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/sci/ - Science & Math

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File: 16 KB, 490x346, envolope.png [View same] [iqdb] [saucenao] [google]
9868062 No.9868062 [Reply] [Original]

Alternatively what if you don't open and see the $10? What's the EV of switching?

>> No.9868070

>>9868062
If you see the $10 then the EV for switching is ($20 + $5)/2 = $12.5.
If you don't see the $10 then you don't have enough information to calculate an EV.

Where is the paradox?

>> No.9868074

>>9868070
>If you see the $10 then the EV for switching is ($20 + $5)/2 = $12.5.

This is wrong, and it's the whole point of the paradox.

>> No.9868082

>>9868074
>This is wrong
Not an argument.

>> No.9868084

The "paradox" comes from the faulty reasoning of switching back to the original envelope after having already switched once. If you start with envelope A which contains X dollars, and envelope B has 50% chance of 0.5X and 2X, of course B has the higher expected value of 5/4X. But switching back to A is just X, not (5/4)^2 X.

If instead you were given the opportunity to switch with new envelopes every time, each of which is 50% 0.5X and 2X of the current envelope, then you would want to keep switching with new envelopes forever since it has infinite expected value.

>> No.9868089

>>9868082
Thanks Stefan. You might try to get a cursory familiarity with what you're talking about (e.g., at least read the wikipedia to see if you're making a retarded mistake that is well documented) before embarassing yourself.

The point of the paradox is that according to your incorrect calculations, one always expects to gain money by switching, regardless of how much money one sees in the envelope. If that's the case, then what you see is irrelevant, and you should switch even *without* seeing how much is in the envelope, which is clearly incorrect.

>> No.9868091

>>9868089
>incorrect calculations
Incorrect why?

>which is clearly incorrect.
Incorrect why?

>> No.9868092

>>9868074
>>9868074
It's right for the exact same reason why the Monty Hall problem works

>> No.9868096

>>9868062
What paradox? It makes sense that you'd always switch. The other envelope can have 5 or 20 bucks. Risking 5 bucks to gain 10 is worth it.

>> No.9868097

>>9868062
This is great. Think about this, what would you say the prior probability of the lesser envelope containing a given amount of money is, i.e. how would you model this from your intuition if this situation was put in front of you?

>> No.9868100

>>9868092
No, it isn't. The Monty Hall problem works because Monty introduces nonrandom information.

>>9868091
I'll start with your second question: The two envelopes are indistinguishable a priori, yet the conclusion is that one is preferred because it isn't the first one you randomly chose, even without looking in it. Thus, two indistinguishable envelopes have different expected values, a contradiction.

The correct calculations are as follows: There is a fixed amount of money, 3X, in the envelopes, with X in one and 2X in the other. Switching between envelopes has 1/2 chance of reducing your money by X, or increasing it by X.

>> No.9868110

>>9868092
False, because in this case you don't know if the $10 is the good prize or not and you don't know what the chance is.
>>9868091
Not him, but this assumes that the chance that $10 is the good prize is the same as it being the bad one, which is an unwarranted assumption.

>> No.9868121

>>9868100
>No, it isn't. The Monty Hall problem works because Monty introduces nonrandom information.
>How much money can you expect to gain or lose, on average
While doing the experiment the amount would become clear
Or is that breaking the un-written rules?

>> No.9868127

interesting stuff

>> No.9868131

>>9868062
what if the first envelope was empty
would you open the other?

>> No.9868134

>>9868121
You can't do the experiment because the money is semi-random, in that only the $10 is fixed. In how many of the experiments would you make in 10, 20 and in how many would you make it 10, 5?

>> No.9868139

>>9868131
oh fuck

>> No.9868141

You haven't defined probability distribution of the amount of money in the envelope, that's where the paradox come from.

>> No.9868145

>>9868100
You might try to get a cursory familiarity with what you're talking about (e.g., at least read the wikipedia to see if you're making a retarded mistake that is well documented) before embarassing yourself.
> If the player was aware of this mechanism, and knows that they hold Envelope A, but don't know the outcome of the coin toss, and doesn't know a, then the switching argument is correct and he/she is recommended to switch envelopes.

>> No.9868149

>>9868121
>While doing the experiment the amount would become clear
You are confusing yourself. Here, you are treating X as a random variable that you gain information about. If you want to reason this way, you must specify a prior distribution on X. When you do this, it becomes clear that the odds are not always 50:50 that the other envelope contains more.

Alternatively, you could treat X as fixed, in which case, the expected value of switching is 0, because you either gain or lose X with equal probability.

>> No.9868152

>>9868110
>hich is an unwarranted assumption.
Are you retarded? There's only 2 envelopes.

>> No.9868154

this is possibly one of the worst wikipedia pages on a mathematical topic

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

it looks like some people from the lesswrong community raided it

>> No.9868155

>>9868145
Your reply is citing a situation in which in the player has additional information about the envelopes, and envelope A and B are not indistinguishable a priori anymore--we know which envelope is A, and we know the mechanism by which the contents of B are generated. Now read the OP again, and describe where this information comes from.

>> No.9868156 [DELETED] 

>>9868100
>Thus, two indistinguishable envelopes have different expected values, a contradiction.
No,
E(Envolope 2|Envolope 1 contains $10) = $12.5
E(Envelope 1Envolope 1 contains $10) = $10
and E(Envelope 1) = E(Envelope 2)
It's impossible to calculate the exact value of E(Envelope 1) without knowing how the envelopes were filled.
>The correct calculations are as follows: There is a fixed amount of money, 3X, in the envelopes, with X in one and 2X in the other. Switching between envelopes has 1/2 chance of reducing your money by X, or increasing it by X.
No, you're talking about a fundamental different problem here because in that case no envelope is fixed to contain $10.

>> No.9868161

>>9868152
Yes, and if you chose an envelope at random, the chance that you choose the one with $10 is 1/2 but you have no information on the chance that the other envelope has more or less than that.
The average embarrassment from you calling other people retards should make you reconsider ever doing it.

>> No.9868168

>>9868156
>No, you're talking about a fundamental different problem here because in that case no envelope is fixed to contain $10.

Look at your argument above, and substitute any constant for 10--it applies to all possible observations of the envelope you chose. Before we open the envelope we choose, we can reason according to your argument that the other envelope contains more, conditioned on *any* possible observation of our envelope.

>> No.9868172

>>9868154
It can't be lesswrong, because there are actual citations in this article.

>> No.9868179 [DELETED] 

>>9868062
Here's my solution:

Say one of the envelopes contains X dollars, and the other contains 2X dollars. If your strategy is to stick with the envelope you pick initially, the expected value of the trial is obviously (X + 2X)/2 = 1.5X. If you instead decide on switching as your strategy, the expected value is the same, because if you initially picked the envelope with X dollars in( it you would end up with the 2X envelope, and if you picked the 2X envelope at first you would get the X envelope. Thus, (2X + X
)/2 = 1.5X, and the strategies are equivalent.

>> No.9868187

>>9868062
Here's my solution:

Say one of the envelopes contains X dollars, and the other contains 2X dollars. If your strategy is to stick with the envelope you pick initially, the expected value of the trial is obviously (X + 2X)/2 = 1.5X. If you instead decide on switching as your strategy, the expected value is the same, because if you initially picked the envelope with X dollars in( it you would end up with the 2X envelope, and if you picked the 2X envelope at first you would get the X envelope. Thus, (2X + X)/2 = 1.5X, and the strategies are equivalent.

