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The Two envelope Paradox

Lex
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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.

unenlightened
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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

DMeister
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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?

Lex
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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

swstephe
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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.
"There are only two industries that refer to their customers as 'users'." -- Edward Tufte

DoctorInWaiting
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Posted 05/02/08 - 12:02 PM:	#26
This is a variation of the Monty Hall Problem. Probability wise without any information (like the monty hall problem instigated) it is a 50% chance. There is no real way of instigating which is the right envelope I feel. DMeister, It depends on what cards you had and exactly how many cards could help you within the implied odds. You probably already knew that though! haha

PeteSF
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Posted 05/04/08 - 06:52 PM:	#27
Brian Bosse wrote: 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. Hi Brian, I'm considering a slight variation on the original problem, in which after selecting an envelope, you look inside to find $100, and I'm getting stuck. Consider Model 1: Model 1 If you happen to pick the envelope that is A and switch, you will have 2X. X = $100 If you happen to pick the envelope that is A and stay, you will have X. X = $100 If you happen to pick the envelope that is B and switch, you will have X. X = $50 If you happen to pick the envelope that is B and stay, you will have 2X. X = $50 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. Now we have a problem, because the value of X is not constant among the four possibilities. Switching averages out to $125, staying averages out to $100. What happened?

Lex
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Posted 05/04/08 - 10:33 PM:	#28
No problem. How in the world did you get something ridiculous like that? x = 50, always, and it can't be any other way. If you switch from A to B you now have 50 dollars, not 100, which is x (consistent). If you switch from B to A you now have 100 dollars, which is 2x (also consistent).

PeteSF
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Posted 05/05/08 - 03:23 AM:	#29
Lex wrote: No problem. How in the world did you get something ridiculous like that? x = 50, always, and it can't be any other way. If you switch from A to B you now have 50 dollars, not 100, which is x (consistent). If you switch from B to A you now have 100 dollars, which is 2x (also consistent). In the scenario I'm considering, you've looked in your envelope and seen $100. If you've picked envelope A (containing X dollars), then X is clearly $100. If you've picked envelope B (containing 2X dollars), then X is $50.

swstephe
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Posted 05/05/08 - 04:42 AM:	#30
Try to avoid multiplying or considering the amount in hand. You should only be considering the value you gain or lose by and nothing else. Consider this variation: You have 2 envelopes, one is empty, the other has $100. What is the expected value of switching verses standing? If you start saying X -> 2X or X -> X/2, then you see the obvious problem that you are treating both envelopes as empty, (2X = 0 and X/2 = 0), when they are not. Instead, it is +$100 if you switch to* the envelope with the money, and -$100, if you switch from it. The expected value is .5(+$100) + .5(-#100) = $50 - $50, meaning the expected value of switching/swapping is both $0. Maybe this is the same kind of fallacy that advertisers use when they claim something like "double your savings", (makes it sound like a lot, even if it is saving 2 cents instead of just 1 cent)? There may be another bias in play, as well -- the "grass is greener", which makes it seem reasonable to overestimate the value of switching???
"There are only two industries that refer to their customers as 'users'." -- Edward Tufte

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