Which envelope would you choose? How much will you win?

Question

There are two sealed envelopes-one contains twice as much money as the other. You open one and find $100. Should you keep it or trade it for the other envelope?

-j's math answer is nice, but heres a curious thing: Before you open the first envelope, you know that whatever amount you find, your reasoning will be the same, so the rational thing to do is to trade your envelope for the other one immediately, without even bothering to open it. BUT, had you originally chosen the other one, then, by the same reasoning, you should trade it for the one you are now holding. Which is it?

-j
There is no slip up in my logic. If you choose an envelope with N dollars in it, then the other one has either 2N or N/2 dollars in it. The expected value of the other one is therefore .5*(2N) + .5*(N/2) = 1.25N. The expected value of the one you have is N. Therefore you should switch. But the same logic could apply if you picked the other one, which would lead you to choose the one you now have. How do you resolve the paradox?

Answer 1

I would keep the hundred bucks because I am broke so $100 is way good money!!!

Answer 2

Which Envelope WOULD I choose: THE OTHER
How much will I win: EITHER $50 OR $200 because I am choosing the other which WILL BE EITHER $50 or $200
Firstly, the probability formula applications proposed here are TOTALLY - that is 100% - WRONG. This is an ALGEBRA question NOT A PROBABILITY QUESTION! You have a KNOWN variable which is $100. YOU HAVE TWO EQUATIONS to work with:
EITHER : T=1/x
OR: T=2x
Where x is the amount in the OPENED ENVELOPE AND, T is the TOTAL which could be had based on the given variable x ONCE x is known. Once an envelope is opened one KNOWS by the offer that the other will either be twice as much or half as much. The correct algebraic conclusion would be met with two formulae: 1/2 x = y
x = 2y
Substitution proves the formula as y=y AND x=x
therefore the formulae are correct.

x is the amount in one envelope and y is the amount in the other. Once you open one envelope, you simply plug in the number and find the two possible amounts in the other. in this case 100.
IF x = 100 then y = 50 & IF y = 100 then x = 200
If you are wondering WHY I chose: THE OTHER, it is because I wouldn't miss $100 so I would go for the max. If it were $1000000, I would keep it because $2000000 is NOT as great an increase as $500000 is a decrease of value RELATIVE to my wants.

Answer 3

If you trade it, you have a 50% chance of winning $50, and a 50% chance of winning $200. Statistically, your expected winnings are

(0.5) * $50 + (0.5) * $200 = 25 + 100 = $125. The expected winnings of staying put are $100, obviously. So you should trade envelopes.

ADDED in response to your details: I think you've made a slip-up in logic. If you have an envelope, but don't know what's in it, you have no idea what your expected winnings are, so it doesn't matter which envelope you take. If you know one envelope has $100, but you don't know which one, your expected winnings are

(1/3) * 100 + (1/3) * 50 + (1/3) * 200 = $116.67, no matter which envelope you have.

Answer 4

I would KEEP it. By keeping it I am assured to leave with 100 bucks. There is a 50-50 chance that the other envelope contains ONLY 50 bucks

Answer 5

I would trade it in. Either way you are going to get more money than you started out with. Whether it is $50 or $200.

Answer 6

trade it in.

You could possibly make 100 more dollars, which is double the 50 you'd be losing. with a 50:50 chance, it's worth taking it.

Answer 7

keep it cuz the other one can be $ 50 but it could also be $200, but $100 is nice.

Answer 8

well id keep it. double 100 is only 200. if i traded it, might only get 50 bucks.

Answer 9

trade it what would you really lose you didn't have the 100. to begin with

Answer 10

I'd keep it, but I'm not a gambler....$100 would come in handy right about now...