>> No.9868190

>>9868179
You're correct, and you've restored a little bit of my faith in /sci/.

>> No.9868193

>>9868134
>You can't do the experiment because the money is semi-random, in that only the $10 is fixed. In how many of the experiments would you make in 10, 20 and in how many would you make it 10, 5?
I was thinking of it as an experiment that was being presented to me x number of times
As the attempts approach infinity I would know the amount

>> No.9868205

>>9868193
>I was thinking of it as an experiment that was being presented to me x number of times
>As the attempts approach infinity I would know the amount

What are you trying to say here?

>> No.9868208

>>9868084
Yes, but then the question is why is the one that's originally give to you special? They should both have the same EV.

>> No.9868209

>>9868100
>The two envelopes are indistinguishable a priori, yet the conclusion is that one is preferred because it isn't the first one you randomly chose, even without looking in it. Thus, two indistinguishable envelopes have different expected values, a contradiction.
They're distinguished once you pick one.

>> No.9868212

>>9868062
Let's say both envelopes were filled by rolling a d100 each and then putting money inside equal to the rolled number.
Let
X be the amount of money in the envelope you picked.
Y be the amount of money in the envelope you didn't pick.
Then the EV of switching is
E(Y| X=10, (Y=5 OR Y=20))
= 5 P(Y=5|X=10, (Y=5 OR Y=20)) + 20 P(Y=20|X=10, (Y=5 OR Y=20))
= 5 P(X=10, Y=5)/P(X=10, (Y=5 OR Y=20)) + 20 P(X=10, Y=20)/P(X=10, (Y=5 OR Y=20))
= 5*(1/10000)/(1/5000) + 20*(1/10000)/(1/5000)
= 12.5

So it's better to switch in this case.

But if you had a different fill algorithm where one envelope was guarrenteed to contain $10 and the other $5 then switching from $10 is always a bad idea.
Without knowing how they were filled you can't say anything for certain.

>> No.9868213

>>9868096
You switched. Now use the same reasoning again. Apparently you should switch again.

>> No.9868216

>>9868213
>Apparently you should switch again.
The original image doesn't seem to give this as an option

>> No.9868219

retarded use of math

>> No.9868222

>>9868208
>Yes, but then the question is why is the one that's originally give to you special?
Because you know how much is in it.

>They should both have the same EV.
Why?

>> No.9868226

>>9868205
That the $10 amount is just 1 attempt
If we ran the same experiment again and I got $20, I would know that the $10 is the lesser amount and could then answer the question
Since there's a chance that I got infinite $10 envelopes I added the
>As the attempts approach infinity I would know the amount

>> No.9868231

>>9868209
That's not what "distinguished" means. Your having picked it does not reflect any information about the envelope.

Before you say, "But looking in it gives information about the envelope," (which is true) remember that the entire point of the paradox is that the observation of any value in the envelope leads (by the faulty argument) to the conclusion that the other envelope contains more, and thus that the other envelope is therefore always preferable, without even needing to look. That is the paradox.

>> No.9868236

>>9868231
>That's not what "distinguished" means. Your having picked it does not reflect any information about the envelope.
But you know it has 10$ in the OP's formulation.

>Before you say, "But looking in it gives information about the envelope," (which is true) remember that the entire point of the paradox is that the observation of any value in the envelope leads (by the faulty argument) to the conclusion that the other envelope contains more
Leads (by the correct argument) to the conclusion that the other envelope *probably* contains more, and hence the correct decision is to always switch.

>and thus that the other envelope is therefore always preferable, without even needing to look. That is the paradox.
Why is this a paradox?

>> No.9868239

>>9868212
This post is a mess.

>> No.9868240
File: 238 KB, 720x452, TMMfc.png [View same] [iqdb] [saucenao] [google]
9868240

post more faulty uses of probability

>> No.9868244

>>9868240
No fault here

>> No.9868245

>>9868240
But this picture is correct.

>> No.9868246

>>9868236
>Why is this a paradox?

Two envelopes are identical, one is chosen at random, and then other's expected value is greater than the one chosen at random. How does a random process give you information about which envelope is better? Answer: It doesn't.

>> No.9868253
File: 26 KB, 720x879, pete_1.jpg [View same] [iqdb] [saucenao] [google]
9868253

Answer is symmetry.

>> No.9868254

>>9868187
Right, this is the objective way of viewing it. However, where is flaw with the other line of reasoning? I have $10, switching gives an apparent EV of $20*0.5+$5*0.5=$12.5. Is it in the statement one contains twice as much as the other?

>> No.9868256

>>9868246
>How does a random process give you information about which envelope is better?
The process doesn't, it's the given knowledge that one contains twice the amount of the other.

>> No.9868259

>>9868254
You're assuming that the probability that the other envelope contains $5 is equal to the probability that is contains $20.

>> No.9868264

>>9868062
The first choice of picking the envelope is no different to the second choice of swapping the envelope, neither choice will tell you if you have the maximum amount, unless you picked an envelope that had more money than exists on earth.

>> No.9868265

>>9868216
Let's entertain it as an option then

>> No.9868266

>>9868254
The flaw in the other line of thinking is that they treat the total amount in the envelopes (3X) as being random, and simultaneously assert that the odds of the other envelope being greater is 50:50 independent of the particular amount you observe.

They are switching between interpreting X as random and X as fixed.

>> No.9868273

>>9868259
I am. What's wrong with that? What's the actually probability it should be $5 or $20?

>> No.9868274

>>9868265
>Let's entertain it as an option then
Even if you can switch again, you shouldn't, since the envelope you originally switched to has a higher EV.

>> No.9868277

>>9868256
That knowledge exists prior to the choice. No new information is gained by randomly choosing an envelope, yet the random choice completely determines which envelope you choose.

Consider the following scenario:

The envelopes are exactly as described before. One envelope is given to you at random, the other to me. We each privately observe how much money is in our envelope, but we aren't allowed to tell each other. We are allowed to trade envelopes. What's the expected value, for each player with respect to the amount they privately observed, of making this trade?

>> No.9868282

>>9868277
>What's the expected value, for each player with respect to the amount they privately observed, of making this trade?
if the first player sees X and the second player sees Y, then the first player expects 5X/4 and the second player expects 5Y/4

>> No.9868287

>>9868282
So, just to be clear, you believe the players should always trade?

>> No.9868288

>>9868287
>So, just to be clear, you believe the players should always trade?
If the player understands that 5X/4 > X, then yes.

>> No.9868291

>>9868273
>What's wrong with that?
It leads to the "paradox".
>What's the actually probability it should be $5 or $20?
There is not enough information given to know.

>> No.9868292

7.5

>> No.9868294

>>9868266
>The flaw in the other line of thinking is that they treat the total amount in the envelopes (3X) as being random, and simultaneously assert that the odds of the other envelope being greater is 50:50 independent of the particular amount you observe.

Ok, so the total amount in the envelopes and the odds of the other envelope being greater are not independent. So is the question simply nonsensical? If not, given that I see $10, what is the probability that the other one will have $5 v.s. $20? If we want to force the EVs to be the same we get P($5)=2/3 and P($20)=1/3, but how else can we get these probabilities?

>> No.9868295

>>9868291
>It leads to the "paradox".
There is no paradox, or even a "paradox".

