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Topic: HARD: ENVELOPE GAMBLE (Read 9815 times) |
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towr
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Re: HARD: ENVELOPE GAMBLE
« Reply #25 on: Dec 24th, 2002, 1:10am » |
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on Dec 23rd, 2002, 7:59pm, Kozo Morimoto wrote:Q: Is Paul’s expected value also positive for the same bet?
Yes, but see below.
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if paul hasn't opened his envelope and knows you have 100 he could have 50 or 200. So his expected value is 125. If he wins he has 50, and gets 100, so that's 150, if he looses he has 200, and looses it. So after the bet his expected value is 75. His expected gain is -50. Not at all positive..
Perhaps it's a matter of perspective, who expects what for whom..
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S. Owen
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Re: HARD: ENVELOPE GAMBLE
« Reply #26 on: Dec 24th, 2002, 8:01am » |
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After thinking for a while, I think I get this one. I'm not sure I fully appreciate the subtlety, but here's a first attempt.
The hidden twist here is that we're sort of saying that X and Y are randomly chosen positive numbers, chosen from (0, infinity). Keeping that in mind, the contradictions go away, in the sense that one realizes that we're talking about infinities.
The expected value for each of them before the bet is really P(X is in envelope)*E(X) + P(Y is in envelope)*E(Y) = 1/2(E(X) + E(Y)) - but this is infinity since the expected value of each is infinite, given what we know.
If it's then revealed that you have 100 (and Paul has Z, where Z is the other value; it's positive and could be more or less than 100), then your expected gain from the bet is P(Z < 100)*(-100) + P(Z > 100)*E(Z). But this is again infinity, since P(Z < 100) is infinitesimal and P(Z > 100) and E(Z) are infinite.
So, that I think solves the mystery of why just knowing how much money you have suddenly makes the bet attractive, no matter how much the value actually is.
Now as for Paul though... I can only assume that the post is talking about the original bet or something, because he has "everything to lose" - following the previous reasoning, his expected value of switching is infinitely negative, since his expected value before knowing how much he has is infinite.
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BNC
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Re: HARD: ENVELOPE GAMBLE
« Reply #27 on: Dec 25th, 2002, 9:40am » |
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OK, I hope I won't repeat too much of what was already posted.
I heard this one long time ago, and thaat ver was in two stages:
1. You open the envelope and find $100. Thus, the other envelope contains either $50 or $200 (note that you do not know in advance what is X, so the argument that you have in your envelope either X or 2X is invalid -- you always, by definition, got X). So, you should switch. No wonder there.
2. Now, you don't open the evelope. You figure: since X is unbounded, no matter what value I would find in the envelope, I would still choose to trade. So, you trade. But then, given the option to trade back -- you wil again, for the same reasoning!
Well, I don't know if it's true, but from what I heard back then, this is an unsolved paradox. There are attempts to bypass this paradox by limiting the higher amount by "the total of money in the world = H" (so, if you get > H/2, so don't switch) and the lower amount by "the lowest currency unit in the world - L" (so, if you get L you switch).
IMHO, these are unsatisfactory solutions. And, as I said, don't think an actual solution exists. I will be happy to learn otherwise, though.
See ya' around,
BNC
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towr
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Re: HARD: ENVELOPE GAMBLE
« Reply #28 on: Dec 25th, 2002, 12:15pm » |
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on Dec 25th, 2002, 9:40am, BNC wrote:| 2. Now, you don't open the evelope. You figure: since X is unbounded, no matter what value I would find in the envelope, I would still choose to trade. So, you trade. But then, given the option to trade back -- you wil again, for the same reasoning! |
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You may not know what X is, but it is a set value..
The envelope you have has expected value X, you trade because the expected value of the other is (0.5X+2X)/2 = 1.25X.
