Envelope paradox

How's that a paradox? is retarded.

Because it seems like switching is better

Of course switching is better. How is that a paradox?

The paradox is that switching is not better

about tree fiddy

It is better. If I give you 2 envelopes, you open one and find that it contains $10, and I tell you the other one contains $1,000,000, how is it a paradox that switching is better?

Thats not what the question says

The other envelope doesn't contain $1000000?

It's the exact same scenario as the original question. How is it a paradox that switching is better?

What if you don't open the first envelope, is switching still better?

give me both envelopes and no one gets hurt

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It's not the same scenario. Where does the question say that the other envelope contains $1000000?

The question baits you into thinking you should always switch, but the real answer is that the problem statement is nonsensical.

You want the average value, but you can't use the specific number $10 to compute the average. If the first envelope you picked always had $10, then that's a contradiction because if you picked it at random, and they don't have the same contents.

The problem doesn't say what the distribution of cash amounts is inside the envelopes.
Do they always put a $10 and a $20? Or is it always 5/10? Or random amounts between 1 and 100 every time?

If you don't know that, you can't tell what your average gain will be if you always switch.

If (You) don't understand that, you may be a brainlet.

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So the other envelope is 50/50 for $5 or $20

So 0.5 * 5 + 0.5 * 20 = $15.

That means you should switch as your expected value is better than $10.

It is pretty much the exact same scenario, except switching $20 for $1,000,000. How is it a paradox that switching after obtaining knowledge gives a better expected return?

It's always better to switch due to exponential growth.If the first envelop has $10 and the second one has only $5, you are only losing 5 dollars by switch.

If it is $20 in the second envelope, you gained $10, twice as much as the potential loss you faced (only 5).

Do your maths again

0.5 * 5 = 2.5

>0.5 * 5 + 0.5 * 20 = $15.
I'm sorry user, you failed 5yo math.

20/2 + 5/2 != 10 + 5

Consider

You only know that one envelope has two times the money the other one has, but you don't know which one. So the other one could contain either 20$ or 5$. Nowhere is stated that the unopened envelope contains 20$.

That's not the same scenario.

Ok, now instead of one envelope containing 2x the money, make it so one envelope contains 1000x the money. How is it any different?

Why does seeing the $10 make switching better? How does that give you any information? What dollar amount would you need to see to make switching not better?

The math changes when you already have the value out of one envelope and are choosing whether to switch or not.

If you don't open the envelope, you are still choosing 1 out of 2 choices both unknown. Your potential loss is 0 because you have 0 in that moment.

Other envelope either contains $5 or $20.

-5 * .50 + 20 * .5 = $7.5 expected value if you switch. So you're better off holding onto the $10 envelope.

Then it's not different. But in that case switching is still not better

checked
and the right answer

The question is not nonsensical. Out of 2 envelopes. Given that the envelope you have contains $10, what is the expected value of switching. This is a conditional probability

>Why does seeing the $10 make switching better?
It gives you more information.
>How does that give you any information?
It gives you the expected values of switching and not switching.
>What dollar amount would you need to see to make switching not better?
$0, or anything negative if it makes sense.

What if I it's given that both envelopes contain positive amount of money? Then is switching always better regardless of the amount you see?

That depends on perception of value. If you need those $10, you shouldn't switch. If you can afford to gamble, switch. Safety is worth something. It's the reason we have insurances.

The only objective is to maximize your expected return

You don't have that information. The envelopes could always be $10 and $5, in which case you would always lose $5. The envelopes could always be $10 and $20, in which case you would always gain $10. How is this a paradox?

There are multiple strategies you can choose to reach that objective. (minmax, maxmax, …)
None is objectively "better".

The only objective is to maximize your expected return. Maximizing expected return is completely objective

You're just assuming a bunch of shit. Given the information presented in the question and the question itself, the answer is you can expect to gain $2.50 on average by taking the other envelope. There is no more to it than that, if you start saying oh they didn't tell us this or that then the whole thing is moot and a pointless exercise because there's a million variables they didn't provide.

You can not expect to gain $2.50 by switching, that's the point of the paradox. Seeing the $10 should not give you any information as to which envelope is better

>Seeing the $10 should not give you any information as to which envelope is better
It's literally middle school tier probability at work. Knowing one value you know the two possible values in the other envelope.

If you see N dollars, you could get either 0.5N or 2N by switching, so your expected value for switching is 1.25N.

1.25N is better than N, so you should switch.

Ok, if you ignore the benefit of a certainity equivalent and utility functions, then (and only then) switching is always better.

Easy
$7.50

That's only true if they have equivalent probabilities. If the 2 envelopes were always $10 and $5 switching will always give you a negative return.

> Seeing the $10 should not give you any information as to which envelope is better
It doesn't.

If seeing any dollar amount makes switching better, then why did you need to see the dollar amount? Is switching better if you don't see the dollar amount?

THERE IS NO FUCKING ANSWER TO THIS QUESTION. YOU DON'T HAVE THE INFORMATION NEEDED TO KNOW THE EXPECTED VALUE OF SWITCHING.

Then is switching without seeing the dollar better?

Except you don't know which of the envelopes (higher or lower) you picked, so the chance is 50/50.

With no envelope opened, you have no information so you can't make any calculation of expected values for switching vs not switching. You just know there are two envelopes with money in, nothing else.

What dollar amount do you need to see to make switching not better? If your answer is that any dolalr amount makes switching better, then why did you need to see the dollar amount?