>> No.9868297

>>9868277
>>9868288
Why make this so complicated? The original "paradox" is that you are expecting the second envolope to contain more or any less than the first envelope

>> No.9868301

>>9868297
>Why make this so complicated?
What part is confusing you?

>The original "paradox" is that you are expecting the second envolope to contain more or any less than the first envelope
Why is that a "paradox"?

>> No.9868306

there's enough retards in this thread who think it's a bad idea to switch to make me consider hustling idiots with this

>> No.9868307

>>9868062
If it's half $10 then it's $5 and you would lose $5 by switching. If it's double $10 then it's $20 and you would gain $10 by switching. So it's a good idea to switch if you're averaging it out over multiple rounds, if not, it depends on your risk preference.

>> No.9868308

>>9868277
It's not a trade to begin with, a trade requires that both traders know what each other have, in order to make the trade.

>> No.9868309

>>9868301
If you expect that the 2nd envelope contain any more or any less than the first envelope, then you somehow distinguished the 2 envelopes without any information

>> No.9868311

>>9868297
>The original "paradox" is that you are expecting the second envolope to contain more or any less than the first envelope
It would be more paradoxical to expect it to contain the same amount, considering you're told it doesn't.

>> No.9868313

>>9868309
>If you expect that the 2nd envelope contain any more or any less than the first envelope, then you somehow distinguished the 2 envelopes without any information
"one envelope has twice as much money as the other" is information.

>> No.9868314

>>9868307
Wait am I retarded? I'm reading the thread and now I think I may have taken a wrong turn.

>> No.9868319

>>9868288
>If the player understands that 5X/4 > X, then yes.
Lol.

Let us consider the following simulation: We choose 100 million different values of X according to any probabilistic process you want, then place X and 2X in 100 million respective pairs of envelopes.

We have players A and B, and for each of these hundred million pairs, we assign them a random envelope. We then generate two counterfactual outcomes: one where the players always switched, and one where the players always did not switch. Do you believe that both players with have approximately 5/4 as much money in the switching scenario as they did in the non-switching?

>> No.9868320

>>9868309
You can't expect, you can only hope.

>> No.9868324

>>9868319
This breaks part of the symmetry though. Putting an upper limit on X changes it.

>> No.9868325
File: 227 KB, 396x382, Homegrowmen_All_Year_Long.png [View same] [iqdb] [saucenao] [google]
9868325

Do you guys never get bored?

>> No.9868326

>>9868311
>>9868313
Nothing distinguished the 1st envelope from the 2nd envelope

>> No.9868327

>>9868326
>Nothing distinguished the 1st envelope from the 2nd envelope
The 2nd envelope is the one you can switch to, the 1st envelope is the one you opened (10$).

>> No.9868329

>>9868319
>any probabilistic process I want
The probabilistic process I want is a deterministic distribution where P(X=10)=1. Then the players should always switch if they get 10 and not switch when they get 20.

>> No.9868337

>>9868324
>Putting an upper limit on X changes it.
The support of the probability distribution does not have to be bounded above. In any case, the players are not told about what the process was and are not aware of it. If you fear they will infer it, you can make it a complicated mixture model with many different modes.

>> No.9868338

>>9868327
If you switch now, you go into an infinite loop

>> No.9868340

>>9868338
>If you switch now, you go into an infinite loop
Why?

>> No.9868344

>>9868340
What if I presented you with the same problem but with the envelopes switched?

>> No.9868346

>>9868329
>The probabilistic process I want is a deterministic distribution where P(X=10)=1. Then the players should always switch if they get 10 and not switch when they get 20.

I agree. Note that the expected value of switching in this case is *not* 5/4 of whatever you observe.

>> No.9868353

>>9868344
>What if I presented you with the same problem but with the envelopes switched?
Then I open the other envelope, see X dollars, and switch with expected value 5/4 * X.

>> No.9868354

it doesn't matter whether you switch or not

and lmao at that retard calculating some nonsense expected value for this problem

>> No.9868358

Are the envelopes on a conveyor belt? If so then both would take off.

>> No.9868359

>>9868062
Monty Hall only works for three things
It's definitely 50/50 here

>> No.9868360

>>9868353
I don't know what you're talking about with someone else

>> No.9868361

>>9868353
Please see >>9868319

and answer the question at the end

>> No.9868362

>>9868360
>I don't know what you're talking about with someone else
What do you mean?

>> No.9868370

>>9868362
I don't know where you got 5/4 from

>> No.9868375

The problem is you're trying to make the claim that the other envelope is in a superposition of containing both $5 AND $20 at the same time but this isn't the case. The other envelope must contain a concrete dollar amount and once I decide to switch I must know that dollar amount. You are trying to ask a question without a solution and that's just retarded. Either you've given me a choice to switch and you WILL TELL ME what the other envelope contains so that I can run the experiment repeatedly and the results converge on a solution or you aren't giving me the option to switch because you refuse to divulge whether the envelope contains $5 or $20 and you've presented me with a false dilemma because I cannot reasonably switch because you have no idea what the dollar amount is in the second envelope.

This whole 'paradox' is just retarded.

>> No.9868381

>>9868375

thats because you dont understand probability theory

>> No.9868383

>>9868381
>thats because you dont understand probability theory
Speak for yourself.

>> No.9868384

>>9868370
>I don't know where you got 5/4 from
Read the thread before posting.

>> No.9868385

>>9868383

the solution is that you don't know whether to change or not and you have no way of probabilistically making any better decisions

>> No.9868389

>>9868385
>you don't know whether to change or not and you have no way of probabilistically making any better decisions
Speak for yourself.

>> No.9868390

>>9868384
Why do I need to read the whole thread?

>> No.9868394

>>9868089
>you should switch even *without* seeing how much is in the envelope
Yes.
>which is clearly incorrect.
No.

What you aren't understanding is that doubling your money is the same value as losing half your money.

You should be indifferent if the situation was "50% chance to gain $5 and 50% chance to lose $5", but it's "50% chance to gain $10 and 50% chance to lose $5".

>> No.9868397

>>9868390
>Why do I need to read the whole thread?
So you can make informed replies.

>> No.9868399

>>9868394
How do you not see the infinite loop?

>> No.9868401

>>9868394
Can you respond to >>9868319 ?

>> No.9868402

>>9868399
>How do you not see the infinite loop?
How do you?

>> No.9868405

>>9868401
Stop using numbers

>> No.9868406

>>9868399
>What is a limit?

>> No.9868407

>>9868405
wut

>> No.9868408

>>9868402
Why would you switch between 2 indistinguishable envelopes?

>> No.9868411

>>9868408
Why wouldn't you?

>> No.9868414

>>9868408
>Why would you switch between 2 indistinguishable envelopes?
Because the one you can switch to has a higher expected value than the one you first pick.

>> No.9868415

>>9868411
But you're saying switching is better. I'm saying switching is neither better nor worse

>> No.9868420

>>9868414
How can that be if the are indistinguishable?

>> No.9868421
File: 109 KB, 944x539, mathbtfo.jpg [View same] [iqdb] [saucenao] [google]
9868421

>>9868062
Looks like Javascript>Math. Switching wins

>> No.9868425

>>9868415
I would always switch because tails never fails

>> No.9868426

>>9868420
>How can that be if the are indistinguishable?
They're distinguished since one is picked first.