Given the chance to trade back you don't, because the expected values don't change till your information changes, so your current envelope keeps expected value 1.25 X
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BNC
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Re: HARD: ENVELOPE GAMBLE
« Reply #29 on: Dec 25th, 2002, 11:44pm » |
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Yeah, you don't have new information. None, whatsoever. Including the value of X (it is set, yes, but still unknown).
Lets define your new evelope as having value Y (=1.25X, but X is artitrary, so Y is arbitrary as well).
The value of the first envelope, although X originally, may now be represented as 1.25Y (and not 0.8Y) -- I think. I'll be happy to hear an explaination as to why the other value is not 1.25Y -- in terms of Y, not X.
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GRAND_ADMRL_THUORN
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Re: HARD: ENVELOPE GAMBLE
« Reply #30 on: Feb 12th, 2003, 12:34pm » |
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ive read every entry in this thread and like always you guys break it down incredible well, anyway after reading everything and some thinking on my own, i have to conclude what i thought when i first read this riddle, changing is irrelavent
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James Fingas
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Re: HARD: ENVELOPE GAMBLE
« Reply #31 on: Feb 12th, 2003, 2:51pm » |
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Grand admiral Thuorn,
I definitely do not agree with you. Paradoxically, you do expect to gain money by switching. Let me remodel the question like this:
1) I pick Y uniformly between 0 and 100 dollars.
2) Now I fill two envelopes, one with Y dollars, and one with 2Y dollars. I then tell you that Y is between 0 and 100.
3) Now open one of the envelopes. Here's where it gets tricky:
4) As per this question, suppose that you open the envelope and discover less than 100 dollars. Computing your payoff, you expect to win money by switching. But how can that be?
5) The catch is that to get a positive expectation, you must find an amount less than $100. If you find an amount larger than $100, you expect (in fact, you are guaranteed) to lose money.
6) Now, I choose Y between 0 and 1000, or 0 and 1 000 000, or 0 and Z (a finite number). The expectation for switching is always positive if X is smaller than Z, the maximum possible value for Y. Of course, 50% of the time, X will be larger than Z, but that's another matter.
7) Here's where the trick is played. You state that Y is chosen between zero and infinity. Then, you say that you find X dollars in the envelope, implying that X is finite. If X is finite, then it must be smaller than Z, which is infinite. Therefore, you have a positive expectation.
8) The problem is that there is absolutely zero chance of finding a finite number in the envelope if Z is infinite. It's not just unlikely that you'll get a finite number, it cannot happen.
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TimMann
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Re: HARD: ENVELOPE GAMBLE
« Reply #32 on: Feb 13th, 2003, 12:46am » |
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Picking some nits:
on Feb 12th, 2003, 2:51pm, James Fingas wrote:| Now, I choose Y between 0 and 1000, or 0 and 1 000 000, or 0 and Z (a finite number). The expectation for switching is always positive if X is smaller than Z, the maximum possible value for Y. Of course, 50% of the time, X will be larger than Z, but that's another matter. |
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Don't you mean 25% of the time? In the case where X = 2Y (50% of the time), it's still not necessarily true that X > Z, as in 50% of those cases, Y was in the low half of its range.
Quote:| The problem is that there is absolutely zero chance of finding a finite number in the envelope if Z is infinite. It's not just unlikely that you'll get a finite number, it cannot happen. |
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I think you meant to say "The problem is that there is absolutely zero chance of finding an infinite number in the envelope if Z is infinite. It's not just unlikely that you'll get an infinite number, it cannot happen."
If that's what you meant, I follow your reasoning. Let's take it a bit further to be more explicit:
9) So the fact that you found a finite number (i.e., a number less than Z) tells you nothing. Unlike in the finite case, the fact that the number you saw was less than Z no longer tells you that this is a good time to switch. Also, it's no longer true that in 25% of the cases you'll find a number larger than Z in the first envelope you open.
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James Fingas
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Re: HARD: ENVELOPE GAMBLE
« Reply #33 on: Feb 17th, 2003, 12:15pm » |
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Tim,
Good point on the 50%/25% thing.
