it's defined. No matter whether we picked the smaller one or the larger one, it's always X, never 2X or .5X.

#43 LongLiveYorke

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Posted 15 April 2008 - 12:07 PM

Anyway, here's what I think the resolution of the problem is so we can all move on:When we look into the envelope, we see exactly how much money is in it. We have figured out what X is. I claim that it is therefore wrong to say that we have picked the smaller one and the larger one with a 50/50 chance. By knowing how much money is in the envelope, we have been given some information as to which one we more likely picked. The problem is that there's really no way to fill the envelopes with all pairs of a number and it's double with equal probability. This sort of distribution could never be normalized and would never add up to 1. We need to fill the envelopes with a real probability distribution.In other words, there must be a nearly 0% chance that we fill the envelope with $1,000,000,000,000 and $2,000,000,000,000, so if I open the envelope and find $1,000,000,000,000, it's more likely that the two envelopes were filled with 500,000,000,000 and 1,000,000,000,000 than $1,000,000,000,000 and $2,000,000,000,000.I think it's neat and quite subtle. So, problem is really ill posed. We have to apply some workable probability to how the filler of the envelopes decides to fill them, and when we do so, the problem becomes solvable and the paradox is eliminated.Imagine that he only fills it with 1, 2, 4, 8, 16, 32, 64, and 128. Then if you get 128, you know you can't switch because the two pairs are 64 and 128, so you can only lose.

#44 keith crime

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Posted 15 April 2008 - 12:11 PM

View PostLongLiveYorke, on Tuesday, April 15th, 2008, 12:07 PM, said:

Anyway, here's what I think the resolution of the problem is so we can all move on:When we look into the envelope, we see exactly how much money is in it. We have figured out what X is. I claim that it is therefore wrong to say that we have picked the smaller one and the larger one with a 50/50 chance. By knowing how much money is in the envelope, we have been given some information as to which one we more likely picked. The problem is that there's really no way to fill the envelopes with all pairs of a number and it's double with equal probability. This sort of distribution could never be normalized and would never add up to 1. We need to fill the envelopes with a real probability distribution.In other words, there must be a nearly 0% chance that we fill the envelope with $1,000,000,000,000 and $2,000,000,000,000, so if I open the envelope and find $1,000,000,000,000, it's more likely that the two envelopes were filled with 500,000,000,000 and 1,000,000,000,000 than $1,000,000,000,000 and $2,000,000,000,000.I think it's neat and quite subtle. So, problem is really ill posed. We have to apply some workable probability to how the filler of the envelopes decides to fill them, and when we do so, the problem becomes solvable and the paradox is eliminated.Imagine that he only fills it with 1, 2, 4, 8, 16, 32, 64, and 128. Then if you get 128, you know you can't switch because the two pairs are 64 and 128, so you can only lose.
that's new info you never said there was a known max or a possible max

#45 ShakeZuma

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Posted 15 April 2008 - 12:12 PM

so you're saying your friend's an asshole?

View PostAmScray, on 30 August 2010 - 12:41 PM, said:

one cannot possibly ascribe themselves to the larger (D) philosophy without first being a poon

#46 LongLiveYorke

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Posted 15 April 2008 - 12:39 PM

View Postkeith crime, on Tuesday, April 15th, 2008, 4:11 PM, said:

that's new info you never said there was a known max or a possible max
Right. I'm saying the original question was ill posed and therefore impossible. To ask a question that makes sense, there needs to be some probability distribution that sums to 1, so in a sense there is a limit higher numbers, for instance.

#47 Napa_Don

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Posted 15 April 2008 - 12:40 PM

View PostLongLiveYorke, on Tuesday, April 15th, 2008, 3:39 PM, said:

Right. I'm saying the original question was ill posed and therefore impossible. To ask a question that makes sense, there needs to be some probability distribution that sums to 1, so in a sense there is a limit higher numbers, for instance.
I still think you're smarter with your old avatar.

#48 ShakeZuma

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Posted 15 April 2008 - 12:41 PM

View PostLongLiveYorke, on Tuesday, April 15th, 2008, 4:39 PM, said:

Right. I'm saying the original question was ill posed and therefore impossible. To ask a question that makes sense, there needs to be some probability distribution that sums to 1, so in a sense there is a limit higher numbers, for instance.
but what if it was a check??? AAAAAHHHHHH.....

