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The Two envelope Paradox
O so simple...

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The Two envelope Paradox

unenlightened
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Posted 03/31/08 - 10:37 AM: Subject: The Two envelope Paradox	#1
I have lifted this from elsewhere for the amusement and bafflement of anyone interested. You are presented with a choice between two envelopes. You know one envelope contains twice as much money as the other, but you don't know which contains more. You arbitrarily choose one envelope -- call it Envelope A -- but don't open it. Call the amount of money in that envelope X. Since your choice was arbitrary, the other envelope (Envelope B) is 50% likely to be the envelope with more and 50% likely to be the envelope with less. But, strangely, that very fact might make Envelope B seem attractive: Wouldn't switching to Envelope B give you a 50% chance of doubling your money and a 50% chance of halving it? Since double or nothing is a fair bet, double or half is more than fair. Applying the standard expectation formula, you might calculate the expected value of switching to Envelope B as (.50)½X [50% chance it has less] + (.50)2X [50% chance it has more] = (1.25)X. So, it seems, you ought to switch to Envelope B: Your expected return -- your return on average, over the long run, if you did this many times -- would seem to be 25% more. But obviously that's absurd: A symmetrical calculation could persuade you to switch back to Envelope A. Hence the paradox. Where have we gone wrong? What's the flaw in the reasoning?
The observer is the observed. J Krishnamurti "Philosophy, to the Philistine, is an evolutionary process, watched over by some sort of brisk dynamic Providence, and culminating in the supreme insight of modern thought." John Cowper Powys

Brian Bosse
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Posted 03/31/08 - 12:51 PM:	#2
Hello Unenlightened, Model 1 Envelope A contains X dollars and Envelope B contains 2X dollars. You do not know what envelope is A or B. If you happen to pick the envelope that is A and switch, you will have 2X. If you happen to pick the envelope that is A and stay, you will have X. If you happen to pick the envelope that is B and switch, you will have X. If you happen to pick the envelope that is B and stay, you will have 2X. This exhausts all possibilities. Switching averages out to 1.5X dollars, and staying averages out to 1.5X dollars. As such, it makes no difference whether you stay or switch. If you pick an envelope, then it either has X or 2X. Model 2 Envelope A contains X dollars and Envelope B contains X/2 dollars. You do not know what envelope is A or B. If you happen to pick the envelope that is A and switch, you will have X/2. If you happen to pick the envelope that is A and stay, you will have X. If you happen to pick the envelope that is B and switch, you will have X. If you happen to pick the envelope that is B and stay, you will have X/2. This exhausts all possibilities. Switching averages out to .75X dollars, and staying averages out to .75X dollars. As such, it makes no difference whether you stay or switch. If you pick an envelope, then it either has X or X/2. --------------------------------------------------------- Where you went wrong in your analysis is that you conflated the two models. In essence, if an envelope is X and you state that the remaining envelope could be 2X or X/2, then you are speaking of two different models - two different ways of looking at the problem. You then went and did an analysis that acted as if model 1 and model 2 where both possible at the same time. In essence, you equivocated. X has a different meaning (value) depending upon which model you are using. For instance, if the money in the envelopes is $5 and $10 respectively, then in model 1 X is $5 and in model 2 X is 10$. You can't treat the X's in both models as if they stand for the same value. They so not. Hence, the equivocation. Does this help? Sincerely, Brian Edited by Brian Bosse on 03/31/08 - 01:55 PM