However, in terms of the finite/infinite thing, I actually meant what I said.
Consider the cdf of the uniform random distribution over all the reals. That cdf is identically zero. Therefore, picking any finite value R, it is absolutely impossible that we can find a number less than R in the envelope (just as impossible as getting -54 when picking a random number from 0 to 10).
So when we do find a finite number, we know that the envelope-stuffer didn't use the uniform distribution over all the reals.
Another corrolary of this is that it's actually impossible to choose a number uniformly at random from the set of all real numbers, but that's not important for this question.
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TimMann
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Re: HARD: ENVELOPE GAMBLE
« Reply #34 on: Feb 20th, 2003, 2:28am » |
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By "it's impossible to choose a number uniformly at random from the set of all reals," do you mean to say that "the uniform random distribution over all the reals" doesn't exist? That is, by assuming that it exists, you've reasoned your way to a contradiction, and therefore you have reduced the idea of a uniform random distribution over all the reals to absurdity. That may be correct (I'm not quite certain), but your previous post with the 8 steps didn't really explain the reasoning in steps 7-8, so it was not clear that's what you were doing.
Backing up a bit, suppose that we naively assume there is such a thing as a uniform random distribution over all the reals. Then it's clear that if we choose a number from this distribution, it will be finite, because all the reals are finite. It's also clear that the probability of getting any particular real number must be 0, but of course that does not mean that it's impossible.
The difficulty is that no function exists that is either the density function or the cumulative distribution function of this distribution. Certainly f(x) = 0 is not the cdf (despite what you said), because f(x) does not approach 1 as x goes to infinity, so it is not a cumulative distribution function at all. Equally, f(x) = 0 is not the density function, because the integral of f(x) from -oo to oo is 0, not 1. Neither is any other function, as the function must be constant to be uniform, but the integral of any other constant function from -oo to oo diverges.
That suggests that indeed there is no such distribution. However, I don't claim to know enough about the foundations of probability theory to say whether that's a valid conclusion or not. It wouldn't seem to make much practical difference, as a distribution for which you can't define either a density function or a cdf doesn't seem useful for much other than as a pathological case.
It certainly can't be true that the distribution exists, but that you get an infinite number, a blank piece of paper, or a rhinoceros when you choose from it, as that would be a contradiction too.
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| « Last Edit: Feb 20th, 2003, 6:34am by TimMann » |
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James Fingas
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Re: HARD: ENVELOPE GAMBLE
« Reply #35 on: Feb 20th, 2003, 12:41pm » |
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Tim,
That's a good question. To be honest, I don't know whether the uniform distribution over all the reals "exists" or "does not exist". It's certainly not a well-behaved distribution, and that's why the cdf is zero for all finite real numbers. You could argue that the cdf still goes to 1 near infinity, but that's getting a little too fuzzy for me.
Whether or not the distribution exists seems to me to be more of a philosophical problem than a mathematical one. The question implies that such a distribution does exist and that you can pick a finite number out of it, and so I assumed that it does exist.
You are correct in saying that a zero chance of picking a number doesn't necessarily mean it's impossible. For instance, when you choose a random number from 0 to 1, there's zero chance of getting any specific number in the distribution, but you always do get one specific number. However, in the case of the unbounded uniform distribution, we are talking about a whole different kettle of fish. The chance of getting a specific finite number is zero, but the chance of getting a number within any finite range is also zero.
I think the best way to avoid this conundrum is just to assume that the envelope-stuffer is using a different distribution. Although the question doesn't tell you what that distribution is, it can't be the unbounded uniform distribution because the envelope-stuffer has no good way to choose a number from that distribution. The properties of such a distribution, and whether or not it exists, can be removed from the scope of the problem.
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Icarus
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Re: HARD: ENVELOPE GAMBLE
« Reply #36 on: Feb 20th, 2003, 4:28pm » |
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Does a uniform distribution over all the reals exist? No.