>Except you don't know which of the lottery tickets (winner or loser) you picked, so the chance is 50/50.

The dollar amount does not fucking matter. There is not enough information presented in the problem to answer with the expected value of switching.

What if you use x to represent an unknown amount of money, retard. Then you can still compare expected value

>2 choices = 3.0 million possibilities0
Kill yourself retard

Switching is not always better, but if you repeat this 100 times and switch every time you're expected to make more.

No.
The moment you get your envelope, those $10 are yours. Congratulations, you've got $10.
Then you are offered to switch. If you decide to switch you have a fifty percent to gain another $10 and a fifty percent chance to loose $5.
Or you could just keep your $10.

Not looking at your envelope doesn't change this. You just won't know any of it.

Nope. What if you didn't open the envelope, then all you know is that the envelope contains some amount of money, x. Is switching better?

No, you'll make an average which is $10 per envelope

The expected value is somewhere between $-5 and $10. You cannot figure out anything more from the information in the problem.

Ok, I give you an envelope, and say it either contains $0 or $1 trillion. What is the expected value of the envelope?

You can't lose anything. You can only gain money.

YOU DO NOT HAVE THE INFORMATION TO DETERMINE THAT. How many times do I have to repeat this to get it through your thick skull?

What dollar amount do you need to see to make you not want to switch?

You dont need to determine the number you dumbfuck. You just need to compare them. Do you not know algebra?

Me personally, I'd take any positive amount.
I don't gamble.

Viewpoint 1: Switching is good.

The envelope I look at contains $10. This means the other envelope has either $5 or $20. If I switch, half the time I lose $5, half the time I gain $10. The expected value of staying is $10 and the expected value of switching is $12.5, therefore you should always switch.

More generally, if the envelope you look at has $X, then the expected value of switching is (2.5x / 2) or (1.25x). Therefore, you should always switch.

Viewpoint 2: Switching doesn't matter.

Let's say Alice and Bob are both presented with the problem. Alice looks inside the left envelope and decides to switch because the EV is 1.25x. Bob looks at the right envelope and decides to switch because the EV is 1.25x. Each had a "rational" plan to switch, but something is off because they each switched to the "hold" plan of the other person. It doesn't really matter if you switch or not because the entire thing is arbitrary and you never got any information in the first place.

Not sure if I explained this well, but something weird is going on and it is a paradox of some sort.

In this scenario we always open the envelope. No need to ask what if.

If you see that one envelope contains x, the other envelope can either contain .5x or 2x. YOU CANNOT FIGURE OUT IF SWITCHING IS BETTER FROM THIS INFORMATION ALONE.

statistics makes people retarded

What information do you need?

What kind of math are you doing where the average is 10? Are you a woman?

This is correct

Whether the values of the envelope are x and .5x or x or 2x.

You don't get that information by opening the envelope

There are 2 envelopes. 1 contains x, the other contains 2x. Which one is which, we don't know (they look identical on the outside). You are randomly given 1 of the 2 envelopes, is switching better?

That's why there is not enough information to answer the question.

You cannot know from that information alone.

There is enough information. You're just asking for extra information

You don't get to know whether you got the the high amount or the low amount

Ok then, if you're so confident, what's the answer?
Is switching better? What if the 2 envelopes are always $10 and $5. Switching is obviously not better in that case.
Is switching the same? What if the 2 envelopes are always $10 and $5. Switching is obviously not better in that case.
Is switching worse? What if the 2 envelops are always $10 and $20. Switching is obviously not worse in that case.

This.

It's almost like quantum physics: looking at one envelope changes the probability of the other envelope.
Except even just THINKING about the probabilities changes the probabilities.
If you never thought about it it wouldn't matter which envelope you picked but once you think about it you always pick the wrong envelope.
Weird man.

This explains the paradox correctly

>half the time I lose $5, half the time I gain $10
No. You don't know that. The value of the 2 envelopes could always be $5 and $10.

here's a better paradox (try to find out how it works before googling the anser):

two people toss a fair coin lots of times.

Person A does this until he gets the sequence heads-tails-heads, then he notes down how many coin tosses it took to get that sequence. Then he repeats this very often (millions of times, say), and in the end notes down the average number avg_HTH of coin tosses needed to get the sequence heads-tails-heads

Person B does the exact same but for the sequence heads-tails-tails. After he is finished he writes down the average number avg_HTT of coin tosses needed to get the sequence heads-tails-tails

will avg_HTH and avg_HTT be (approx) same size? why?
will one be bigger? which one? why?

No it can't, you randomly chose 1 of the 2 envelope

Or the value of the 2 envelopes could always be 10$ and 20$. If you open one of the letters it's always worth to switch to the other.

zero.
You are always winning money, whichever envelope you choose.

avg_HTT would be lower. Because if you are at HT and you roll H, you fail but start over with a H, already partway through with the sequence.

You are given 2 envelopes and told at one envelope contains $10 and one envelope contains $5. After randomly choosing one, you find that it contains $10. How much money do you expect to gain or lose, on average by taking the other envelope instead?

>don't google the answer
>too dumb to even come up with question so he had to google it

Is that really a paradox?

I think "HTH" is (slightly) harder because if you throw "HTT" you have to start over and throw a new "H".
Whereas with "HTT" if you throw "HTH" you're already on the next HTT sequence and only need to throw "TT"