April 20th, 2017, 04:06 PM  #21  
Senior Member Joined: Jun 2014 From: USA Posts: 238 Thanks: 10  Quote:
In the formal setting, if you see 1 in your envelope you always swap because you know you'll get 2. Otherwise, you are tricked into thinking you have a better shot by swapping by incorporating an impossible outcome. Here is a clearer exposition: Traditionally, the paradox arises from comparing what appears to be someone’s odds when they are allowed to see what’s in their envelope vs. their odds when they aren’t. Without loss of generality, let x = 10 and 2x = 20. Assume the person’s envelope to be selected uniformly at random from those two options. Without seeing what is in the envelope, the person makes a choice that is true to reality. They realize that it doesn’t matter whether or not they trade envelopes because they can’t distinguish. Half the time they get 10 and half the time they get 20 regardless of whether or not they swap. They are ok with it, and they average 15. If someone sees what is in the envelope, they incorporate an impossible outcome. If the person sees 10 in the envelope, they think they have a 50% chance of swapping for 20 and a 50% chance of swapping for 5, but they don’t. They have a 100% chance of swapping for 20. Alternatively, if they see 20 in the envelope, they think they have a 50% chance of swapping for 40 and a 50% chance of swapping for 10, but they don’t. They have a 100% chance of swapping for 10. The equation that is given to construct the paradoxical sense of needing to swap to improve one's odds…: $$0.5( \,0.5x) \, + 0.5( \,2x) \, = 1.25x$$ … is just a red herring. It does not properly establish the one and only one true average outcome of $1.5x$. That said, I see no true paradox here. A sample space is not dependent upon someone’s frame of reference as the paradox suggests. Last edited by AplanisTophet; April 20th, 2017 at 04:08 PM.  
April 20th, 2017, 04:57 PM  #22 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,699 Thanks: 2177 Math Focus: Mainly analysis and algebra  You have two different formulations of the problem with a nontrivial difference in the definition of $x$. Your $\frac32$ is only the "one and only one true average outcome" for one of them. The calculation you took from the other thread is fine, but there $x$ represents not the minimum "win" but the amount you found in your envelope. Thus, whatever you do, the average win is $\frac32$ times the minimum "win". But, if you change every time, the average win is $\frac54$ times the amount in your envelope. The real challenge is to show that these to formulations mean the same thing. I don't suppose it's awfully difficult. 
April 20th, 2017, 05:56 PM  #23  
Senior Member Joined: Jun 2014 From: USA Posts: 238 Thanks: 10  Quote:
Does taking a peek really change the probability of what that person will get? I assume you are saying "yes" to that question, so what is this nontrivial difference in the definition of x that you speak of? Either way, one envelope has x dollars in it and another 2x dollars in it. There are two possibilities, no more and no less, and they average to 1.5x. The folly is to assume your envelope contains x and therefore there are three possibilities depending on whether you swap or not: 0.5x, x, and 2x, which you assume by thinking you have a 50% chance of swapping for 0.5x, a 100% chance of keeping x if you don't swap, and a 50% chance of swapping for 2x, when in truth you have a 100% chance of swapping for either x or 2x depending on whether your envelope contains 2x or x, respectively. Quote:
http://www.franzdietrich.net/Papers/...opeParadox.pdf It states that "the expected value of the other envelope is higher than the given amount in yours, more precisely:" $E( \,YX = x) \, = \begin{cases} \frac{11}{10}x & \text{for } x > 1 \\ 2 & \text{for } x = 1 \end{cases}$ I just gave the usual "folly" of... $0.5( \,0.5x) \, + 0.5( \,2x) \, = 1.25x$ ... in the OP as taken from the stackexchange question I linked to, which just goes to show that there are all kinds of crazy opinions out there for this. Quote:
If it's not awfully difficult, then by all means, show that the two formulations (whatever you mean by that) mean the same thing. This "paradox" (if it is one) has been around for a while, so good luck.  
April 20th, 2017, 06:16 PM  #24  
Math Team Joined: Dec 2013 From: Colombia Posts: 6,699 Thanks: 2177 Math Focus: Mainly analysis and algebra  Quote:
Quote:
No. The folly is in misinterpreting the results of the calculation. Quote:
Given that others have written papers on it, I see no reason to work it all out here.  
April 20th, 2017, 06:57 PM  #25  
Senior Member Joined: Jun 2014 From: USA Posts: 238 Thanks: 10  Quote:
That's not the same problem as the two envelopes in reality though, is it? In the two envelopes version, either 0.5x or 2x does not exist!! Incorporating a nonexistent solution is no way to set up your sample space. IMHO. Final answer.  