Being uniform, the probability density is constant. But the integral of any non-zero constant over all the Reals is infinite, while the integral of zero is of course zero. In neither case can you get the required integral of 1.
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James Fingas
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Re: HARD: ENVELOPE GAMBLE
« Reply #37 on: Feb 24th, 2003, 10:05am » |
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Icarus,
I agree with you, but it seems to me that the problem is essentially one of limits.
Certainly no pre-defined "uniform distribution over the reals" can exist, but when we look at it as the limit of the uniform distributions over [0,Z), then we can still have properties which hold as we take the limit. For instance, half of the values you choose from this distribution will be less than Z/2. These properties are what allow the epectation to remain zero for all finite uniform distributions.
I think the expectation still exists as long as we are taking the limit, but if we start from the "uniform distribution over the reals", then we have nothing to work with. It's just like trying to find the value of 0/0. Depending on how you got to 0/0, you can find the value in different ways (eg using L'Hospital's rule). But if somebody just gives you "0/0", you can't work backwards.
In the same way, working from the finite uniform distributions, we expect a certain payoff for switching envelopes, which we can take the limit of. Working from the other direction, we get nonsense. It would be nice if we could do this:
0.75*integral( x=0, x=inf/2, 1.25X ) + 0.25*integral( x=inf/2, x=inf, 0.5X )
However, mathematics can't get an answer out of this any more than it can get an answer out of 0/0.
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pedronunezmd
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Re: HARD: ENVELOPE GAMBLE
« Reply #38 on: May 31st, 2004, 10:10am » |
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Over a year since last post on this riddle, but I personally don't think that the real answer to this riddle has yet been posted, so here goes. (Hope I don't inflame anyone.)
First of all, I hope that everyone agrees with S.Owen's very first post which is the correct way to analyze the question as to whether or not you should switch. The expected gain from switching is zero. If anyone truely believes that the the expected gain (by switching) is actually 0.25x then I would be more than happy to write a Visual Basic program that simulates the game, and play it with you, where you agree to pay 1.125x (where x is the result of your first opened envelope) each time and I pay you whatever is in the second envelope (which has already been determined before you picked your envelope), and we run the number of trials at around 1 million or so. (If the expected gain from changing is truely .25x, then you will gain in the long term by 1.25x - 1.125x = .125x, and if it is 0, obviously you will lose in the long term 1.125x - 1x = .125x)
The reason I believe the riddle is still unsolved, however, is that no one has been able to point out the actual falacy of the reasoning stated in the riddle that ultimately leads to the (incorrect) assertion that the expected return is 1.25x
The error in the logic of the riddle is more fundamental than all this talk about "random distribution" and "infinity". If anyone cares anymore, please post here and I'll provide the correct answer to why the logic in the riddle is not correct, but consider this my hint to the solution of the riddle.
(I can't imagine why people I know in real life have called me narcissistic...)
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rmsgrey
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Re: HARD: ENVELOPE GAMBLE
« Reply #39 on: May 31st, 2004, 11:12am » |
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I'm always willing to hear new thoughts on classic paradoxes like this one.
I'm sure other people are also curious.
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Grimbal
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Re: HARD: ENVELOPE GAMBLE
« Reply #40 on: May 31st, 2004, 12:51pm » |
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I think it is a distribution problem.
Whatever the method you use to choose the amounts to put in the envelopes, the probability of (x,2x) being in the envelopes must tend to 0 when x goes to infinity. If you choose real numbers, the density function must tend to zero.
So, at least for some values, the probability of (x,2x) is lower than the probability of (x/2,x). For instance, in the case where there is a maximum, M, then if you see x>M/2, you certainly don't switch. The expecatation would by x*0.5. If not, you are twice as likely to have choosen the low number. Switching gives you an expectation of x*1.5.
If you don't know the number, either you have a high number in range M/2..M averaging 3/4*M and swiching brings you down to 3/8*M, or you have a low number in range [0..M/2], averaging M/4, and switching brings you 1.5*M/4 = 3/8*M. Either way is the same.
