

Go to page [Previous] 1~2~3~4~5~6~7~8 [Next]  
Steve Martin Inner circle 1119 Posts 
I think you've expressed there my problem with the puzzle. I don't know the full explanation for the paradox (which I believe is pretty complex stuff) but it has always seemed to me that if you say "I hold X and therefore the other one contains either 2X or X/2" you are actually describing two different siutations 
envelope 1: X envelope 2: 2X or envelope 1: X envelope 2: X/2 whereas in reality only one of these situations exists. I think there is a counterargument to this line of thinking, but I can't remember what it is.
Any man who reads too much and uses his own brain too little falls into lazy habits of thinking.
Albert Einstein 

Daegs Inner circle USA 4283 Posts 
But still, both of these statements are true AND fit the problem exactly, though still result in different outcomes which is a paradox:
1: Your envelope contains X. By switching you either lose X/2 or gain X, so the amount you gain is greater than the amount you lose. 2: The amounts in the Envelopes are Y and 2Y, by switching you either lose Y or gain Y, so the amount you gain is the same as the amount you lose. 

Jonathan Townsend Eternal Order Ossining, NY 27143 Posts 
Maybe if divided the problem in half we might get a better answer.
you walk into one room and have a choice of two envelopes. you take that envelope into another room and have a choice of exchanging it for an envelope that is on the table there. does this make for less of a paradox?
...to all the coins I've dropped here


Daegs Inner circle USA 4283 Posts 
Depends if we know that new envelope is relational to the envelope we have.
I *believe* the problem comes from believing that we have more information about our envelope than we have, by assigning it X when we don't know if its X or 2X(or X or X/2). It's assuming that its X and then basing the other envelope off it that is the problem, and yet it seems foolish to not be able to assign a variable name to our envelopes money especially when we know its a constant. The more you think about it and come up with explanations, the more it is clear that this actually *IS* a paradox. The only way to make this not a paradox, would be to explain how we could have an X amount in our envelope and NOT state that the other envelope has either X/2 or 2X inside it, however given that we know X and know the other envelope is either half or double, that is impossible.... so its a paradox. I believe a whole new branch of mathematics would need to be invented to explain this away... I predict this will remain a paradox for at least a couple hundred years.... 

magicjohn2278 Special user Isle of Man UK 538 Posts 
Quote:
On 20060327 00:27, Daegs wrote: Wrong! Read on.... No paradox…. Just Tomas’s faulty maths! In his calculation of expected outcome, where X is the value of the envelope that you start of with: (0.5*2X)+(0.5*X/2) = 1.25X Because the expected gain 1.25X is greater than 1 we assume that this shows that it is more favourable to switch. – It doesn’t! The maths is flawed! The expression (0.5*X/2) = 0.25X is in error. It gives a positive expectation of winning on a bet giving 1 to 2 odds (and you don’t get your stake back.) That is to say, for each bet of 2 units, if you win, you only get paid 1 unit, which obviously has a negative expectation, which is the equivalent of X/2 but the negative outcome isn’t reflected in the calculation. It shows a 0.25X positive outcome. … and before you say, “Well now the calculation shows that it is less favourable to switch – so that is just another paradox!”, it doesn’t! It is just the maths that are at fault! …pleased that I cleared that one up! And as for his essay The TwoEnvelope Paradox: A Complete Analysis? David J. Chalmers Department of Philosophy University of Arizona Tucson, AZ 85721 Well! Posted: Mar 27, 2006 6:17am  The fundamental error in the calculation is saying if my envelope contains X then the other contains EITHER X/2 or 2X. If I hold the envelope with the higer amount in, then the other never contains 2X, likewise if I have the lower amount, the other never contains X/2. 

Steve Martin Inner circle 1119 Posts 
Well, as I said I above  that is what I have always had a problem with in this puzzle. But I have been told that this argument does not get to the root of the paradox.
Any man who reads too much and uses his own brain too little falls into lazy habits of thinking.
Albert Einstein 

magicjohn2278 Special user Isle of Man UK 538 Posts 
Perhaps:
Probability of making a profit by switching = Possible gain  Possible loss P=((2X)X)(X/2) = 0.5 ? 

Sergey Smirnov New user Belgium 70 Posts 
As has already been mentioned here, the paradox is resolved by noting that the probabilities of there being 2X or 0.5X in the other envelope are not 50%50%. The thing is that you may not know what they are, but it's irrelevant.