April 21st, 2017, 05:44 PM  #26  
Senior Member Joined: Jun 2014 From: USA Posts: 238 Thanks: 10  Quote:
... Su, Francis, et. al. have a short description of the paradox here: https://www.math.hmc.edu/funfacts/ff...0001.68.shtml I used that link because it concisely sets forth the paradox both in the basic setting but also given the version where the two envelopes contain $( \,\$2^k, \$2^{k+1}) \,$ with probability $( \,\frac{\frac{2}{3}^k}{3}) \,$ for each integer $k \geq 0$. Where the paradox is formulated by considering one person’s odds when choosing to swap an envelope, my question is whether the paradox might be resolved by considering the paradox from both swapper’s perspectives instead of just one (i.e. for one person to swap, there must be another person for the original to swap with). From a single person’s perspective, the paradoxical odds are traditionally given by the equation: $$0.5( \,0.5x) \, + 0.5( \,2x) \, = 1.25x$$ To incorporate a twoperson perspective, the equation would be one person’s odds for gain less “their opponent's” odds for gain: $$[ \,0.5( \,0.5x) \, + 0.5( \,2x) \,] \,  [ \,0.5( \,0.5x) \, + 0.5( \,2x) \,] \, = 0$$ The result is that neither person improves their odds by swapping. Paradox resolved. Comments, suggestions, agree, disagree… I’m just fishing here. Thank you?  
April 21st, 2017, 07:39 PM  #27 
Senior Member Joined: Jun 2014 From: USA Posts: 238 Thanks: 10 
Regarding the above post: In being concerned only with what you make, you must factor into account what you lose as well because what your opponent stands to make comes at your expense. Where you stand to make just as much as you stand to lose to your opponent, there is no incentive to switch. That is what the equations say to me. 
April 22nd, 2017, 05:25 AM  #28 
Senior Member Joined: Jun 2014 From: USA Posts: 238 Thanks: 10 
We can also break it down into the four possible scenarios when swapping where, when ambiguously starting with $x$, we have 25% of the time you’ll lose $0.5x$ to your opponent, 25% of the time you’ll gain $0.5x$ from your opponent, 25% of the time you’ll lose $x$ to your opponent, and 25% of the time you’ll gain $x$ from your opponent. The same is true for your opponent. $$x  0.25( \,0.5x) \, + 0.25( \,0.5x) \,  0.25( \,x) \, + 0.25( \,x) \, = x  0.25( \,0.5x) \, + 0.25( \,0.5x) \,  0.25( \,x) \, + 0.25( \,x) \,$$ $$x = x$$ In addition to what is noticed in posts 26 and 27, the above shows that there is no incentive to swap. The same general logic works for all consistent versions of the paradox that prey on the ambiguity of the variable $x$ (see the link in post 26 for more insight into that statement). I assert that the paradox is resolved. 
April 30th, 2017, 12:56 PM  #29  
Senior Member Joined: Apr 2015 From: Planet Earth Posts: 121 Thanks: 23  Quote:
The numbers you used, 1/2 in each case, are based on the Principle of Indifference. It says that when you have N random outcomes that occur under functionally identical circumstances, they each should have a probability of 1/N. So when you flip a coin, each side lands faceup under circumstances that are functionally equivalent to the other, and has probability 1/2. When you roll a sixsided die, each side has a 1/6 chance. But when you ask a 3year old girl what if her favorite color is, from among {red, orange, yellow, green, blue, purple, pink}, you'd be foolish to say each has a 1/7 chance. They don't occur under equivalent circumstances. Almost cerytainly, neither do the possibilities that the other envelope was filled with half, or twice, what you have. To see this better, consider two variations of your problem:
I haven't looked at your later approach closely, but it a similar one applied to my problems should say that the expected value of every sealed envelope in the first problem is \$11.25, and in the second problem is \$8.25. So why switch? But now suppose you open your envelope, and see \$10. In the first problem, there is a 50% probability that the other envelope contains \$5, and another 50% probability that it contains \$20. So the expected value for switching is indeed 1.25*x=\$12.50, just like the paradoxical solution you showed said. But note that these probabilities are 50% because we know how the envelopes were prepared, not because there was a 50% chance that you would pick the higher, or lower, envelope of your pair. This becomes clearer in the second problem: now there is a 90% chance that the other envelope contains \$5, and only a 10% chance that it contains \$20. So the expected value of the envelope you would switch to is only \$6.50. The point is, when you try to treat your envelope as having a certain amount of money, even if that amount is called X, then the probability you have the lower amount is not the same as the probability of having the lower amount when you ignore the amount. The latter is always 50%; the former depends on the probabilities that the pair contained (X/2,X) or (X,2X). There are envelopefilling strategies where it is almost certain that switching will provide the expectation of a gain[1] for most values you could have; but the rare losses will be such large amounts, that the expected gain is zero. There are even strategies where the first switch is guaranteed to be an expected gain, but they require an infinite amount of money. I don't know why many mathematicians don't understand this error, and so feel the need to write papers on it. +++++ [1] Fill N pairs with (X,2X), (2X, 4X), ..., and (2^N*X,2^(N+1)*X) There is a (N1)/N chance that the original "paradoxical" solution is correct.  

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