If the set of values is discreet, obviously, there are some values that can not been divided. If you see an amount like $6.11, and there is no half-cent, it must be the low value. These are cases where (x/2,x) is actually less likely than (x,2x).
If you use continuous values, and you choose the low value with density f(x) and the high value as 2x, to say that (x/2,x) and (x,2x) is as likely, is like to say that f(x)=2*f(2x). One solution of this is f(x) = c*1/x. Obviously, it can not be integrated. f(x) must decrease faster than that.
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| « Last Edit: May 31st, 2004, 12:54pm by Grimbal » |
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towr
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Re: HARD: ENVELOPE GAMBLE
« Reply #41 on: May 31st, 2004, 1:27pm » |
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After you've opened the first envelop
you've either got x, and by switching go to 2x, a gain of x.
Or you have 2x, and by switching go to x, a 'gain' of -x
(x+-x)/2 = 0 no mystery..
Total expected gain from the start remains 1.5 x
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Re: HARD: ENVELOPE GAMBLE
« Reply #42 on: May 31st, 2004, 4:30pm » |
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But suppose you open the envelope, it is $42. You can either win $42 or loose $21, right? That is +$10.5, right?
In fact, after you open the envelope, whatever the value, you should switch, right?
So the best thing to do is open and switch. Whatever the value. No?
So you know that you are going to switch anyway, right?
Why not switch immediately?
If you switch immediately, before opening, do you want to switch again?
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pedronunezmd
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Re: HARD: ENVELOPE GAMBLE
« Reply #43 on: May 31st, 2004, 9:00pm » |
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on May 31st, 2004, 1:27pm, towr wrote:After you've opened the first envelop
you've either got x, and by switching go to 2x, a gain of x.
Or you have 2x, and by switching go to x, a 'gain' of -x
(x+-x)/2 = 0 no mystery... |
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Like I said before, I think everyone agrees that this way (which is how S.Owen explained it in the first post of this thread) is the correct way to analyze the problem. However, it still does not explain why the reasoning in the riddle is not correct.
Now I'm going to try to explain the falacy in the logic of the riddle which results in the expected return of 1.25X. I don't have an easy way to explain in words what I am thinking, probably because I lack a strong mathematical background, so I'll try to explain the best I can, and someone can put the final result into better words.
First of all, the rules stated that the envelopes will have a non-zero sum of money, meaning no negative values for any envelope. This should be obvious, but I have to state it outright. Then...
Once you've picked your first envelope, if you call the monetary amount you have in there X, you cannot say that there is a 50% chance that you are looking at X of the (X/2,X) possibility versus a 50% chance that you are looking at the X of the (X,2X) possibility. The reason you cannot do this is: then what you are stating is that you have the higher of the 2 numbers if you have the (X/2,X) possibility, and the lower of the 2 numbers if you have the (X,2X) possibility. You are now looking at only a subset of the (X/2,X) versus (X,2X) set of possibilities. You are basically stating that "I'll know X when I see it" before you have picked your envelope.
Basically, by defining X as "whatever amount is in the envelope I picked", you are playing the game where you will always have the highest of the (x/2,X) possibility or the lowest of the (X,2X) possibility. You have now defined a new game.
You are now calculating the expected payoff for an entirely different game, namely the game where you pick an envelope, but it is guaranteed to be either the higher-valued of the two envelopes if you have the (X/2,X) possibility, and the lower-valued of the 2 envelopes if you have the (X,2X) possibility. In this (different) game, it would make sense to always switch. But this different game has no relevance to how the actual game is being played.
After re-reading everything I just wrote just now, I think that many will still disagree with me. Hopefully at least someone smarter than me will understand what I'm trying to say and restate it better.
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ThudnBlunder
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Re: HARD: ENVELOPE GAMBLE
« Reply #44 on: May 31st, 2004, 9:21pm » |
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Quote:| However, it still does not explain why the reasoning in the riddle is not correct. |
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As with the d(x2)/dx = x fallacy, I think it is case of confusing constants with variables.
