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Help with a logic puzzle

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I sometimes like to read logic puzzle books and I was reading a puzzle in one the other day that I still can not understand and it has been eating away at me. I am wondering if anyone here smarted and wiser than I can explain the solution better than the book did. The book is called "World Class Puzzles" by Erwin Brecher. The puzzle is called "Double of Half" Here is the puzzle exactly as written in the booK:


DOUBLE OR HALF

Here is an amzing paradox.
A friend hands you an unmarked envelope (E1) and tells you that it contains a number of $100 bills. You are about to open it, pleased with your friend's generosity, when he produces a second envelope (E2). He then tells you that E2 contains either haff or double the amount contained in E1. Furthermore, he gives you the choice of sticking or swapping. What would you do?
You would probably switch, arguing as follows:
Suppose E1 contains $1,000. By definition, E2 contains either $2,000 or $500. In other words, I stand to gain $1,000 dollars but risk losing $500. As the odds are equal, I should swap.
Is this line of reasoning correct?


When I read the puzzle, I can find no fault with the above reasoning and the solution confused me. Here is the solution again exactly as written from the book:




No. Surprisingly, it makes no difference whether you switch or stick. Here is the proof:
Suppose you have twenty envelopes on the table, ten of which contain $1,000 each and the other ten contain $2,000 each. This arrangement satisfies the conditions of the paradox, inasmuch as if you pick an envelope and then switch, you could either double or halve the your money. Yet, once you pick ten envelopes , whether you switch or stick, your average take will be $15,000.
The essential difference between a decision based on your reasoning and the above proof is that the envelopes are unmarked. If for instance, E1 were marked $1,000 and you would then know that E2 contained $2,000 or $500, you should switch.


I have read and reread it, and thought it over again and again and am still confused. I do not understand how knowing the amount of money in E1 affects the decision. Lets say that amount of money in E1=X. This would mean the amount of money in E2 either equals 2X or .5X. In the long run, half the time E2 equals 2X the other half it equals .5X. Now, lets do this the way book does. If I decide not to switch every time, I end up with 20X, but if I decide to switch every time, I end up with 25X. I can not figure up how my reasoning is wrong and the book's is corret. Is there anyone here who is better at these puzzles than I am? I know that is stupid to let a puzzle annoy you like this, but puzzles like this get under my skin. Can anyone help?

Yes I know, I am a nerd/geek and have no life.



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Well, from what I understand, the answer is that it makes no difference if the envelopes are switched.

e1 = x (money you might get by picking the first envelope)
e2a = 2x
e2b = x/2

Okay? So, what are the odds of each event?

There is 100% of chance that the envelope 1 contains x dollars.

There is 50% of chance that envelope 2 contains x/2 or 2x.

The universe you have is: e1-e2a and e1-e2b, 50% of chance for each.

On e1-e2a if you pick the first envelope, you lose x. If you pick the second envelope, you gain x.
On e1-e2b if you pick the first envelope, you gain x/2. If you pick the second envelope, you lose x/2.

There is 25% of each choice be made.

Pick E1, lose x
Pick E1, gain x/2

Pick E2, gain x
Pick E2, lose x/2

Hmn... Seems to me it's more logical to switch, as you can lose the less and gain more dollars if you pick correctly.

BUT since NO ONE is going to give 100 dollar bills into envelopes to someone like that, WHO CARES?!

“Voice or no voice, the people can always be brought to the bidding of the leaders. That is easy. All you have to do is tell them they are being attacked and denounce the pacifists for lack of patriotism and exposing the country to danger. It works the same in any country.” — Nazi Reich Marshal Hermann Goering
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yeah, I know this puzzle isn't important or anything, but it is eatting away at me. Ric, you have seemed to have come to the same conclusion that I did. Yet, the solution the book gives, says otherwise. I hope someone can explain why.
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you shouldnt worry about averages cause the bottem line is that you get one chance not 10 or 20, so the question is whether you want to gamble or not and that is up to you. and so it doesnt matter which is the right logic its personal preference. get it.

the whole question changes when you have more then one trial because that is when averages start to count and so by switch every time you wil average out with more money. i beleive.
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The way the solution is written, it seems to work like this: The envelopes either contain $1000 or $2000. However, they are not marked so you cannot know how much is in the envelope. You friend hands you one (E1) and then offers to switch the second (E2). If E1 has $2000 in it, then E2 has half that, $1000. If E1 has $1000 in it, then E2 has double that, $2000. But, since the envelopes are unmarked, you cannot tell which is which using logic, so it does not matter which one you take. And quite frankly, if you have a friend who has $3000 to fiddle around with in this way, just kick him in the nuts and take both envelopes. His money would be better spent by you, anyway, apparently.


