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Steve Martin Inner circle 1119 Posts 
A good friend of mine (a mathematics/logic professional) presented this puzzle to me about ten years ago. It is a wellknown paradox. It may have been discussed here before  I can't remember. So here it is again:
There are two identical envelopes lying on the table. You are told that they each contain a sum of money, and that one of them contains exactly twice as much money as the other. You cannot tell from the appearance or weight of the envelopes (or by any means at all) how much money is involved or which contains the greater sum. You are invited to select one of the envelopes and keep the contents. You pick up one of the envelopes. Before you open it, you are asked if you would like to swap the envelope you hold for the one on the table. What is the best course of action for you? Do you keep what you have, or do you swap?
Any man who reads too much and uses his own brain too little falls into lazy habits of thinking.
Albert Einstein 

TomasB Inner circle Sweden 1143 Posts 
I do know the reasoning that causes the discussion, so I guess I can present it here.
Let's say the envelope you hold has X dollars in it. That means that the other envelope has 2X dollars or X/2 dollars in it with equal probability. Your expected winnings of switching would be 0.5*2X + 0.5*X/2 = 1.25X while if you keep your envelope your expected winnings is X. So you are better off switching. That of course makes no sense, so what is the error in the reasoning? /Tomas 

magicjohn2278 Special user Isle of Man UK 536 Posts 
It just shows that probability sucks!
You should reason that the average amount in the envelope that you hold is (1X+2X)/2 = 1.5X  as the probability is that the other envelope contains 1.25X then you would be better off keeping your envelope! 

Nir Dahan Inner circle Munich, Germany 1390 Posts 
Quote:
On 20060316 02:41, TomasB wrote: this is a very deep paradox! there are many explanations on the web for it. the one I am sticking with is the fact that there is no PDF (probability districution function) that supports the above condition. this is just on one leg, I will try to find a better explanation phrased by another. nir 

TomasB Inner circle Sweden 1143 Posts 
For me the error is in saying that the expected value of the one you hold is X and yet use that as a 100% fact that it contains X. You can't use X in both ways since the expectation of what you hold is not the same as what's actually in the envelope. (Throw and cover a die before looking at it. The expected value is 3.5 but I assure you that it is not showing that.)
/Tomas 

Greg Arce Inner circle 6173 Posts 
I would point suddenly behind my friend and shout, "What is that!!!!?" As he turns around, I'd pick up the other envelope and run off with both.
Greg
One of my favorite quotes: "A critic is a legless man who teaches running."


magicjohn2278 Special user Isle of Man UK 536 Posts 
Quote:
On 20060315 19:04, Steve Martin wrote: The puzzle becomes much more interesting if you are allowed to open your envelope before deciding whether to switch. You know one envelope contains twice as much as the other. You open your envelope and see $20  now you know that the other envelope contains either $10 or $40, so if you switch you either lose $10 or gain $30. So obviously the odds are in your favour if you switch.  Regardless of which envelope you are holding??? 

TomasB Inner circle Sweden 1143 Posts 
That's a wonderful and a bit annoying idea, John. So if you switch, your expected gain is 0.5*200.5*10=5 dollars since you know with 100% certainty what amount you hold at the moment. That reasoning would always tell you to switch and I can't really see where the error in the reasoning is but there of course has to be one.
/Tomas 

landmark Inner circle within a triangle 4763 Posts 
Well, here's some infobut my math education isn't good enough to make a whole lot of sense out of it:
http://consc.net/papers/envelope.html Is it possible to simplify this explanation? My limited understanding of this article is this: if the amount of money in each envelope is chosen from a restricted range, then it is not necesarily true that the chance of getting the higher amount is equal to the chances of getting the lower amount. For example, if we knew the amount of money in the envelope was chosen randomly from the numbers between 0 and 1000, then if one envelope held an amount around 100, then there is actually a greater chance that the other envelope has a number around 200 than a number around 50. Hence, no paradox. This much I think I understand. But I don't get what happens when the range is not restricted. The argument in the article seems to imply that, once again, it is not necessarily true that the chances of expecting the higher number is the same as expecting the lower number. At least I think that's what the article is saying. But I don't understand the reasoning here. Anyone here able to write Probability for Dummies? (BTW I had the television on in the background when I was writing this. The moment I typed the word Probability, a person on the television simultaneously said Probability! What's the probability of that? Interesting to try to guesstimate) Jack Shalom
Gerald Deutsch's Perverse Magic: The First Sixteen Years available now.
All proceeds to nonprofit charity Open Heart Magic. You can read my daily blog at Musings, Memories, and Magic 

Nir Dahan Inner circle Munich, Germany 1390 Posts 
Quote:
On 20060322 20:43, landmark wrote: that is exactly what I meant by saying that the PDF has to be fixed before. N. 

