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Logic/Math Puzzle

D

desteffy

Golden Member
Suppose I have two envelopes full of money and let you choose one. You open your envelope and find X dollars. Then I tell you that one envelope contained exactly twice as much money as the other. You are allowed to either keep the X dollars you have, or switch and keep the money from the second envelope.

It seems switching to the second envelope should be the best bet since it either doubles or halves your current amount of money (you will get 2X or X/2 dollars with equal probability). However your initial choice was arbitrary, so making a switch should not increase your expected outcome. What is wrong?


EDIT: for some good correct solutions see the links people posted below.


 
I don't get it. Switching to the other envelope is just a gamble...it could cause you to lose or gain money. And I'm assuming you blindly picked an envelope. So what's the question?

What if X=0? 😉

err...now i'm just confused
 
Originally posted by: desteffy
Suppose I have two envelopes full of money and let you choose one. You open your envelope and find X dollars. Then I tell you that one envelope contained exactly twice as much money as the other. You are allowed to either keep the X dollars you have, or switch and keep the money from the second envelope.

It seems switching to the second envelope should be the best bet since it either doubles or halves your current amount of money (you will get 2X or X/2 dollars with equal probability). However your initial choice was arbitrary, so making a switch should not increase your expected outcome. What is wrong?
Click to expand...

If the envelope you first picked had an odd number of $$, then the second envelope will be twice as much.

Only problem is that you didn't state if all monies were paper, so both could have coins thus making my argument null.
 
It never mentions anything about paper money or coins...it just says "money", so it could be a check, a gift certificate from Circuit City, or even a debit card with a positive balance. Who knows?
 
Originally posted by: AnonymouseUser
Originally posted by: desteffy
Suppose I have two envelopes full of money and let you choose one. You open your envelope and find X dollars. Then I tell you that one envelope contained exactly twice as much money as the other. You are allowed to either keep the X dollars you have, or switch and keep the money from the second envelope.

It seems switching to the second envelope should be the best bet since it either doubles or halves your current amount of money (you will get 2X or X/2 dollars with equal probability). However your initial choice was arbitrary, so making a switch should not increase your expected outcome. What is wrong?
Click to expand...

If the envelope you first picked had an odd number of $$, then the second envelope will be twice as much.

Only problem is that you didn't state if all monies were paper, so both could have coins thus making my argument null.
Click to expand...

It's assumed that you can't inspect the envelope. Why not pick the fatter of the two envelopes? Why not pick the skinner one (larger bills)? How you picked it doesn't matter. It only matters that a choice was made.

I don't think it matters if you switch or don't switch. Go with your gut, not your head. You gut has more nerve endings than in your head.
 
This is not the monty hall problem, this one is harder.

Think about a "double or nothing" bet, if you bet X dollars you could either get 2X or 0 back, and that is a "fair bet", if you do it over time you will come out even.

But this you could get either 2X or X/2, which is better than the double or nothing bet.


If you dont understand the solution to monty hall problem or why .9999.... = 1 you probably wont even understand what the question is here...
 
This sounds like they're going for a Monte Hall type situation, but the way it's worded doesn't allow the same type of math.

In the Monte Hall problem, you're initially faced with a 1/3 chance, then he shows you an empty window and asks if you want to change your guess. You should, since now you'll have a 1/2 chance of being right, as opposed to 1/3 before.

But in this problem, you start with a 1/2 chance of it being the 2x amount, and you still have a 1/2 chance after he asks if you want to change your guess. I can see people trying to argue that you didn't know what your chances were when he asked you the first time, but now you do know the chances. But that doesn't change the probability since you'd always end up with a 50/50 shot.
 
Originally posted by: desteffy
This is not the monty hall problem, this one is harder.

Think about a "double or nothing" bet, if you bet X dollars you could either get 2X or 0 back, and that is a "fair bet", if you do it over time you will come out even.

But this you could get either 2X or X/2, which is better than the double or nothing bet.


If you dont understand the solution to monty hall problem or why .9999.... = 1 you probably wont even understand what the question is here...
Click to expand...


I know that Einstien. I thought your problem may have a similar solution. But thanks for being condescending.
 
Man I'm dumb at statistics. It makes sense in that using A/2 and 2A in the same equation doesn't work, I think, but the rest lost me.
 
Originally posted by: desteffy
This is not the monty hall problem, this one is harder.

Think about a "double or nothing" bet, if you bet X dollars you could either get 2X or 0 back, and that is a "fair bet", if you do it over time you will come out even.

But this you could get either 2X or X/2, which is better than the double or nothing bet.


If you dont understand the solution to monty hall problem or why .9999.... = 1 you probably wont even understand what the question is here...
Click to expand...

That's about the simplest explanation I've seen... not bad!

Another way to think of it is this: You've got $20 in your hand. You're offered the opportunity to bet $10. If you win, you'll actually win 20 more. If you lose, you're only out 10. 50/50 chance of winning/losing.
 
This is kinda just sad that people here are so dense they can't even see far enough to see the supposedly obvious paradox, let alone the solution. The paradox of course is that the expected value of switching is higher than what you currently have, but logically this does not make since the envelope you ended up with is completely random.

EDIT: on another note, I think its funny how people here just post random paradoxes off of wiki's list of them and the laugh as dozens of people fall for them completely and then are to prideful to admit they are wrong and instead argue the semantics of the problems even though most like this one are decades old and very well formed.
 
You should always switch, here's why.

Assume your envelope has $10. This means that the other has either $5 or $20.

If you always keep your envelope, you'll end up with $10 on average.

If you switch, assuming a fair game, youll average (5+20)/2, 12.50 each time on average.



 
Originally posted by: Praxis1452
it doesn't matter. That's the answer. it's been explained in the links.
Click to expand...

heh, well if everyone believed this than the .999=1 and airplane treadmill threads wouldn't have lasted past the first page.
 
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