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Posted 04/01/08 - 02:09 AM:	#3
Another way of looking at it. Call the one amount "$1" and the other amount "$2". The expected value of the amount you have in your hand is .5($1)+.5($2) = $1.50 for keeping your first choice. If you switch, you could consider a new independent game: .5($2)+.5($1) = $1.50. Or if you take the switching into account: .5($1-$2)+.5($2-$1)=.5-.5=$0! That is, you will get $1.50 whether you switch or not. The second choice becomes 100% after you make the first choice. I think the 1.25 was calculated wrongly. The values are 2X and X, not 2X and 1/2X -- which would make the first four times the value of the second. But it doesn't matter the proportion of the values, switching doesn't affect the outcome. If the problem is consistent, it doesn't change the outcome. This is similar to the "Monty Hall problem", where Monty tips his hand by revealing one of the losing choices and shifts the odds a bit.
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unenlightened
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Posted 04/01/08 - 11:52 AM:	#4
No prizes yet I'm afraid. excellent analysis from Brian, but I didn't even use your two models, so how have I conflated them. I know the answer is wrong, I even sort of know why it is wrong, but where is the flaw in the reasoning? It's here somewhere: You arbitrarily choose one envelope -- call it Envelope A -- but don't open it. Call the amount of money in that envelope X. Since your choice was arbitrary, the other envelope (Envelope B) is 50% likely to be the envelope with more and 50% likely to be the envelope with less. But, strangely, that very fact might make Envelope B seem attractive: Wouldn't switching to Envelope B give you a 50% chance of doubling your money and a 50% chance of halving it? Is there something wrong with saying 'Let X be the amount of money in this envelope.' ? Surely not, but then it must be the case that the other envelope has 2X or X/2... and so on. Again, for swstephe, I know you are right, and yet I have clearly defined X to be the amount in envelope A so the other envelope does indeed contain either 2X or X/2. If it contained X, they would both have the same, but we know they don't.
The observer is the observed. J Krishnamurti "Philosophy, to the Philistine, is an evolutionary process, watched over by some sort of brisk dynamic Providence, and culminating in the supreme insight of modern thought." John Cowper Powys

Brian Bosse
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Posted 04/01/08 - 07:17 PM:	#5
Hello Unenlightened, I didn't even use your two models, so how have I conflated them...Is there something wrong with saying 'Let X be the amount of money in this envelope.' ? Surely not, but then it must be the case that the other envelope has 2X or X/2... and so on. There is nothing wrong with allowing X to be the amount of money in the envelope you picked. However, you are equivocating when you say that the remaining envelope has either 2X or X/2. Allow me to be more explicit. There are two envelopes we will call A and B. Envelope A contains half the amount envelope B contains. As such, there is a clear relationship between the two envelopes, namely, 2A=B. You do not know which envelope is which. You pick an envelope. At this point, you assign the value of X to the envelope you picked. You then concluded that the remaining envelope contains either X/2 or 2X. STOP!!!* What is the basis for concluding that the remaining envelope contains either X/2 or 2X? The reasoning process probably went something like this: Since one envelope is twice as much as the other envelope, and if I picked the more expensive envelope and assign it the value X, then the remaining envelope is X/2. However, if I picked the less expensive envelope and assign it the value X, then the remaining envelope is 2X. In other words... Assumption 1: If I picked B and assign it the value X, then A is X/2. Assumption 2: If I picked A and assign it the value X, then B is 2X. My guess is that you thought because assumptions 1 and 2 exhaust all possibilities, then you are left with either X/2 or 2X. But this is where you go all wrong. In assumption 1, the value of X is B, and in assumption 2 the value of X is A. That is to say, the 'X' in assumption 1 means something different than the 'X' in assumption 2! So, when you assert the conclusion that the remaining envelope is either X/2 or 2X, the X's have different values. They are not the same. Yet, you treat them as if they are the same. This is the fallacy of equivocation. What you should have said is... (1) If I assign the value of X to the envelope I just picked, then the remaining envelope is either A or B. (This is a completely trivial consequence.) Because we know that 2A=B, then we can say ... (1A)* If I assign the value of X to the envelope, then the remaining envelope is either A or 2A. Or, if one prefers... (1B) If I assign the value of X to the envelope, then the remaining envelope is either B/2 or B. In no case is it valid to say that the remaining envelope is 2X or X/2. Sincerely, Brian Edited by Brian Bosse on 04/01/08 - 07:32 PM

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Posted 04/08/08 - 08:49 PM:	#6
Hello, I believe the flaw in this particular reasoning is assuming that the money contained within each envelope has any real significance to the probability of actually choosing between the two envelopes. The money is only an incentive to choose a particular envelope out of the set of two, that, of course, is alien to you. This means that the money is semantically meaningful and therefore of no probabilistic significance.