Now, here's another one for you:

Three men get a hotel room to play poker all night. The day manager tells them the room is $30, so each man pays with a $10. $10 + $10 +$10 = $30. Later, the night manager comes in and goes over the books and finds that the men were overcharged. The room cost $25, not $30. So he calls a bellboy over and pulls five $1 bills out of the till and sends them up to the men with his apologies. $30 - $5 = $25. In the elevator, the bell boy realises that the men may fight over who will get shorted the dollar, so he "selflessly" puts two of the $1 bills in his wallet and returns the three $1 bills to the men. So now the men paid $10 - $1 = $9, times three men, $9 x 3 = $27, Plus the two $1 bills in the bell boy's wallet. $27 + $2 = $29. Where's the last dollar?

Do not say taxes.
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The problem is that the equation is supposed to work out to be $25 dollars.
Its a sign convention error. As far as the manager is concerned, he HAS $25 dollars.

As far as the guys are concerened they GAVE 9 dollars.

Here's the crux. The boy TOOK 2 dollars.

Therefore the equation is:
9*3 - 2 = 25.

Want to make the above trick even worse?

Have those men give their three dollars back.

That way they each paid 9*3 + 3. which adds up to thirty. But then where did the boy's two dollars come from?!?! Its the PERPETUAL MONEY MACHINE!!!
gtfo
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Let X represent the money in envelope one.
Let Y represent the money in envelope two.

Y = 2 X OR Y = X / 2

Therefore:

X = Y / 2 OR X = 2 Y


So as you can see, without knowing which envelope is which, both envelopes are equally likely to be the one containing more money.
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I think the first reasoning is correct. The second argument is flawed.

Your friend gives you an envelope containing an unknown amount "x". He says the second envelope contains either double or half the amount in the first envelope.

Your friend has not stated what the odds are, but it would seem a reasonable assumption that you have a 50/50 chance of either result.

The expected value of cash in the second envelope is 0.5*(2x) + 0.5*(x/2) = 1.25x

So if this process was repeated 10 times, and x was $1000 each time, if you switched every time you would expect to come out with $12500, as opposed to $10000 if you stuck every time.

It makes no difference what the value of x is, if you switched every time, you would in the long run always come out with 25% more than if you didn't switch.

Now, here's where the second argument falls down:
You are no longer dealing with a random value x. The problem has been limited to only 2 possible values, a and b. You have a 50% chance of picking either.

If the first envelope you pick contains a, you can't switch to get 2a or 0.5a, you can only get b.

If you pick b and switch, you can only get a.

The expected value is 0.5a + 0.5b.

In this case, as b=2a, the expected value is 0.5a + 0.5 * 2a = 1.5a. So if a=$1000 and the situation was repeated 10 times, you would expect to come out with $15000 regardless of whether you switched or not.

It is placing this limitation on the initial possibilities that changes the problem, and the final result.

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No I think jack Spencer Jr has come to a good conclusion. Let me see if I got it straight.

Let's say that the e1 has 2000 dollars. That means that the other envelope could have 1000. And what if the first envelope has 1000? The second one could have 2000. Same thing, but with different envelopes! In an infinite number of possibilities for the contents on the envelopes, it dosen't matter which one you pick, as there are equal odds of anything from 100 to, theorically, infinite dollars being in any envelope.

This reminds me of that mathematical paradox in which a contestant in a game show is asked to pick a door from three doors, two of them have nothing behind, the other has a prize. After the contestant picks a door, the game show host opens yet another door, showing it to be empty. Now we have two doors, the one the contestant picked, and the other one, still unopened. Should he change his mind and choose the other door? And why?