Jonathan Townsend Eternal Order Ossining, NY 27143 Posts 
Anyone here want to do the tableaux demonstration?
false( it is to your advantage to switch envelopes )  false (The other envelope has more money inside )  and break it down from there? What one expects from this tree structure are the suppositions and presuppositions which might make a "world" where the original statement in parentheses could be "true"
...to all the coins I've dropped here


Sergey Smirnov New user Belgium 70 Posts 
Quote:
On 20060325 16:13, magicjohn2278 wrote: This is exactly the reasoning that leads to the paradox. In the above situation the pobability of coin coming down tails is 50%, while with envelopes it's not the case. This apparent similarity tricks us into admiting that the paradox exists, while it does not. In the above situation, a rational player should take the bet, but this situation is not identical to the one after choosing the first envelope. 

magicjohn2278 Special user Isle of Man UK 538 Posts 
Ok going back to the "paradox" (Thanks Tomas)
Let's say there are X in the chosen envelope. Then there is 50% chance that there is 2X in the other envelope and 50% chance that there is X/2 in the other envelope. The expected value (calculated according to the definition of expected value) is 0.5 * 2X + 0.5 * X/2 = 1.25X which means that you are better off switching and taking the other envelope. This can't be true so the puzzle is to find the error. That's "all". /Tomas Bad Maths! The Expected Value has to be your expected winnings LESS your expected losses. This is fine in the first half of the calculation, you start with X and end with 2X. But the second half is wrong. You are "winning" X/2 for a stake of 1X, which you don't get back, so the expected value of this bet is your "winnings" X/2 less your loss of X/2.  and let's face it if you bet 1 unit at 1:2 odds often enough, eventually you will end up with nothing! So the Expected Value is: (0.5 * 2X) + (0.5 *(X/2X/2)) = 1 How's that? 

Steve Martin Inner circle 1119 Posts 
Sergey,
You say that if I hold an envelope with X dollars, it is not a 5050 bet that the other envelope contains less or more than X dollars. Why do you say that? You haven't explained why you think that. I don't agree with you on that point (unless you are referring to the consideration in my footnote 1 below). Here is what I think: Let's say the envelope you are holding contains 20 dollars. Then there are two possible scenarios: 1. One envelope contains 20 and the other contains 10 2. One envelope contains 20 and the other contains 40 That cannot be disputed, as it is stated in the original question. The thing is, you don't know which scenario actually applies. IF WE ASSUME (*** see footnote 1) that each scenario is equally likely then without question, it is equally as likely that the other envelope (the one you don't hold) contains 40, as it is that it contains 10. Therefore, if you swap, you stand to either gain 20 (going from 20 to 40) or lose 10 (going from 20 to 10) with equal probability. If we define a good bet as being one in which you stand to gain more (if you win the bet) than you lose (if you lose the bet), then clearly in this case it is a good bet to swap envelopes. So you swap, and now hold the other envelope. You know that this contains either 10 or 40, and that both are equally likely (which is true). And you know that the other envelope contains 20. You are now invited to swap again. If you swap, you stand to either gain 10 (going from 10 to 20) or lose 20 (going from 40 to 20) with equal probability. In swapping, you would therefore stand to lose more than you would win, so clearly it is a good bet to KEEP the envelope you now have. So we know that the best thing to do is to choose one envelope and then swap ONCE. But because it is just as likely that you would initially choose one envelope as the other, this strategy could leave you holding EITHER envelope after the swap (+++ see footnote 2). Therefore, what we are really saying is that it is no better to choose one envelope or the other, and no better (given the opportunity) to swap or not. I believe this solves the apparent paradox (at least, in nonmathematical language). ==== *** footnote 1: As we know that there is not an unlimited amount of money in the world, then clearly it is not strictly speaking always true to say that it is equally likely that the other envelope contains more money or less money than the one you hold. e.g. If you know the one you hold contains $1000,000,000,000,000,000,000 (or some appropriately huge sum) this may affect the probability that the other contains twice that amount. +++ footnote 2: Let me clarify that point: Let's say the envelope you initially choose contains 10 dollars. BECAUSE YOU DO NOT KNOW WHAT THE AMOUNTS INVOLVED ARE, there are two possible scenarios: 1. One envelope contains 10 and the other contains 5 2. One envelope contains 10 and the other contains 20 ... and therefore all the above argument applies equally as well to this situation, as it does to the 10, 20, 40 situation. In the above, I have used the figures 20, 40 and 10 for clarity. We could just as easily use X, 2X and X/2. The argument holds for any value of X.
Any man who reads too much and uses his own brain too little falls into lazy habits of thinking.
Albert Einstein 

magicjohn2278 Special user Isle of Man UK 538 Posts 
Steve the problem with defining the flaw in the logic is trying to define everything in relation to the envelope you have (X).
In the cases where you swop, in half of them X is the "bigger value" and in the other half, X is the lesser. the only time you win, is when X is the lesser, and you win another X. on the other hand, you lose only in the cases where X is the greater value, and here you only lose "half of X"  but in these cases X is worth double what it was in the first case, so you are actually losing exactly the same! ...but try expressing that as a mathematical equasion! 

