April 20th, 2017, 10:24 AM  #1 
Senior Member Joined: Jun 2014 From: USA Posts: 306 Thanks: 21  Solution to the Two Envelope Paradox Statement of the Paradox: Two envelopes exist, one containing $x$ dollars in cash and the other containing $2x$ dollars in cash. You are given one envelope and are told that you may either keep it or swap it for the other envelope. You want to maximize your cash. You have no idea how much cash is in the envelope you are given or how much is in the envelope that you may swap for. The paradox is the supposed result of the probability when swapping: 0.5(0.5x) + 0.5(2x) = 1.25x The above suggests that you should always swap envelopes, but clearly this is absurd. Assertion: You are guaranteed $x$ dollars, so each choice can be expressed as a gain of $x$ or breaking even. We have four possibilities: 1) You have 2x and keep 2x (gain x) 2) You have 2x and swap for x (break even) 3) You have x and keep x (break even) 4) You have x and swap for 2x (gain x) If you want just the probabilities when swapping, the correct equation is…: 0.5(0) + 0.5(x) = 0.5x …but this is equal to the probabilities when not swapping, which are: 0.5(x) + 0.5(0) = 0.5x. The full equation is written as [probabilities when swapping] + [probabilities when not swapping] = total probability: Probability when swapping = 0.25(x) + 0.25(0) Probability when not swapping = 0.25(0) + 0.25(x) Full Equation = 0.25(x) + 0.25(0) + 0.25(0) + 0.25(x) = 0.5x As a result, swapping has no impact on maximizing value as, no matter what you do, you have a 50/50 shot of gaining $x$ more dollars so your average gain is $0.5x$. The equation associated with swapping is often incorrectly stated as 0.5(0.5x) + 0.5(2x) = 1.25x. That is incorrect because it’s ignoring the other two options associated with not swapping. In that context (where 0.5x represents a loss and 2x a gain), we have... 0.25(2x) + 0.25(0.5x) + 0.25(2x) + 0.25(0.5x) = 1.5x ...for an average gain of 0.5x. Conclusion: As far as I can tell, there is nothing paradoxical about any of this. It appears to stem from peoples’ apparent confusion when it comes to distinguishing between two distinct things: the gain (equal to an average of 0.5x whether you choose to swap envelopes or not) and the possibility of choosing “correctly” which is 50/50. Nevertheless there are papers on this (type of) paradox. Here's one for example: The Two Envelope Paradox: An Axiomatic Approach So, as always forum, am I missing something? 
April 20th, 2017, 10:32 AM  #2 
Senior Member Joined: Sep 2015 From: CA Posts: 1,301 Thanks: 664 
what you've calculated is the expectation of the gain. There's no requirement that this total 1. 
April 20th, 2017, 10:42 AM  #3  
Senior Member Joined: Jun 2014 From: USA Posts: 306 Thanks: 21  Quote:
So yeah, I focused on the potential gain for swapping. Is that wrong?  
April 20th, 2017, 10:45 AM  #4 
Senior Member Joined: Sep 2015 From: CA Posts: 1,301 Thanks: 664  I don't know. Once I saw you calling expectation probability I stopped reading.

April 20th, 2017, 11:39 AM  #5 
Math Team Joined: Dec 2013 From: Colombia Posts: 6,876 Thanks: 2240 Math Focus: Mainly analysis and algebra  Some understanding of probability theory? Your first calculation looks wrong whatever you are trying to do. I think the 0.5x should be just x to get the expected value of swapping  or not swapping (they are equal). The fact that they are equal tells you that neither strategy is better than the other. It doesn't say that there isn't a better strategy though. Expected value is not probability. 
April 20th, 2017, 12:02 PM  #6  
Senior Member Joined: Jun 2014 From: USA Posts: 306 Thanks: 21  Quote:
https://math.stackexchange.com/quest...243872#2243872 I'm saying the first calculation is wrong, so um, try again (and don't be rude)? The paradox is that it is (supposedly) always better to swap.  
April 20th, 2017, 12:06 PM  #7 
Senior Member Joined: Aug 2012 Posts: 1,434 Thanks: 352  
April 20th, 2017, 12:11 PM  #8 
Senior Member Joined: Sep 2015 From: CA Posts: 1,301 Thanks: 664 
two envelopes with equal probability of being given to you first. let's say you don't swap $E[X] = xP[\text{given x envelope first}] +2xP[\text{given 2x envelope first}] = 0.5(x+2x) = \dfrac 3 2 x$ let's say you swap $E[X] = 2xP[\text{given x envelope first}] +xP[\text{given 2x envelope first}] = (0.5)(2x+x) = \dfrac 3 2 x$ As expected the two expectations are equal. The whole business of saying you're always going to gain at least x etc. is just a red herring. 
April 20th, 2017, 12:54 PM  #9 
Senior Member Joined: Jun 2014 From: USA Posts: 306 Thanks: 21 
The paradox lies in whether you know the value of your envelope. If you know the value of the envelope, you're inclined to always swap because if you're lucky you get 2x whereas if you aren't you get 0.5x. You stand to gain x but only lose 0.5x with 50/50 odds. If you don't know the value of your envelope, you're indifferent. That makes more sense. 
April 20th, 2017, 12:56 PM  #10  
Senior Member Joined: Sep 2015 From: CA Posts: 1,301 Thanks: 664  Quote:
If you know the value of your envelope then if it is $x$ you will swap, if it is $2x$ you will not. You will always end up with $2x$  

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