>> No.9868428

>>9868062
This is out of the realms of any logical "choice" from the beginning.

There is nothing you can do to increase your chances of getting the higher amount. Your chances are the same from the beginning to the end.

The paradox arises by conflating a pure gamble with a logical choice.

>> No.9868433

>>9868426
Any of the 2 envelopes could've been picked and designated as "1st"

>> No.9868435

There are two possibilities:
1. the second envelope has $20. (You have picked the smaller envelope)
So, if you switch you gain $10, which is the amount in the smaller envelope.
2. the second envelope has $5. (You have picked the larger envelope).
So, if you switch you lose $5, which is again the amount in the smaller envelope.

No matter what you pick, you are either gaining or losing the amount that is in the smaller envelope so the expected value is the same for both envelopes.

>> No.9868436

>>9868433
>Any of the 2 envelopes could've been picked and designated as "1st"
And?

>> No.9868437

>>9868436
If I designate an envelope as 1st and asked if you want to switch to the 2nd envelope, you would say yes

>> No.9868440

>>9868437
>If I designate an envelope as 1st and asked if you want to switch to the 2nd envelope, you would say yes
Yes, because the expected value is higher for switching

>> No.9868441

>>9868440
Actually since you designate it as 1st it's a different situation so ignore that post.

>> No.9868442

>>9868440
Jesus Christ you trolls are ridiculous

>> No.9868444

>>9868442
>Jesus Christ you trolls are ridiculous
see >>9868441

>> No.9868450

>>9868440
What if after you switch I designate the 2nd envelope as 1st?

>> No.9868451

>>9868450
>What if after you switch I designate the 2nd envelope as 1st?
see >>9868441

>> No.9868452

HURR DURR USING MEAN INSTEAD OF MEDIAN, RETARDEDLY.

Let me tell you about "average" incomes in different countries.

>> No.9868454

>>9868451
So we agree now? Switching is a neither better not worse

>> No.9868456

>>9868454
>So we agree now?
What gave you that impression?

>Switching is a neither better not worse
It is better when I get to randomly choose the 1st envelope, by the expected value.

>> No.9868464

>>9868456
I give you an envelope, do you switch?

>> No.9868470

>>9868464
>I give you an envelope, do you switch?
Did you choose randomly?

>> No.9868471

>>9868470
Yes

>> No.9868472

>>9868456
If the envelopes are assigned to two people at random, does trading increase the expected value for both people? Across many trials, will both end up with more, so long as they switch?

>> No.9868477

>>9868471
>Yes
Then I switch, by the expected value.

>> No.9868485

>>9868472
>If the envelopes are assigned to two people at random, does trading increase the expected value for both people?
see >>9868282

>> No.9868486

>>9868477
Does that sound right?

>> No.9868488

>>9868486
>Does that sound right?
Yes, by the expected value.

>> No.9868492

>>9868485
Okay, you're trolling.

>> No.9868493

>>9868488
How did you distinguish the 2 envelopes?

>> No.9868496

>>9868492
>Okay, you're trolling.
What do you mean?

>> No.9868498

>>9868493
>How did you distinguish the 2 envelopes?
You only gave me one.

>> No.9868500

>>9868488
I have an envelope and you have an envelope. Why do you want to switch to my envelope?

>> No.9868501

>>9868500
>I have an envelope and you have an envelope. Why do you want to switch to my envelope?
The expected value of your envelope is higher than mine.

>> No.9868502

>>9868501
What makes that the case?

>> No.9868505

>>9868496
I've already responded to the argument that both players expect 5/4 of their observed value from switching here >>9868319, and it has gone entirely ignored, because it directly refutes the claim.

>> No.9868506

>>9868502
>What makes that the case?
It either has twice the amount of mine, or half. I either gain X or lose X/2.

>> No.9868509

>>9868505
>I've already responded to the argument that both players expect 5/4 of their observed value from switching here >>9868319, and it has gone entirely ignored, because it directly refutes the claim.
What part of that post refutes the claim? It just appears to be a hypothetical and no conclusions are drawn.

>> No.9868511

>>9868062
You fucked up the prompt OP, you're not supposed to open the envelope before deciding to switch.

By opening the envelope, you're gaining information and thus changing the expected value of the envelopes. There's a 50/50 chance that the total amount of money is either 15$ or 30$, so the expected value of both envelopes combined becomes 22,5$. Do the math and tell me the expected value of the other one.

>> No.9868512

>>9868506
Why can't the same argument be made by me for wanting to switch to your envelope?

>> No.9868518

>>9868512
>Why can't the same argument be made by me for wanting to switch to your envelope?
It can be, see >>9868282.

>> No.9868523

>>9868518
I don't want to read the whole thread. How come I want to switch to your envelope and you want to switch to my envelope?

>> No.9868526

>>9868509
From my post:
>Do you believe that both players with have approximately 5/4 as much money in the switching scenario as they did in the non-switching?

This question is rhetorical. If the A's non-switching total is X, and B's non-switching total is Y, then when they adopt the switching strategy, A's total is Y and B's total is X. Either both of these totals are equal, or one of them is smaller than the other. There are no possible circumstances under which switching benefits both players, even in the aggregate.

>> No.9868527

>>9868523
>I don't want to read the whole thread.
You don't have to, just see >>9868282.

>> No.9868530

>>9868527
I shouldn't have brought "my" and "your" into this

>> No.9868531

>>9868530
>I shouldn't have brought "my" and "your" into this
What do you mean?

>> No.9868534

>>9868531
There are 2 envelopes. We have 1 of the 2 envelopes, do we switch?

>> No.9868535

>>9868526
> If the A's non-switching total is X, and B's non-switching total is Y, then when they adopt the switching strategy, A's total is Y and B's total is X. Either both of these totals are equal, or one of them is smaller than the other.
This is all trivial. What is the intended relevance?

>There are no possible circumstances under which switching benefits both players, even in the aggregate.
Switching is always beneficial since the potential gain is always greater than the potential loss.

>> No.9868538

>>9868534
As in switch our envelope for the other

>> No.9868539

>>9868534
>We have 1 of the 2 envelopes, do we switch?
Who chose the first envelope, and how? If it was chosen randomly, then switch, by the expected value.

>> No.9868540

>>9868535
>Switching is always beneficial since the potential gain is always greater than the potential loss.
And yet for one of the players A and B, it was not beneficial, even after 100 million trials. How mysterious!

>> No.9868548

>>9868540
>And yet for one of the players A and B, it was not beneficial
For which X>0 does X/2 become greater than X?

>> No.9868552

>>9868539
So which envelop should we pick?

>> No.9868553

>>9868552
>So which envelop should we pick?
The one we don't have.

>> No.9868554

>>9868553
Ok, now we have that envelope. Should we switch?

>> No.9868556

>>9868554
>Ok, now we have that envelope. Should we switch?
No, because the envelope we switched to has a higher expected value.

>> No.9868562

>>9868548
You noted it was "trivial" that either neither has more money, or one of them loses money by adopting the switching strategy. Just stop posting.

>> No.9868564

>>9868556
How did you distinguish the 2 envelopes just now?

>> No.9868572

>>9868562
>You noted it was "trivial" that either neither has more money, or one of them loses money by adopting the switching strategy.
Yes, if X > Y, and two people (hypothetical or not) trade X for Y, then X > Y. Real life trades (hypothetical or not) don't change the ordering of numbers.