Surprisingly, the answer is YES he should. I'll have the answer here later.
“Voice or no voice, the people can always be brought to the bidding of the leaders. That is easy. All you have to do is tell them they are being attacked and denounce the pacifists for lack of patriotism and exposing the country to danger. It works the same in any country.” — Nazi Reich Marshal Hermann Goering
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I look at it this way:

Eliminate the logic... Your getting $500+!!!
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Yeah I'd say take $1000 dollars, what's the point in risking $500? At least to me this is still a lot of money.
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I didn't think a thread would be worth making around just this, so I'm just going to post it here.
___

A criminal is sentenced to death. He is put in a cell and told that sometime in the next week (Monday through Friday) he will be removed from his cell and hanged. BUT, when the man comes to remove him from his cell, it will be a complete surprise to him.

Obviously the criminal gets no sleep that night, so he thinks about what was said to him. After a while, the criminal is very pleased because he reasons this:

If Thursday has passed, and he has not been hanged, then he knows he will be hanged on Friday. If this is true, then it wouldn't be much of a surprise. So he reasons that he will NOT be hanged on Friday.

He then reasons that if Wednesday has passed, and he now knows that he will not be hanged on Friday, then he will have to be hanged on Thursday. But this wouldn't be a surprise either!

The criminal then concludes that he will not be hanged ANYTIME in the next 5 days because it would never be a surprise, thus conflicting with what he has been told.

The criminal gleefully sits in his cell....Monday passes, nothing. Tuesday passes, nothing. Wednesday comes, and he is taken from his cell and hanged.

Because he was so sure of his reasoning, this comes as a complete surprise to him.

___

The power of a paradox!
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thank you for you're help everyone, I think I understand the puzzle now.
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hahaha dont mind warbler he is still young, and a newly apointed member of the OT council.
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Moth3r's analysis of the original problem is exactly correct.

Also, I think that in Ricarleite's poser about the three doors, the surprising answer is actually NO.
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This is similar to a problem that economics talks about in decision making and game theory. It is called the prisoner's dilemma. I'm doing this from memory, but I think it goes like this...

Two criminals are captured after robbing a bank. The prisoners have the following choices. If neither of them talk and refuse to cooperate with the police, they will only serve one year on another minor charge. If one of them talks, and the other refuses, the police will release the one who talks on time served. The one who refuses to talk will get 20 years in jail. If both of the talk, they will both serve 5 years. What should they do?

According to the theory, whey will both talk. Each prisioner will be only interested in minimizing their own time in jail. Since they don't know what the other prisoner will do, the most logical decision will be to talk so they will serve the lowest time.
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I know believe Ric is right on both his Riddle and mine.

(on my Riddle)
Because there is no limit in E1 (due to the fact that the envelope was unmarked), the amount you make in the long run will just as much if you stick with E1 or switch. If there is a limit on what E1 is, whenever E2 is > E1, the amount in E2 will aways be greater than the amount the would be in E1 when E1 > E2.

Example: let us say that E1=1000. When E2 > E1, E2=2000. When E1 > E2, E1 = 1000. 2000 is > 1000. Therefore when there is a limit on E1, it makes sense to switch.


When there is no limit on E1(as is the case with the Riddle), the above example is not true. Therefore it does not matter wheither switch or not. So the answer to the Riddle is no. It is confusing, but no is the right answer.


(on Ric's Riddle)

The prize will only be behind the door you first chose 1/3 of the time. 2/3 of time, the prize will be behind the other two doors. The door the game show host opens will aways be a door that the prize is not behind. Therefore 2/3 of time, the door that you did not original chose and was not opened by the host, will contain the prize,and 1/3 of the time. So therefore, it would make sense to switch. The answer is yes.

Originally posted by: Shimraa
hahaha dont minf warbler he is still young, and a newly apointed member of the OT council.


what does "minf" mean?
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minf

http://www.ebaumsworld.com/forumfun/misc15.jpg

cause it wasnt in my original post.
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LOL warbler is a bunny with a pancake on his head look look.
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WOW Warbler, you look HAWT!!
gtfo