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Posted 04/09/08 - 05:50 PM:	#7
The value is significant since we are calculating expected value in dollar amounts. If the value is not significant, then we might as well call one "A" and the other "B", and the expected value is always "A/2 + B/2" for staying or switching.
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Posted 04/10/08 - 03:06 AM:	#8
The numerical value of the contents shouldn't be used in the calculation. They are irrelevant to it - this is obvious if you rerun the original post calcs with one envelope containing $1 and the other $75. Then $1 and $1.01. The numerical values chosen seem to affect the probablilities - surely not! Try it with unmeasureable values - one envelope contains a pic of your cat when he was a sweet little fluffy kittten ooooo, the other has one of (shudder) Donald Rumsfeld - smiling. The switch/no-switch choice looks straightforward and as we intuitively grasp... you have equal chances of being revolted or melted at any stage. The problem seems to come from a bit of thinking that prioritises numbers and their importance - which presumably is common to us all? Or just to people who contribute to these boards?
Looks like freedom but it feels like death Its something in between I guess.

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Posted 04/10/08 - 07:38 AM:	#9
Hello Klingor's Goldfish, The numerical value of the contents shouldn't be used in the calculation. They are irrelevant to it... I agree. However, I think I understand why someone would think this. Consider the game of Poker. The amount of money in the pot, the amount of money you have invested in the pot, and the odds of getting a certain hand all play a part in whether or not you stay in the hand or fold. For instance, let's say that you estimate that you have a 15% chance of having a winning hand. You have already invested 10$ in the pot and you are required to invest $10 more. This would bring the pot value to $50. In other words, by investing $20 you have a 15% chance of winning $50. At the same time, by not investing $10 more dollars into the pot you will only lose the initial $10 you put into the pot. So, this is how it breaks down... 1. Stay - [85% lose $20 and 15% win $50] - This equates to an average loss of $9.50. 2. Fold - [100% lose $10] - This equates to an average loss of $10.00. Based on this analysis, you should stay in the game. Now, let's say that the pot value would be $45 instead of $50. 1. Stay - [85% lose $20 and 15% win $45] - This equates to an average loss of $10.25. 2. Fold - [100% lose $10] - This equates to an average loss of $10.00. Based on this analysis, you should fold your hand. The point of this illustration is to show how pot values have a direct correlation on probability. As such, someone might mistakenly be thinking along these lines and figure that the value of the contents (just like the value of the pot) is pertinent to the discussion. Sincerely, Brian

unenlightened
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Posted 04/10/08 - 08:26 AM:	#10
Here is the original article, in case you haven't had enough already. http://www.sorites.org/Issue_20/dever.htm
The observer is the observed. J Krishnamurti "Philosophy, to the Philistine, is an evolutionary process, watched over by some sort of brisk dynamic Providence, and culminating in the supreme insight of modern thought." John Cowper Powys

Brian Bosse
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Posted 04/10/08 - 01:19 PM:	#11
Hello Unenlightened, Here is a quote from the article you referenced. Allowing X to have different expectations in different parts of the formula in this way is like comparing apples and oranges. The «X» in the «2X» just isn't the same as the «X» in the «½X» part. That is what I was getting at when I accused you of equivocation. Here is how I put it in my last post... Brian wrote: So, when you assert the conclusion that the remaining envelope is either X/2 or 2X, the X's have different values. They are not the same. Yet, you treat them as if they are the same. This is the fallacy of equivocation. I think we nailed the solution to this problem. Thanks for the fun puzzle. I enjoyed thinking about it. Sincerely, Brian

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Posted 04/11/08 - 09:12 AM:	#12
Ok, have a gold star, and this round of applause...
The observer is the observed. J Krishnamurti "Philosophy, to the Philistine, is an evolutionary process, watched over by some sort of brisk dynamic Providence, and culminating in the supreme insight of modern thought." John Cowper Powys