View PostAmScray, on 30 August 2010 - 12:41 PM, said:

one cannot possibly ascribe themselves to the larger (D) philosophy without first being a poon

#49 keith crime

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Posted 15 April 2008 - 12:43 PM

View PostLongLiveYorke, on Tuesday, April 15th, 2008, 12:39 PM, said:

Right. I'm saying the original question was ill posed and therefore impossible. To ask a question that makes sense, there needs to be some probability distribution that sums to 1, so in a sense there is a limit higher numbers, for instance.
nah it can be theoretical money and any number can be doubled

#50 LongLiveYorke

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Posted 15 April 2008 - 12:48 PM

View Postkeith crime, on Tuesday, April 15th, 2008, 4:43 PM, said:

nah it can be theoretical money and any number can be doubled
It can be theoretical money, but what determines how much money is put in the envelope? How can you come up with a system that determines what goes into the envelopes? At some point, a person or a machine or something must actually put the money in the envelope, or decide the value, or whatever. If you make a machine that does this, it at some point must have been given some sort of probability distribution that controls what money it puts in the envelope. If it's a human, then we internally have some sort of distribution. Either way, there's no way we can have a flat probability over the entire number line, since it would integrate to infinity.

#51 keith crime

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Posted 15 April 2008 - 12:51 PM

ignore

#52 Shimmering Wang

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Posted 15 April 2008 - 12:59 PM

View PostLongLiveYorke, on Tuesday, April 15th, 2008, 9:59 AM, said:

Okay, my turn.Let's say a friend comes up to me and gives me the following offer:He's going to put a certain amount of money in one envelope and twice that amount of money in another envelope. He then hands me one of the envelopes and lets me open it and see the amount of money inside. I do so. He then gives me the chance to switch envelopes. Should I switch or should I stay?Here's my solution:Well, there's clearly a 50/50 chance that I pick either the smaller one or the bigger one. So, if I switch, I could either double or half the money I have with a 50/50 probability. So, .5 * 2x + .5 * (1/2)x = 1.25xSince 1.25x > x, the amount of money I have, I should switch.But of course this solution makes no sense. Had I picked the other one, I would have come to the same conclusion and switched. What's the resolution?
Let x = the amount of money in the SMALLER of two envelopesOur expected equity, then, is (x + 2x)/2, or 1.5xIf we choose an envelope, it will either contain x dollars or 2x dollars. If we switch, we will either gain 1x (when we pick the envelope with less money, and switch to more) or lose 1x (when we pick the envelope with more money, and switch to less)Case 1: Pick BIG, stay: Money received, 2x Case 2: Pick BIG, switch: Money received, 1xCase 3: Pick SMALL, stay: Money received, 1xCase 4: Pick SMALL, switch: Money received, 2xIn all cases, there is a 50% chance we'll gain a unit, and a 50% chance we'll lose a unit by switching. .5(1) + .5(-1) = .5 -.5 = 0Our switching equity is always neutral.The mistake you made, yorke, is in your definition of x. If x is "the amount of money in the envelope," it is oscillating, and is therefore THE AVERAGE of the two possible amounts. There aren't x$'s in the envelope. There are 1.5x$'s. Because half the time there's 2 bucks, and the other half there's one. I think.

#53 brvheart

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Posted 15 April 2008 - 01:00 PM

View PostLongLiveYorke, on Tuesday, April 15th, 2008, 2:43 PM, said:

LOL at everything about this. I had the old avatar since day 1. I'm digging the new one.
The new one sucks. The straight-on head shot made him look less gay and more real. It's science.

View PostiZuma, on 20 August 2012 - 11:32 AM, said:

napa I was jesus christing suited, you guys just slipped in before me.

View PostEssay21, on 25 February 2013 - 08:32 PM, said:

.