Jonathan Townsend Eternal Order Ossining, NY 27143 Posts 
For some reason I've bee having images of Shrodinger's cat hiding in one envelope and Alice's Cheshire Cat hiding in the other.
Is that about right?
...to all the coins I've dropped here


magicjohn2278 Special user Isle of Man UK 538 Posts 
... no, there is a male dog in one and a dog of indeterminate sex in the other!


Daegs Inner circle USA 4283 Posts 
All that is fine and dandy, but please explain these two statements then:
1: Your envelope contains X. By switching you either lose X/2 or gain X, so the amount you gain is greater than the amount you lose. 2: The amounts in the Envelopes are Y and 2Y, by switching you either lose Y or gain Y, so the amount you gain is the same as the amount you lose. They both have true premises, yet have contradictory results.... a paradox. 

Steve Martin Inner circle 1119 Posts 
Daegs,
Well, exactly  that is indeed the paradox. My post above is an attempt to explain it in terms of what is really going on in the process of selecting and swapping envelopes. It demonstrates that, despite the apparent paradox, there is no advantage to choosing either envelope, or choosing to swap or not (which, of course, we all know instinctively anyway). But as I say, it is a nonmathematical explanation. I believe the mathematical explanation goes rather deeper  way beyond our considerations (so far) of X's and Y's.
Any man who reads too much and uses his own brain too little falls into lazy habits of thinking.
Albert Einstein 

magicjohn2278 Special user Isle of Man UK 538 Posts 
Daegs,
Statement 2 is almost correct. 2: The amounts in the Envelopes are Y and 2Y, by switching the amount that you GAIN would be: (Y2Y)=Y or (2YY)=Y You already have one envelope containing either Y or 2Y and have to give it away to get the other. If you give 2Y your "gain" is 1 so you have 1Y ... if you give 1Y your gain is 1Y so you have 2Y. (As you said, the amount you gain is equal to the amount that you lose.) Statement 1 is only half way there. 1: Your envelope contains X. By switching you either lose X/2 or gain X, so the amount you gain is greater than the amount you lose. Wrong! The amount you gain (profit) is the same as the amount you lose! If you switch… It costs the contents of your envelope to play –1X Half the time you win…. How much? Your winnings are 2X… but… But you still owe the contents of YOUR envelope to play (1X) … so your expected profit is 1X If you switch… It costs your envelope to play –1X Half the time you lose…. How much? Your “winnings” are half of X … but….. If you have 1X and win half of X, you have also LOST half of X. (You started with 1X and now have X/2 – winning half of X AND losing half of X is the same as not winning or losing anything… AND you haven’t yet paid to play! (1X) …. so your expected profit = X/2 (win)X/2 (loss)  1X (bet) = 1X So the expected gain by switching is 0 ! 

Steve Martin Inner circle 1119 Posts 
John  your maths is all over the place
If I give you £1 to keep and tell you you can then swap it for the coin in my hand (which you have not seen) which I tell you is either £2 or 50p... If you swap and I had a 50p coin, you LOSE 50p. If you swap and I had a £2 coin, you GAIN £1. As each of the above is equally likely, your EXPECTED gain in pence is (100 + (50)) /2 = 25p This is not the same as an expected gain of zero, as you claim.
Any man who reads too much and uses his own brain too little falls into lazy habits of thinking.
Albert Einstein 

Daegs Inner circle USA 4283 Posts 
Ok... so let's say there are 3 envelopes, $50, $100 and $150.(this obviously is a worse deal than $50, $100, $200).
Would not the offer to switch here (to $150 or $50) be even? By your own formula's: $150  $0  $100 = $50 $50  $50  $100 = $100 Using your formula that means that a $100 to $50 or $150 switch is a actually unfavorable, but its simple to see that: If you take $100 away, you start with 0 and have $50 and $50. We all know if you flip a coin for even money, you'll come out even in the long run, but by your math it says your a loser.... I have to disagree with what you've posted, it is illogical! 

The Magic Cafe Forum Index » » Puzzle me this... » » Two envelopes (0 Likes)  
Go to page [Previous] 1~2~3~4~5~6~7~8 [Next] 
[ Top of Page ] 
All content & postings Copyright © 20012021 Steve Brooks. All Rights Reserved. This page was created in 0.2 seconds requiring 5 database queries. 
The views and comments expressed on The Magic Café are not necessarily those of The Magic Café, Steve Brooks, or Steve Brooks Magic. > Privacy Statement < 