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Posted 04/21/08 - 11:24 AM:	#13
Brian Bosse wrote: Hello Klingor's Goldfish, I agree. However, I think I understand why someone would think this. Consider the game of Poker. The amount of money in the pot, the amount of money you have invested in the pot, and the odds of getting a certain hand all play a part in whether or not you stay in the hand or fold. For instance, let's say that you estimate that you have a 15% chance of having a winning hand. You have already invested 10$ in the pot and you are required to invest $10 more. This would bring the pot value to $50. In other words, by investing $20 you have a 15% chance of winning $50. At the same time, by not investing $10 more dollars into the pot you will only lose the initial $10 you put into the pot. So, this is how it breaks down... 1. Stay - [85% lose $20 and 15% win $50] - This equates to an average loss of $9.50. 2. Fold - [100% lose $10] - This equates to an average loss of $10.00. Based on this analysis, you should stay in the game. Now, let's say that the pot value would be $45 instead of $50. 1. Stay - [85% lose $20 and 15% win $45] - This equates to an average loss of $10.25. 2. Fold - [100% lose $10] - This equates to an average loss of $10.00. Based on this analysis, you should fold your hand. The point of this illustration is to show how pot values have a direct correlation on probability. As such, someone might mistakenly be thinking along these lines and figure that the value of the contents (just like the value of the pot) is pertinent to the discussion. Sincerely, Brian I don't think this is quite right. In Poker, once you have put money into a pot, that money is no longer yours. In the situation you described, there is $50 in the pot, and you are required to put $10 to stay. You should not consider the $10 that you put into the pot previously, because it is irrelevant at this point. Consider the following extreme example: You are playing No Limit Texas Hold 'Em. You are dealt KK, pocket kings. You make a very large raise pre-flop, say 50% of your stack, or $1,000. Everyone folds, except for the big blind. To make things simpler, let's ignore the value of the blinds in the pot. So you have a pot of $2,000. The flop comes A27, rainbow. The big blind makes a very poor bet of $400, or 20% of the pot. Right now you have a strong pair with no flush or straight draw on the board, but an overcard has hit. Based on your knowledge of the big blind's pre-flop strategy, you estimate that 75% of the hands that he'd call such a large bet with will contain an Ace (say, AK, AA, AQ, KK). So right now you have a 25% shot at winning the pot. The important thing to remember is that it doesn't matter how the money that's currently in the pot got there. If you mistakenly give me a thousand dollars, and I tell you that your only two options are either to let it go or to give me $100 for a 15% chance of getting your thousand back plus the hundred, you should do it every time. You're getting good odds, regardless of where the thousand dollars came from. Back to the Hold 'Em hand: you at least call the $400, and you might even raise. You don't think of it as putting $1400 into a $2400 pot for a 25% chance of winning it, you think of it as putting $400 into a $2400 pot for a 25% chance of winning it. The $1000 is already gone.
You down with OPP(Original Poster's Prerogative)?

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Posted 04/26/08 - 12:23 PM:	#14
Hello Thoughtless, Thank you for the correction. Sincerely, Brian

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Posted 04/28/08 - 09:44 AM:	#15
No problem.
You down with OPP(Original Poster's Prerogative)?

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Posted 04/29/08 - 06:05 PM:	#16
You have not solved the problem if the person is allowed to look into the envelope. For suppose the person opens up the envelope and sees the amount A, then the other envelope has either 2A or A/2, with even probability for either one. Therefore, you're expected value is greater if you switch. However, you will always reason like this, no matter what the value of A is. Therefore, you will opt to switch without even looking into the envelope.

unenlightened
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Posted 04/30/08 - 10:49 AM:	#17
esmoothie wrote: Therefore, you will opt to switch without even looking into the envelope. And by the same reasoning, you will at once switch back. Et cetera. Until you jump out of the cycle and decide that there must be something wrong with the reasoning. But I'm still really waiting for a well justified and clear rule to tell me under what conditions I am not justified in assigning variables in this way.
The observer is the observed. J Krishnamurti "Philosophy, to the Philistine, is an evolutionary process, watched over by some sort of brisk dynamic Providence, and culminating in the supreme insight of modern thought." John Cowper Powys

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Posted 04/30/08 - 07:28 PM:	#18
It's not a paradox. You are not in any way justified in using the equation you used in this situation. That is exactly why you get screwed up results, If you wanted to get an accurate representation you would have to calculate the actual outcomes of each choice for 4 possibilities: 1) You get x and stick with it 2) You get x and switch to 2x 3) You get 2x and stick with it 4) You get 2x and switch to x 1) You lose x 2) You lose nothing 3) You lose nothing 4) You lose x Where loss is the fluctuation from the maximum amount you could gain. Symmetry. This is strictly a probability problem. Using your equation is the inherent mistake in your argument (you used it on "instinct", which is not allowed in any sort of mathematical analysis).