TomasB Inner circle Sweden 1143 Posts 
It sounds like the perfect problem to simulate. I do however think a simulation would prove them wrong. According to that statement you _should_ switch when you see 100 in the envelope, right? I don't have time to make any simulations now but if someone does, it could look like this:
1. Chose a number from 0 to 1000 with equal probability and store in A. 2. Put half of that number in B. 3. Flip a cybercoin to decide if you look in A or B. 4. Upgrade a counter the times you see 100 and upgrade another counter if at the same time there is 200 in the other envelope. 5. Repeat millions of times and check the two counters. If the second counter is more than half the first counter we know that it's a higher probability to see 200 in the other envelope. /Tomas 

magicjohn2278 Special user Isle of Man UK 536 Posts 
10127447 trials gave 10149 envelopes with 100 in, and in 5069 of these cases the other envelope contained the 200.
(Not sure how good the random number generator is though!) _________________________________________________________________ I think the solution to this problem lies in the fact that you can't lose  you haven't paid anything to play. If we change the deal to "After you have picked and exchanged envelopes, if you wish, you keep the envelope you are holding, but pay me the value of the lesser envelope", then it is pointless switching....? Or... If one envelope is empty, then it is pointless switching (Which is the same thing really.) I think the mathematical answer is going to be something on the lines of: You have to assume that at the end of the game you should pay the ACTUAL VALUE for an envelope  in fairness the ACTUAL VALUE is the value of the total contents of the envelopes divided by 2. Even when you know what your envelope contains ($20), you don't know the ACTUAL VALUE of the envelope that you hold. Your envelope contains $20. If the other envelope contains $10 then the ACTUAL VALUE of each envelope would be $15  stay with what you have and you are $5 up, switch and you are $5 down  a 50/50 bet. Your envelope contains $20. If the other envelope contains $40 then the ACTUAL VALUE of each envelope would be $30  stay with what you have and you are $10 down, switch and you are $10 up a 50/50 bet. Nevertheless, in the situation given, it still seems advantageous to switch? 

TomasB Inner circle Sweden 1143 Posts 
Quote:
On 20060323 08:09, magicjohn2278 wrote: So you would win by switching those times and it's only 0.499 of the times according to the simulation. IF the reasoning in the article above was correct it should have been noticably larger than 0.5 but it isn't. My guess is that they were wrong. Thanks for doing the simulation, John! /Tomas 

Jonathan Townsend Eternal Order Ossining, NY 27032 Posts 
Adding some conditions to the original might make it paradoxical. As it stands, regardless of which you pick, or even if the person offering can tell which is which, they make the offer, hence nothing to compute. Just take one, and of course remember to say thank you before opening your gift envelope.
Add in one or both conditions and you get something else.
...to all the coins I've dropped here


magicjohn2278 Special user Isle of Man UK 536 Posts 
... so why is the expression "one contains TWICE as much as the other" so significant?
If you are told one envelope contains MORE than the other, then switching looks like a 50/50 proposition. If you were told that one envelope contains 1.5 times the other, then it's a bad bet. If you are told one contains twice the amount or three times the amount, (or more) than the other, it becomes a good bet. Why? If you are told that one envelope contains MUCH MORE than the other, then you would be wary of switching, as you stand to lose quite a lot if you already hold the winning one, but the situation is very much the same? (by MUCH MORE, then I think we could infer that one contains more than twice as much as the other.) What's going on? 

Jonathan Townsend Eternal Order Ossining, NY 27032 Posts 
Simplification:
You are given permission to walk into a room and pick up one of two envelopes on a table. You pick up one envelope and hear a speaker in the room announce: If it's Tuesday, the other envelope contains more money And then you... Correct me if I'm missing something here, it seems to me that the added bit about either being offered or forced to switch envelopes is impertinant. Till you have some data to work from, there is nothing upon which to base an analysis. Is there some data I've overlooked in this analysis?
...to all the coins I've dropped here


magicjohn2278 Special user Isle of Man UK 536 Posts 
You are told that one envelope contains twice as much as the other, then if your envelope contains $20, then the other contains either $10 or $40.
Switching appears to be a $10 bet at 3 to 1 odds  switch and risk losing $10 or gaining $20. ... But having done so, you are in exactly the same position... 

Daegs Inner circle USA 4283 Posts 
Using my own logic:
2 identical envelopes, 1/2 of getting higher money. knowing what is inside one of the envelopes doesn't affect it since you have no idea if its low or high and that doesn't change the contents of the envelopes... you are still left with 1/2 of getting either. Anything you add to it is going to be wrong imho.... 

Jonathan Townsend Eternal Order Ossining, NY 27032 Posts 
If your analysis leads you to a paradox
At least one of our axioms or premises must be faulty. So pick them out one at a time And find out which lead you into the trap. Above, I removed the "let x be the dollars in the envelpe in my hand" and replaced it with "at the end, when I open my envelope and enjoy the money, I will not know whether the other one contains half or twice as much". Ie limited information. That seemed to greatly simify the analysis and remove the paradox.
...to all the coins I've dropped here


Daegs Inner circle USA 4283 Posts 
To Simplify and Expand:
What we have is 2(X) envelopes with different amounts of money, we pick one, are shown the contents and then offered to switch. Removing the double, Let's say $1 and $100 (but we don't know if they are 50 cents or $1000 in other) Let's say we always switch: 1/2 we pick $1, see it and switch> WIN  $100 1/2 We pick $100, see it and switch> LOSE  $1 Let's say we always stay: 1/2 we pick $1, see it and stay> LOSE  $1 1/2 we pick $100, see it and switch>WIN  $100 So 2 out of 4 we lose, 2 out of 4 we win. Extrapolated: We have 100 envelopes with varying amounts of money, we pick one, are shown the contents, and are given the option to switch to another single envelope.... If we know they are in a certain range than obviously if you are under average of total you should switch.... however because whether the envelope is half or double(and in this case we have no idea of its value in relation to others), then you have the same chance of picking highest on your first 1/100 try than your second. This means that knowing the value of your envelope is MEANINGLESS to the problem, since you don't know the total range. IE you could already have the highest with a $5 if the rest are $1's, or you could have the highest with $1 if the rest are 50 cent pieces.... If any of this is doubted just go through it again above, where it is seen its always 1/2 chance of winning or losing. Adding anymore math or logic to a solved problem just creates problems. 

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