>Just stop posting.
Why?

>> No.9868575

>>9868564
>How did you distinguish the 2 envelopes just now?
You said we switched, so now we have the second envelope (i.e. the envelope is different than the first one we opened), and by the expected value is beneficial to keep.

>> No.9868576

>>9868572
Can I ask how old you are?

>> No.9868578

>>9868576
>Can I ask how old you are?
What is the relevance?

>> No.9868581

>>9868575
What if we didn't know what our original envelope was? How would you know whether or not to switch?

>> No.9868583

>>9868578
Personal curiosity

>> No.9868585

>>9868581
>What if we didn't know what our original envelope was?
Do you mean if we forgot which was 1st and 2nd or we never opened the 1st envelope?

>> No.9868587

>>9868583
>Personal curiosity
Old enough.

>> No.9868589

>>9868585
We find 2 envelopes on the ground. Which one do we pick?

>> No.9868596
File: 7 KB, 300x168, oldenough.jpg [View same] [iqdb] [saucenao] [google]
9868596

>>9868587
Can you give me a formal breakdown of the problem, explicitly defining what is a constant and what is a random variable, and what distributions the expectations are taken over?

>> No.9868597

>>9868589
>We find 2 envelopes on the ground. Which one do we pick?
If your input can be ignored and I'm the one picking, then either one, randomly.

>> No.9868602

>>9868597
So it doesn't matter which 1 we pick, right?

>> No.9868606

>>9868596
>explicitly defining what is a constant and what is a random variable
Not really worth the time for such a trivial problem, but here, have a read, this is probably a good enough place to start if you're unfamiliar: https://simple.wikipedia.org/wiki/Random_variable

>> No.9868610

>>9868597
>>9868602
So we expect that the 2 envelopes contain equal amounts

>> No.9868611

>>9868602
>So it doesn't matter which 1 we pick, right?
"Matter" in terms of what?

>> No.9868613

>>9868610
>So we expect that the 2 envelopes contain equal amounts
No, the problem specified:
>one envelope has twice as much money as the other

>> No.9868615

>>9868611
If our goal is to maximize the value

>> No.9868624

>>9868615
>If our goal is to maximize the value
Then as long as we choose randomly, switching maximizes the value.

>> No.9868626

>>9868624
There are 2 envelopes. Which one should we pick?

>> No.9868628

>>9868606
Lmao. Thanks

>> No.9868629

>>9868626
>There are 2 envelopes. Which one should we pick?
Either one.

>> No.9868630

>>9868628
>Lmao. Thanks
No problem, let me know if you're still confused.

>> No.9868632

>>9868629
There are 2 envelopes. We picked either one by your input. Now do we switch?

>> No.9868634

>>9868632
>Now do we switch?
Yes, by the expected value.

>> No.9868637

>>9868634
And switching will be beneficial, why?

>> No.9868639

>>9868634
The envelope we have now was already picked by your input

>> No.9868646

>>9868637
>And switching will be beneficial, why?
Because the 1st envelope contains X, so the potential gain is X, while the potential loss is X/2, and X > X/2 (assuming X>0).

>> No.9868648

>>9868639
>The envelope we have now was already picked by your input
No, as I said it was chosen randomly, not by my input.

>> No.9868652

>>9868506
You gain X in both scenarios in terms of beginning with 0, and ending with a higher value, you don't lose anything by swapping the higher value envelope with the lower one because you can't and haven't claimed ownership of it in the first place.

When choosing the first envelope, you are not claiming ownership of the value the envelope contains, only the envelope itself. Once you have opened the envelope and seen the value, you can either claim ownership of this value and not swap, or not claim ownership and swap.

>> No.9868656

>>9868648
There are 2 envelopes. You said we can pick either, right?

>> No.9868657

>>9868652
>You gain X in both scenarios in terms of beginning with 0, and ending with a higher value
The envelopes don't contain the same amount, so you can't possibly gain X by switching and gain X by not switching.

>> No.9868662

>>9868656
>You said we can pick either, right?
I thought you we're the one formulating the hypothetical here, as far as I could tell you haven't made any rule against picking either.

>> No.9868665

>>9868662
I mean to maximize value

>> No.9868669

>>9868652
>Once you have opened the envelope and seen the value, you can either claim ownership of this value and not swap, or not claim ownership and swap.
Yes, and if the first envelope contained X, then swapping has the potential gain of X and the potential loss of X/2.

>> No.9868671

>>9868662
>>9868665
There are 2 envelopes. Which one to pick? Doesn't matter? Ok, you get to pick our envelope. Now that you have chosen our envelope, should we switch?

>> No.9868672

>>9868665
>I mean to maximize value
You switch to maximize value. If the first envelope contains X, then the potential gain (X) is greater than the potential loss (X/2) (assuming X>0).

>> No.9868675

>>9868671
>Now that you have chosen our envelope, should we switch?
Yes, because the potential gain (X) is greater than the potential loss (X/2) (assuming X>0).

>> No.9868677

>>9868675
But you chose the 1st envelope

>> No.9868679

>>9868677
>But you chose the 1st envelope
Well someone had to, it didn't seem like you did in this hypothetical.

>> No.9868684

>>9868679
You could have just as easily chosen the other envelope, so why is switching better?

>> No.9868688

>>9868684
>You could have just as easily chosen the other envelope, so why is switching better?
Because it has an expected value of 5/4 times what we saw in the first envelope.

>> No.9868691

>>9868084
The paradox is that it always seems sensible to switch, regardless of what you see in the envelope. You thus never actually gain information.

>> No.9868692

>>9868691
>The paradox is that it always seems sensible to switch, regardless of what you see in the envelope.
Why is that a paradox?

>> No.9868697

>>9868692
Tell me how looking in the envelope informs you to switch. What dollar amount would you say, "no, it's not a good idea to switch"?

>> No.9868698

>>9868688
Why do you think that? The first envelope was chosen by you

>> No.9868700

>>9868698
>Why do you think that?
Because we either gain X (finishing with 2X) or lose X/2 (finishing with X/2).

>The first envelope was chosen by you
What's the relevance of this?

>> No.9868702

>>9868697
>Tell me how looking in the envelope informs you to switch.
It doesn't, the knowledge that the other envelope either contains twice or half as much informs me to switch.

>What dollar amount would you say, "no, it's not a good idea to switch"?
A negative dollar amount, since then I would have larger debt by switching.

>> No.9868706

>>9868702
>since then I would have larger debt by switching.
since then I would probably* have larger debt by switching.

>> No.9868709

>>9868657
>>9868669

You can't gain or lose against something you didn't own in the first place. If the swap takes place, then both parties are not claiming ownership of the value contained in their original envelope.

If, after the game, one of the players accidentally drops their envelope into a fire, then they have now lost money.

>> No.9868710

>>9868700
There are 2 envelopes. You picked our envelope. Now switching is better? Remember, you picked our envelope

>> No.9868712

>>9868709
>You can't gain or lose against something you didn't own in the first place. If the swap takes place, then both parties are not claiming ownership of the value contained in their original envelope.
What's the relevance of ownership here? You either end with X from the original envelope if you don't switch, or either 2X or X/2 if you do switch. The chance of gaining an extra X is worth the chance of losing X/2.

>If, after the game, one of the players accidentally drops their envelope into a fire, then they have now lost money.
This seems completely irrelevant.