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Posted 05/01/08 - 06:44 AM:	#19
Lex wrote: ...You are not in any way justified in using the equation you used in this situation. ... Symmetry. This is strictly a probability problem. Using your equation is the inherent mistake in your argument (you used it on "instinct", which is not allowed in any sort of mathematical analysis). You haven't really pointed out the exact nature of the error. You've only shown that you can arrive at a different conclusion. Why, exactly, aren't we justified in applying the equation? From Expected Value In probability theory the expected value (or mathematical expectation, or mean) of a discrete random variable is the sum of the probability of each possible outcome of the experiment multiplied by the outcome value (or payoff). Thus, it represents the average amount one "expects" as the outcome of the random trial when identical odds are repeated many times. Note that the value itself may not be expected in the general sense - the "expected value" itself may be unlikely or even impossible. So, other than the demonstrations of "accurate representation", why can't the expected value function be applied to the discrete random variable "switch envelopes"? What, exactly, is different here than in, say, a heads-up Hold 'em hand where the opponent is all-in after the flop, and you have a nut flush draw, and you calculate the expected value of calling? The complete answer, that I'm looking for anyway, will have a form similar to the following: (1)The Expected Value function operates only on domains with property X. (2)The variable "call all-in" creates a domain with property X, based on such and such criteria. Therefore, the Expected Value function can be applied to the variable "call all-in." (3)The variable "switch envelopes" does NOT create a domain with property X, based on such and such criteria. Therefore, the Expected Value function cannot be applied to the variable "switch envelopes" and the variable assignment in the paradox is invalid. As far as I can tell, none of the answers provided can be used to create statements 1, 2, and 3. Edited by DMeister on 05/01/08 - 08:59 AM
We are waiting for God only knows what. How about a chocolate soda?

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Posted 05/01/08 - 12:47 PM: Subject: More Hold'em	#20
unenlightened wrote: You are presented with a choice between two envelopes. You know one envelope contains twice as much money as the other, but you don't know which contains more. You arbitrarily choose one envelope -- call it Envelope A -- but don't open it. Call the amount of money in that envelope X. I have X chips on the table. Since your choice was arbitrary, the other envelope (Envelope B) is 50% likely to be the envelope with more and 50% likely to be the envelope with less. But, strangely, that very fact might make Envelope B seem attractive: Wouldn't switching to Envelope B give you a 50% chance of doubling your money and a 50% chance of halving it? My heads up opponent is all in before the flop. There are X chips in the pot. It will cost me X/2 to call. I am certain that my opponent and I each have a 50% chance of wining the hand.(pocket pair vs 2 overs, etc.) Who would fold here? The expected value of calling is in fact 1.25X.
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Posted 05/01/08 - 01:03 PM:	#21
DMeister wrote: You haven't really pointed out the exact nature of the error. You've only shown that you can arrive at a different conclusion. Why, exactly, aren't we justified in applying the equation? If you want to use the expected value function: you say, we have 2 envelopes. In one envelope we have 2x money, in another we have x. What is the expected value of making a choice? (2x + x)/2 = 1.5, (NOT a possible solution by the way). But that does not matter. Now if you want to apply it in your case (PAY ATTENTION) you have say: I am making a decision to change the envelope. If I already have 2x, I will lose x (-x), if I have x, I will gain x (x), so the expected value of making any one choice is (-x + x)/2 = 0. That is the correct way to apply the formula in this case. Yours is kind of like saying sin(A+B) = sinA + sinB because I feel like it.