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| « Last Edit: May 31st, 2004, 9:22pm by ThudnBlunder » |
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towr
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Re: HARD: ENVELOPE GAMBLE
« Reply #45 on: Jun 1st, 2004, 12:28am » |
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I agree,
you either have X from the pair (X, 2X) or Y from the pair (Y/2, Y).
If you think Y=X, then you'd expect 1/4 X gain
But (X, 2X) = (Y/2, Y), because it's the same pair of envelopes. So Y=2X and expected gain is 0
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| « Last Edit: Jun 1st, 2004, 12:29am by towr » |
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Re: HARD: ENVELOPE GAMBLE
« Reply #46 on: Jun 1st, 2004, 6:14am » |
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Having been traumatized by physics courses, I regard the envelope I pick as being in a quantum superposition of X and 2X so having a value of 1.5X. The other envelope has the same value, so there's no point switching.
Of course, once you open an envelope, you have additional data on the actual value of X, which, combined with information on the distribution of X, may affect your decision to switch...
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asterix
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Re: HARD: ENVELOPE GAMBLE
« Reply #47 on: Jun 1st, 2004, 7:30am » |
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Let's try changing the rules to get rid of the infinity confusion.
I'll write 1 and 2 on one piece of paper and 2 and 4 on another and put them in the envelopes. You pick an envelope, and I'll open it and tell you that there's a 2 on it. Now you can either keep the 2 or trade it for whichever number in the other envelope is not a 2.
Well, 1 and 4 have an expected return of 2.5, so you're better off switching. Knowing that, and knowing that my opening of the first envelope will not give you any additional information (since I'll always find a 2), Even before I open the first envelope, you're better off switching because you don't want the 2, you want the 1 or 4. But there would be no reason to switch back unless I change the rules and say you're going to get the 2 in this envelope instead and it's the first envelope that has the variable payout.
The reason the riddle has an expected payout of 1.25 is because the uncertain envelope is worth more than the certain one. But in the original riddle, the certain one becomes certain only after you open it and see a fixed amount. There is no point in switching before you see that amount because, unlike my variation, what you find in one envelope will not change what's in the other; it only changes how much information you have about what's there. (And information changes odds even when it doesn't change reality)
I hope this makes some sense.
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asterix
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Re: HARD: ENVELOPE GAMBLE
« Reply #48 on: Jun 1st, 2004, 7:49am » |
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Of course, the other way to avoid having to find a random distribution over infinity is to say the numbers aren't random. I deliberately pick the values in each envelope. Then, you know I wouldn't put an odd number in either envelope because when you saw it you'd know the other envelope has to have 2x. Therefore, I'd never pick a number which divided by 2 would be odd, because if you picked that one you'd know the other envelope certainly doesn't contain an odd number, so it must have 2x. Therefore I'd never pick a number which divided by 4 would be odd...
So the only logical choice for me would be to put Infinity in one envelope and 2xInfinity in the other. And the payout is the same, so there's no reason to switch.
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Uberpuzzler
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Re: HARD: ENVELOPE GAMBLE
« Reply #49 on: Jun 1st, 2004, 7:54am » |
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on Jun 1st, 2004, 7:49am, asterix wrote:Of course, the other way to avoid having to find a random distribution over infinity is to say the numbers aren't random. I deliberately pick the values in each envelope. Then, you know I wouldn't put an odd number in either envelope because when you saw it you'd know the other envelope has to have 2x. Therefore, I'd never pick a number which divided by 2 would be odd, because if you picked that one you'd know the other envelope certainly doesn't contain an odd number, so it must have 2x. Therefore I'd never pick a number which divided by 4 would be odd...
So the only logical choice for me would be to put Infinity in one envelope and 2xInfinity in the other. And the payout is the same, so there's no reason to switch. |
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Yeah... and the exam could never happen 
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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