>> No.9868714

>>9868710
>Now switching is better?
Yes, by the expected value.

>> No.9868717

>>9868697
Why are you still talking to him? He's trolling you

>> No.9868721

>>9868717
>Why are you still talking to him?
I'm not a "him".

>> No.9868725

>>9868717
What does he gain from this?

>> No.9868726

>>9868712
Because choosing not to switch means you claim ownership of that value, the value in the other envelope is irrelevant because you never owned it. You can't lose money you never owned.

If you choose to swap and the other person does not, then you're forced to claim ownership of the money you hold, you have not lost anything.

>> No.9868729

>>9868725
What do any trolls gain? They are amused at other people's frustration

>> No.9868731

>>9868726
>Because choosing not to switch means you claim ownership of that value, the value in the other envelope is irrelevant because you never owned it. You can't lose money you never owned.
You lose out on the potential* extra X by not switching, but this distinction makes no difference.

>> No.9868734

>>9868714
What expected value? I expect the 2 envelops to have equal value since they are indistinguishable?

>> No.9868736

>>9868729
>What do any trolls gain?
I'm not a "troll".

>They are amused at other people's frustration
I'm not getting much amusement, and I don't know what anyone else would be frustrated about, it's a very elementary problem.

>> No.9868739

>>9868736
>it's a very elementary problem.
True!

>> No.9868740

>>9868734
>I expect the 2 envelops to have equal value since they are indistinguishable?
Why would you expect two envelopes with different values to have equal value?

>> No.9868746

>>9868740
What would you expect their values to be?

>> No.9868750

>>9868746
>What would you expect their values to be?
The 1st envelope is given to contain 10$, and the second has an expected value of 12.5$.

>> No.9868754

>>9868750
Expectation is taken over what distribution?

>> No.9868755

>>9868750
How can that be? You picked which envelope was 1st

>> No.9868759

>>9868755
>How can that be?
see >>9868062

>> No.9868762

>>9868759
I don't understand

>> No.9868764

>>9868754
>Expectation is taken over what distribution?
B(1,1/2), assuming process for choosing the first envelope was random enough (perhaps shuffling them behind my back, or flipping a coin to decide).

>> No.9868768

Look at it this way. Run the experiment 500 times and tell me what the value converges on. You can't because you refuse to decide whether the envelope contains $5 or $20. The other envelope is empty.

>> No.9868770

>>9868762
The 1st envelop contains 10$ because as the image states
>you find that it contains 10$
and the second has an expected value of 12.5$ since
>one envelope has twice as much money as the other

>> No.9868776

>>9868770
Why would you expect 1 envelope to have more value than the other?

>> No.9868781

>>9868731
Extra compared to what? If you are holding $10 and swap it for $5, then you never claimed ownership of the $10, you only held it, therefore this value cannot be considered lost, however you have gained $5.

>> No.9868782

>>9868776
>Why would you expect 1 envelope to have more value than the other?
Because it's a given that one envelope contains twice the other (i.e. has more than the other, as long as X != 0).

>> No.9868786

>>9868781
>Extra compared to what?
Compared to what you get by not switching, those are the only two outcomes are they not?

>> No.9868804

>>9868782
Out of 2 envelopes, which envelope do you pick?

>> No.9868808

>>9868804
>Out of 2 envelopes, which envelope do you pick?
The one that I didn't open first.

>> No.9868810

This is the most retarded thread I've ever seen, is this what brainlet math majors do ??

>1st envelope
>open it
find 10
So you will either win 10$ or lose 5$

WHERE'S THE FUCKING PARADOX ????

>> No.9868812

>>9868399
Because time costs you nothing here. You never need to arrive at an actual decision. IF every time you swapped you had to pay 1 cent, then you would have an entirely different problem.

>> No.9868816

>>9868808
Which one is that?

>> No.9868819

>>9868812
So when I ask you to pick an envelope, you'll just freeze?

>> No.9868821

>>9868816
>Which one is that?
What's ambiguous about it? It's the one that doesn't have 10$ in it.

>> No.9868823

>>9868821
Which out of 2 envelopes is that?

>> No.9868824

>>9868823
>Which out of 2 envelopes is that?
The one I switched to, as per >>9868062.

>> No.9868826

>>9868824
There are 2 envelopes. All we know is that 1 contains twice the other. Which do you pick?

>> No.9868828

>>9868826
>Which do you pick?
Pick randomly, and then switch.

>> No.9868832

>>9868828
Why switch?

>> No.9868835

>>9868832
>Why switch?
The expected gain is greater than the expected loss.

>> No.9868853

>>9868835
Is it?

>> No.9868904

>>9868853
>Is it?
X>X/2 for X>0

>> No.9868906

>>9868904
No

>> No.9868913

>>9868904
Prove it

>> No.9868926

>>9868913
X > 0 => X/2 > 0 => X-X/2 > 0 => X > X/2

>> No.9868930

>>9868926
Wrong

>> No.9868970

>>9868828
>Pick randomly, and then switch
This would be hilarious if it didn't honest to God sound like how schizophrenics reason

>> No.9868990

the act of switching doesn't change anything
this thread is stupid

>> No.9869011

>>9868970
>This would be hilarious if it didn't honest to God sound like how schizophrenics reason
Mathematically speaking it's the optimal strategy.

>> No.9869033

When can you say you've truly chosen an envelope? If I consider taking the left one in my mind but opt for randomly taking right instead have I already made the switch, or will switching this time after tapping the envelope only count as switch?

>> No.9869038

>>9869033
There are 2 envelopes. Which do you pick?

>> No.9869053

>>9869033
Does it matter?

>> No.9869055

>>9869038
I stare intensely at number 1 but actually go for 2, I tap it slightly since I don't need to see what amount of cash is in there to know that switching is always beneficial, but I ponder the nature of switching as I am closing on 1 and actually come to conclusion I would've anti switched if I picked 1 and come back to open 2. Actually I tell the guy observing me that I chose 1 and 2 is my after switch choice and my final answer that I can't switch then open 2.

>> No.9869061

>>9869053
Since switching is beneficial as proven by very smart anons you gotta know when you're actually switching and when just backing out of switching.

>> No.9869063

>>9869061
I see

>> No.9869070

>>9868062
I keep switching envelopes until I have infinite money.

>> No.9869072

>>9869070
You don't get the money until you actually open one

>> No.9869074

>>9869072
OK, keep switching until I get a billion dollars.

>> No.9869078

>>9869074
Then you lose, because had you switched 1 more time, you would've gotten more than a billion

>> No.9869080

>>9869078
I'm a simple man of simple tastes, A billion is fine for me.

>> No.9869082

>>9868062
Wait wut it's 50/50 right? Either $20 or $5.
Don't tell me switching is the way to go here.

>> No.9869083

>>9869082
Why wouldn't you switch for an the other envelope that has higher expected value?

>> No.9869088

>>9869083
10$ is a lower expected value than 12.5$

>> No.9869090

>>9869083
(20+5)/2= 12.5
Ok so the mean and median are 12.5 but is that the same as the expected value?

>> No.9869092

>>9869090
>Ok so the mean and median are 12.5 but is that the same as the expected value?
Yes, (1/2)(10/2)+(1/2)(10*2) = 12.5.

>> No.9869095

>>9869090
I don't know. How much do you expect the other envelope to contain?

>> No.9869098

>>9869088
So you agree that switching is better?