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Posted 05/01/08 - 01:32 PM:	#22
Lex wrote: Yours is kind of like saying sin(A+B) = sinA + sinB because I feel like it. Ah, a proof by reductio ad analogium.
The observer is the observed. J Krishnamurti "Philosophy, to the Philistine, is an evolutionary process, watched over by some sort of brisk dynamic Providence, and culminating in the supreme insight of modern thought." John Cowper Powys

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Posted 05/01/08 - 03:18 PM:	#23
Lex wrote: you have say: Must I Lex? Now (PAY ATTENTION): No one disagrees with what you've written. However, your reason for why the equation: .5(x/2) + .5(2x) = ExpectedValueOfSwitching is wrong is exactly: "This equation doesn't have the same results as what Lex has written." I am calling that a baloney reason. Please demonstrate, in the general case, not by working out each specific case, why this formulation of the Expected Value function is incorrect. Please also indicate clearly why switching envelopes is a completely different situation than calling the all-in bet I outlined in post #20. Again (PAY ATTENTION, pretty please), I'm not in disagreement with you or Brian or anyone else regarding the actual expected value of picking an envelope, nor do I believe there is an inherent profit to switching envelopes. But, if this sentence is TRUE: Since your choice was arbitrary, the other envelope (Envelope B) is 50% likely to be the envelope with more and 50% likely to be the envelope with less. Why does the assignment of variables lead to the incorrect answer when switching envelopes, but the correct answer when I'm calling an all-in bet? How should the values be assigned IN TERMS OF AN ARBITRARY AMOUNT X?
We are waiting for God only knows what. How about a chocolate soda?

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Posted 05/01/08 - 05:08 PM:	#24
BECAUSE 1) Why do you even think "5(x/2) + .5(2x) = ExpectedValueOfSwitching" is right? You have to work with absolute amounts in each case without using percentages multiplied by percentages each time. Why? Actually unenlightened gave me a good idea, I will show you that using your formula is ridiculous by definition via reductio ad absurdum and show you the right formula. 2) Actual proof One of the uses of the expected value is for gambling (I'm sure you know). By betting on black, you get an expected value (per dollar) of (-1 * 18/37) + (+1 * 19/37) = -.027027 (because you have a chance of landing on 0 and losing it all, reducing the expected value), which represents the value of your investment. If you start applying the expected value theorem to a choice already made, you will obviously start getting powers, and the tiny negative amount will go to zero which is of course absurd. Because you no longer longer change the probability when you have already picked ACCOUNTING for the probability of getting zero, changing your choice now involves creating ANOTHER expected value function where the probability is exactly the same in each case: (-1 * 1/2) + (+1 * 1/2) = 0. That is the actual value of making a choice now. You were right when you said expected value is 1.25x, however this is where you can no longer randomly apply the same thing. Your new formula is (1/2x * 1/2) + (-1/2x * 1/2) where 1/2x and -1/2x are the amounts you win or lose by choosing. This adds to 0. (PROOF complete now an aside) If you want to know where you went wrong: you use the expected value on actual values of the envelopes rather than VALUE GAINED (-x + x) in either case which is inherently wrong (SOURCE OF PROBLEM IDENTIFIED). And please, we are not arguing about religion, it's math, you don't just say "OMG paradox prove me wrong!" YOU are the one who has to prove it a paradox and saying "I used this formula because I felt like it" is not proof. Edited by Lex on 05/01/08 - 05:18 PM

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Posted 05/02/08 - 03:04 AM:	#25
Just to keep it brief, the problem is that the "x" in x/2 and 2x are different values. The switching scenarios are: x <becomes> 2x 2x <becomes> x If you use v1 and v2 to show the value of switching, x + v1 = 2x, so v1 = x 2x + v2 = x, so v2 = -x So the expected value is really: .5(x) + .5(-x) = 0 You have a 50% chance of gaining "x" (the smaller of the two values), and 50% of losing the same amount. So it is the same as not switching. Also, the relative value is irrelevant, so even if it were $1 versus $100 the expected value is the same from switching and staying. The only thing that matters is the probability of getting either result.
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