#54 Shimmering Wang

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Posted 15 April 2008 - 01:01 PM

View PostLongLiveYorke, on Tuesday, April 15th, 2008, 4:48 PM, said:

It can be theoretical money, but what determines how much money is put in the envelope? How can you come up with a system that determines what goes into the envelopes? At some point, a person or a machine or something must actually put the money in the envelope, or decide the value, or whatever. If you make a machine that does this, it at some point must have been given some sort of probability distribution that controls what money it puts in the envelope. If it's a human, then we internally have some sort of distribution. Either way, there's no way we can have a flat probability over the entire number line, since it would integrate to infinity.
Smaller amount is 1x, bigger amount is 2x. Total money, on average, is 1.5x. Your math is wrong. Losing half a unit is not the same as gaining a whole unit. Going from 1 unit to .5 units is NOT THE SAME as going from 1 unit to 2 units. We're actually going from 1.5 units to either 1 unit or 2 units. That's where you're messing up.

#55 LongLiveYorke

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Posted 15 April 2008 - 01:02 PM

View PostShimmering Wang, on Tuesday, April 15th, 2008, 4:59 PM, said:

Let x = the amount of money in the SMALLER of two envelopesOur expected equity, then, is (x + 2x)/2, or 1.5xIf we choose an envelope, it will either contain x dollars or 2x dollars. If we switch, we will either gain 1x (when we pick the envelope with less money, and switch to more) or lose 1x (when we pick the envelope with more money, and switch to less)Case 1: Pick BIG, stay: Money received, 2x Case 2: Pick BIG, switch: Money received, 1xCase 3: Pick SMALL, stay: Money received, 1xCase 4: Pick SMALL, switch: Money received, 2xIn all cases, there is a 50% chance we'll gain a unit, and a 50% chance we'll lose a unit by switching. .5(1) + .5(-1) = .5 -.5 = 0Our switching equity is always neutral.The mistake you made, yorke, is in your definition of x. If x is "the amount of money in the envelope," it is oscillating, and is therefore THE AVERAGE of the two possible amounts. There aren't x$'s in the envelope. There are 1.5x$'s. Because half the time there's 2 bucks, and the other half there's one. I think.
I mean, I agree with all of this, this is all very true and it should be since the problem is symmetric. I guess the real question should have been, "What is wrong with the following line of reasoning?" in relation to my original "solution."

#56 runthemover

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Posted 15 April 2008 - 01:03 PM

View PostShimmering Wang, on Tuesday, April 15th, 2008, 1:59 PM, said:

I don't finish reading a thread before posting
I'm a bit sad that Keith Crime got there first too.

#57 LongLiveYorke

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Posted 15 April 2008 - 01:05 PM

Okay, we've beaten this one to death. I'll try to find a new one. I like that one, but it's more of a puzzle in reasoning rather than a real puzzle that could have a real solution instead of just showing where someone is wrong.

#58 JoeyJoJo

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Posted 15 April 2008 - 01:07 PM

View PostLongLiveYorke, on Tuesday, April 15th, 2008, 2:05 PM, said:

Okay, we've beaten this one to death. I'll try to find a new one.
I'm not sure you understand the concept of a Dear LLY thread.Don't stop, I like what you're doing.That's what she said.You wish.
Homer: Moe, I need your advice.
Moe: Yeah?
Homer: See, I got this friend named... Joey Jo Jo... Junior... Shabadoo.
Moe: That's the worst name I ever heard.

#59 LongLiveYorke

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Posted 15 April 2008 - 01:11 PM

View PostJoeyJoJo, on Tuesday, April 15th, 2008, 5:07 PM, said:

I'm not sure you understand the concept of a Dear LLY thread.Don't stop, I like what you're doing.That's what she said.You wish.
Fine then, I'll open myself up and do whatever anyone wants me to do....that's what she said.

#60 Shimmering Wang

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Posted 15 April 2008 - 01:27 PM

View PostLongLiveYorke, on Tuesday, April 15th, 2008, 5:02 PM, said:

I mean, I agree with all of this, this is all very true and it should be since the problem is symmetric. I guess the real question should have been, "What is wrong with the following line of reasoning?" in relation to my original "solution."
Oh.

View Postrunthemover, on Tuesday, April 15th, 2008, 5:03 PM, said:

I'm a bit sad that Keith Crime got there first too.
Haha! Joke's on you! I don't read the thread PERIOD!




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