>> No.9869133

>>9868428
>The paradox arises by conflating a pure gamble with a logical choice.
Thank you, I was trying to figure out how to put this into words

>> No.9869137

>>9869095
Equally 20 or 5 dollars
so it's a straight gamble, no info

>> No.9869147

>>9869137
So you expect the second envelope to contain $20 with a 50% probability and $5 with a 50% probability?

>> No.9869159
File: 56 KB, 191x315, 1465509499023.png [View same] [iqdb] [saucenao] [google]
9869159

>>9869147
Yes. Is that wrong?

>> No.9869165

>>9869159
Yes, then you would expect the second envelope to have $12.5, making switching better

>> No.9869168

>>9869165
But isn't that a non sequitur or some other fallacy?

>> No.9869169

>>9869168
Not if you think >>9869147 is true

>> No.9869179

>>9869169
hmm I thought there was something else to this....

>> No.9869185

>>9869179
Do you understand now?

>> No.9869347

Interesting problem. The misunderstanding largely just comes from poor phrasing of the problem. As presented, what SEEMS to be the question is “I give you an envelope with $10. If you choose, I can give you a different envelope that has equal chance of being $5 or $20. Do you switch?” However, that is NOT what’s being asked. Consider a similar question, “I have two envelopes where one envelope has twice as much money as the other. You pick one of the envelopes at random. Should you switch?” Obviously the answer is that it doesn’t matter if you switch because you chose at random, and choosing at random and the picking the other is equivalent to choosing at random. So in both cases, you’re choosing at random, so how could they differ? Given that, how could observing that the envelope you picked first had $10 possibly change anything? Obviously it couldn’t, but then where’s the intuition coming from? What you’re assuming is that if you pick an envelope and observe it has $10, the other envelope has equal odds of having $5 and $20. That’s wrong. Either we were presented with two envelopes with 5 and 10 dollars, or we were presented 2 envelopes with 10 and 20 dollars. There’s no probability there. The universe either was one way or was the other. The thing that had a 50 percent chance of happening was that we picked the one envelope from those pairs that had $10. The confusion is that we’re supposed to assume that, in every trial of this experiment, we pick the envelope with $10, and that the other envelope adjusts it’s value to be either half or double $10. Instead, there simple are two envelopes with two dollar amounts, and one of them happens to be $10. During different runs of the experiment, then, we have a 50 percent chance of picking the envelope with $10, and a 50 percent chance of picking the other envelope, but the other envelope’s value is fixed at either $5 or $20, depending which universe we live in.

>> No.9869352

>>9869011
"Pick randomly then switch" is an identical strategy to "Pick randomly"

>> No.9869355

>>9869347
How is it poor phrasing? What's a better way to phrase it?

>> No.9869358

>>9869347
Yours is the correct frequentist interpretation. A Bayesian could argue that the total amount in the envelopes is a random variable and give a prior on that value, which would invalidate the claim that the switch has 50% chance of doubling and 50% chance of halving (except for very specific observations of the envelope).

Under both interpretations, the "always switch" argument is wrong.

>> No.9869362

>>9869347
>>9869355
The similar question you bring up uses the same phrasing as the picture

>> No.9869493

>>9868421
>scale that so that env1 = 10
Retard

Try it again without scaling, and just put 1 and 2 in the envelopes

>> No.9869508

>>9868062
They wouldn't give away more money then they needed to, keep the 10

>> No.9869522

>>9869493
Huh?

>> No.9869552

I think the root of the paradox is that you stand to win more than you lose.

Something that befuddles me is two contestants (zero sum) vs two possible worlds (in which one world has twice as much money).

But they seem equivalent in practice. It's a gamble to double your money at the risk of halving. Expectedd value isn't valid here, because after only one trial you know if you got the good envelope or not: If you have a 10 and trade to get a 5 it will be clear that there never was a 20, but it was still worth a shot.

>> No.9869562

>>9868306
What strategy would you use to distribute the money to tip the odds in you're favor.

>> No.9869587

>>9868428
>>9869133
But the problem isn't actually completely random unless you someone divorce it from all meaningful context. If you opened an envelope you were presented IRL and you gained one dollar, it's probably more likely that the second envelope would contain a higher amount compared to if it contained 100 dollars. There's no way to present this problem in a way such that the hypothetical probability distribution selecting the amounts of money is arbitrary and 'not biased' to some distributions over others. Given this, knowing the amount of money on one of the envelops allow you to take a stab at how likely it is that the other envelope contains more, though this can't be done mathematically without creating a prior probability.

>> No.9869589

>>9868606
The fact that you focus on that issue, and ignore the problem of "what distributions the expectations are taken over" tells me all I need to know.

>> No.9869604

Ok this is the answer:
there are 2 strategies: ALWAYS switch and ALWAYS stay (the experiment is only performed once so if you consider the "amount of trials to infinity" to get the expected return you have to do the same every time.
You ALWAYS start with 10$.
10$ is the high or the low number and its the same every time.
Now:
SWITCH and 10 is HIGH:
5
SWITCH and 10 is LOW:
20
SWITCH on uniform distribution:
20/2+5/2=12.5

STAY and 10 is HIGH:
10
STAY and 10 is LOW:
10
STAY on uniform distribution:
10/2+10/2=10

therefore switching is better

>> No.9869679
File: 39 KB, 500x500, 1490302312175.jpg [View same] [iqdb] [saucenao] [google]
9869679

To anyone who believes switching is better, please consider this:

10 000 people are selected for the following test. In this test, we have an envelope that contains 10 dollars and another one that contains 20 dollars. Participants do not know how much each envelope contains, but they do know that one envelope contains double the amount contained by the other.

During the test, roughly half of all participants initially opens the envelope that contains 10 dollars (population A), while the other half opens the envelope that contains 20 (population B).

Among each of these populations, there are two subpopulations: the one that decides to switch envelopes after seeing the content of the first envelope (population 1), and the one that doesn't (population 2).

People among population A1 gain 10 dollars each by switching.
People among population A2 lose 10 dollars each by choosing not to switch.
People among population B1 lose 10 dollars each by switching.
People among population B2 gain 10 dollars each by choosing not to switch.

Now, let's make averages. Surprise-surprise: your chance of gaining or losing money by switching is statistically exactly the same as by choosing not to switch. These are facts.

Don't believe me? Test it yourself at home. Use envelopes whose contents are secretly decided by someone else after each iteration of the test, of course following the rule that one envelope must contain double the amount of the other. By switching 100 times and then by not switching 100 other times, you'll gain/lose the same amount of money statistically.

Your mathematical theory doesn't agree with these FACTS? Then either it is wrong somehow, or you used it the wrong way in a situation that doesn't truly fills its prerequisites.

>> No.9869719

>>9869679
Congratulations on dissecting a completely separate problem unrelated to the OP.

>> No.9869731

>>9868213
What? You have just the two envelopes. It's not like their contents are going to change at this point.

>> No.9869760

>>9869719
I really don't see how the problem is any different.

>> No.9869885

Why this fuck has nobody posted this?

http://rspa.royalsocietypublishing.org/content/465/2111/3309.full

>> No.9869991

>>9869362
The difference is that I don’t mention the dollar amount I observe in my similar question, where the OP mentions $10 specifically. Mentioning that you observe $10 is what I think introduces confusion, because there was really only a 50 percent chance you see $10, with the other 50 percent going to either $5 or $20 depending on the total amount of money involved. If you don’t mention any particular dollar amount and instead just say “We have two envelopes, one with X dollars and one with 2X dollars. You pick one at random. Should you switch?”, the answer becomes obvious. As others have noted, picking at random and switching is exactly the same thing as picking at random.

