/sci/ - Science & Math

[Anonymous No.3425156]

http://en.wikipedia.org/wiki/Two_envelopes_problem

Great mindfuck question. If you're unfamiliar with it, try figuring out the flaw in the reasoning yourself. Here's the gist of it:

You are given two indistinguishable envelopes, each of which contains a positive sum of money. One envelope contains twice as much as the other. You may pick one envelope and keep whatever amount it contains. You pick one envelope at random but before you open it you are offered the possibility to take the other envelope instead.

At this point, you think:
1. This envelope contains some amount of money, X.
2. The amount in the other envelope is either 2X (with a 50% chance) or X/2 (with a 50% chance).
3. So the expected value of the other envelope is (2X)(1/2) + (X/2)(1/2) = 5X/4. This is greater than X, so you switch to the other envelope.
4. Using the same reasoning, you immediately switch back.
5. And on and on forever.

Good one, eh? Yes I know it's all there in the wikipedia page, but I just ran across this for the first time and thought /sci/ might like it.

[Anonymous No.3425191]

So basically, the other envelope always looks more appealing according to this mathematical approach because, while the chance of doubling your unknown amount of money is the same as halving it, doubling would yield a greater difference from the original unknown value?

Cool story bro. The math is bogus and doesn't apply to real life, though. In reality, switching to the other envelope does not increase your chances of getting more money.

[Anonymous No.3425247]

Easy. Weigh the envolope. Swap, weigh the other one. Whichever one is heavier, chose that one. Problem?

[Anonymous No.3425251]

>>3425247
They're cheques.

[Anonymous No.3425253]

>>3425156

But the X's aren't the same. In one case, the "X" is 1/2 the other "X".

[Anonymous No.3425262]

But wouldn't memory of the initial act of changing your envelope to acquire the higher probability of getting the larger amount of money make the changing back invalid?
It's similar to the Monty Hall problem, except that one actually makes sense.

[Anonymous No.3425269]

>>3425247

a $50 bill and a $100 bill both weigh a gram

[Anonymous No.3425270]

your chances are the same. pick an envelope and stick with it.

[Anonymous No.3425276]

>>3425270

but if you used the same logic as the Monty Hall problem, wouldn't it always be best to switch?

[Anonymous No.3425281]

haven't read the wikipedia article
but as OP has written it it's no paradox at all

you start with 2 envelops one containing more money than the other
at the start you have 50% chance at picking the envelope which contains more money

then you're given the choice to switch envelops
the money contained in the envelops is still the same amount
and again the probability of getting more money is 50%
so it would be just a waste of time to switch again...

[Anonymous No.3425285]

>>3425276

No, because you have the act of choice changing the probability of the actual outcome.

[Anonymous No.3425291]

>>3425276
no, because no other choice falls down
which means that the probability of wining doesn't change when swapping the envelope

[Anonymous No.3425298]

>>3425276
the monty hall problem you should switch because of the probabilities of the door being a "good" one - the prob. of the "other" one going up, and of yours not, when the 3rd door is open

you just have 2 envelopes. ain't no same

[ehrenheilman No.3425310]

so uv basically just proven that there is an even chance of either envelope containing more money? that is the paradox for any equal outcomes. Not that mind bending...

Try this one out:
If you were to shuffle a deck of tarot cards and ask "Will this outcome be true" and it comes up false... is it false or true?

A rehash of:
The below statement is false.
The above statement is true

[Anonymous No.3425829]

>3. So the expected value of the other envelope is (2X)(1/2) + (X/2)(1/2) = 5X/4.
The expected value predicts an average value over many trials of a random variable. In this finite discrete case, this prediction has no meaning, because the actual value will never converge to this average.
>This is greater than X, so you switch to the other envelope.
So this bit is a bad idea.
This is an error in semantics, not very different from confusing the use of the word "theory" in layman's terms with that in scientific terms.

[None !/oddSpacew No.3425830]

The problem is that you don't start with an envelope containing x, I think. You either have one containing X/2 in which case a switch would give 300% more or one containing 2x in which a switch would give 75% less.

[Anonymous No.3425845]

>>3425156
>1. This envelope contains some amount of money, X.

Here's your problem. The envelope doesn't contain X, it contains either X or 2X. You're assigning the same variable to two different values, and therefore getting nonsensical results.

[Anonymous No.3427123]

MATH!

[Anonymous No.3427152]

There are two envelopes. One contains X dollars. One contains 2X dollars.

You pick an envelope.
Case 1: You have the one with X dollars. You gain X dollars by switching.
Case 2: You have the one with 2X dollars. You lose X dollars by switching.

Your expected gain/loss is 1/2(X) + (1/2)(-X) = 0

[Anonymous No.3427165]

>>3427123
this made my brain hurt for a second.

>>3425156
op, this is retarded. it assumes a lot, but most of all it assumes the person is a gambler.