>> No.9869998

>>9868074
No it's not. You gain more if the second envelope is 20 than you lose if it is 5. If you want the second envelope to have the same expexted value it should be either 0 or 20, or have different probabilities.

>> No.9870001

>>9869991
>As others have noted, picking at random and switching is exactly the same thing as picking at random.
No it isn't. You get to make a new choice with new information.

>> No.9870051

>>9870001
Except you don’t know the total amount of money in envelopes, so knowing that you have $10 doesn’t tell you anything useful. Either you live in the world where the envelopes were 5 and 10 or you live in the world where the envelopes are 10 and 20. In both of those worlds, switching has no expected gain. Comparing between worlds doesn’t make sense because there’s no transition probability from one world to the other

>> No.9870070

>>9868819
He'll pick the other one

>> No.9870153

>>9868213
>switch again
>only two options
Summer is almost over

>> No.9870189

>>9869493
I already tested that if I don't scale, the value is the same for both strategies.
But that doesn't match the situation in the OP

>> No.9870197

>>9870153
Hypothetically, say you can. Shouldn't you switch again, A.K.A. stay with the envelope you first picked?

>> No.9870257

>>9870189
>the value is the same for both strategies
And what exactly do you think that says?

>> No.9870260

>>9870051
You dont need to gain information on a flip to profit. Why is this being parroted in the thread?

>> No.9870333

>>9870260
What are you talking about? If I pick one of the two envelopes at random and don’t look inside, then picking at random and switching is obviously an equivalent strategy to picking at random, because both are methods of picking at random. The only way it could possibly be different is if somehow gaining the information that there are $10 in the envelope I picked changes my strategy. I was pointing out that, even then, the information I gain isn’t useful information strategically and thus nothing changes. Switching is always a pointless strategy given the question being asked

>> No.9870430

>>9870333
What is instead of picking randomly, I always pick the switch one

>> No.9870445

>>9870333
Let's say you start with an unopened envelope A, and then switch to B. It's true that B has an expected value of 1.25A, since it has an equal probability of either doubling or halving the value of A. So far so good.

However, since the values in the envelopes are fixed, if you switch back to A, you're just undoing whatever operation you did by switching to B by definition. If you've doubled your winnings by switching to B, switching back to A can only halve what you're currently holding. Similarly, if B is only half of A, then you can only double your winnings by switching back to A. In either case, you've simply returned to your original value of A, and there is no advantage in sequential switching.

>> No.9870583

>>9870445
>since it has an equal probability of either doubling or halving the value of A

How do you conclude this? This completely ignores that the person who put the money in can't have just chosen an amount uniformly from the integers/reals and has to in some sense use a distribution that makes higher values of money more unlikely. You don't know how much they can actually afford a priori, but you can at least try to reason about what values are high enough that switching is less likely to give the more valuable envelope on seeing its value.

>> No.9870638

>>9870583
While counter-intuitive, there is no paradox if you just switch once, unopened or no. You calculation of the expected value of the envelope B as 1.25A is correct, so on average, it makes sense to switch. You can run a brute-force simulation of this for any value of A, and blindly picking B every time will get you an average winning of approximately 1.25A. I literally just threw one together in Excel in about 2 minutes, there's no problem with this first part.

>> No.9871176

>>9870333
The question says you have to choose whether to switch after seeing an envelope.

If you don't know the contents of the one you picked, it doesnt even matter because the other one still has a 50% chance to double.

So even in your incorrect interpretation you still switch. Switching is always better in this situation.

>> No.9871392

>>9870445
But it isn’t true that B has equal probability of halving or doubling A. Suppose I hand you two envelopes that I have filled with X dollars and 2X dollars. You pick one and open it and observe that it has Y dollars inside. Now you don’t know if Y is 2X or X, but you know each has equal odds. If Y=X and you switch, you gain X. If Y=2X and you switch, you lose X. Thus, switching has no value. As presented, that’s the problem, and that’s the solution. The problem that you’re solving is a DIFFERENT problem. In your case, you pick an envelope and observe Y dollars inside. Then, the person presenting you the envelopes puts either 2Y or Y/2 dollars in the other envelope and asks if you want to switch. Then you gain Y or lose Y/2, so you should switch, but the problem isn’t the same. Now, you might look back and say “but wait, I gained Y or lost Y/2 in the first problem too.” But those were different Y’s with respect to the fixed underlying X value. The problem is confusing as presented because you’re told that you observe $10 as though that were the thing that’s determined and it’s the value in the other envelope that’s probabilistic, but it’s the reverse. The value in the other envelope is fixed by whoever gave you the envelope. The thing that was probabilistic was that you happened to pick the one of the two envelopes that had $10 in it.

>> No.9871402

>>9871176
That literally makes no sense at all. Suppose I give you two envelopes where one has $5 and one has $10. I tell you that one envelope has twice as much as the other (a true statement, obviously). You pick at random and happen to get the envelope with $10. Despite this example fitting the problem statement exactly, there was NEVER a chance of you getting $20. Just because I tell you a fact about the relationship between the values of the envelopes doesn’t mean EVERY combination of envelopes for which that fact is true is equally likely.

>> No.9871406

>>9870638
But picking at random and switching is exactly the same as just picking at random. If you don’t look, then all I’ve said is that I pick at random one way, or pick at random another way. Since they’re both random, the expectated outcome has to be identical, but you’re claiming that somehow one method of random guessing is inherently better than another method of random guessing because somehow the universe magically knows that you “switched”, thereby duping the universe into creating more money out of nowhere

>> No.9871462

>>9868070
right except its negative $5 not positive $5

($20 - $5)/2 = $7.50

>> No.9871465

>>9868070
>>9871462
I'm retarded actually its

($10 - $5)/2 = $2.50

>> No.9871565

>>9870638
>You can run a brute-force simulation of this for any value of A, and blindly picking B

Then you are not actually simulating the problem. If your simulation is anything like the previous simulation posted in this thread, then you are 'normalizing' the value of the second envelope around the amount contained in the envelope you happened to open. This assumption is exactly why the simulation makes no sense, and this is what unrealistically biases the simulation so that 'always switching' 'wins'.

As I said before, to actually simulate the problem the two envelopes have to be *both* filled in from some distribution first (by selecting a single value and then making the other 2x that amount), *then* one of them needs to be selected at random from the two and opened before offering the choice to switch. When simulated like this 'always switching' will only provide an advantage if there is bias towards the lesser envelope being presented first.

Again, I implore you to read this: >>9869885

If you present a value, and make the other envelope randomly 2x or 0.5x the value of the presented envelope, then this is equivalent to being presented a specific envelope by the other person instead of choosing at random. But here, the problem is that there is nothing preventing the other person doing something like, say, deliberately presenting you the higher valued envelope, knowing you are more likely to switch.

The only way for this to really work the way you assume it would is for some amount of money to be put into an envelope, for the amount in the other envelope to be guaranteed to be determined as either double or half by coin flip, and for the initial envelope to then specifically be presented to you. Then switching really would be the best strategy. But this assumes tons of things that explicitly go against the initial problem statement.