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Envelopes paradox

Anonymous Sun 27 Oct 00:49:29 2019 No.11094480 View Reply Original Report

I don't understand how this is a paradox. Can someone explain what the paradox is?

Anonymous

Anonymous Sun 27 Oct 2019 00:52:25 No.11094489 Report

Quoted By:

>>11094480
is this supposed to be like the monty hall problem with a different win/lose scenario?

Anonymous

Anonymous Sun 27 Oct 2019 00:52:35 No.11094490 Report

Quoted By: >>11097661 >>11097965

>>11094480
the paradox is that you figure it would be a 50/50 chance of either you get some money or lose some, but it is better to switch since you'd be making $10 as opposed to losing $5 (the expected value of the switch is positive)

Anonymous

Anonymous Sun 27 Oct 2019 00:53:33 No.11094494 Report

Quoted By: >>11097569

>>11094480
That's simple: switch when the amount is odd.

Anonymous

Anonymous Sun 27 Oct 2019 00:54:50 No.11094499 Report

Quoted By:

This is better than the Monty Hall problem because it forcibly confronts you with its ambiguity.

Anonymous

Anonymous Sun 27 Oct 2019 00:56:17 No.11094504 Report

Quoted By: >>11094520 >>11095049 >>11095096

>>11094480
If you find $10, then according to the information given, the other envelope contains either $5 or $20. There is no way to discern which of these is the case. A naive expectation value calculation gives (($5-$10)+($20-$10))/2 = (-$5+$10)/2 = $2.50, which would seem to imply that switching to the other envelope is always profitable over many trials. That cannot be the case, however. The actual expectation value of the gain is zero.

Anonymous

Anonymous Sun 27 Oct 2019 01:03:34 No.11094520 Report

Quoted By: >>11094534

>>11094504
let's say over many trials it is 20 or 5 dollars and averages to $25 per every two, E(X) = 12.50 for one, that is definitely profitable

Anonymous

Anonymous Sun 27 Oct 2019 01:08:54 No.11094534 Report

Quoted By: >>11095065 >>11095074

>>11094520
That's the problem. Only one of the two situations is true. Either you have the envelope with value 2x, or the envelope with value x. If you switch from x to 2x, you gain x, and if you switch from 2x to x, you lose x. So the expectation value is ((x-2x)+(2x-x))/2 = (-x+x)/2 = 0.

Anonymous

Anonymous Sun 27 Oct 2019 01:44:27 No.11094630 Report

Quoted By:

>>11094480
There isn't. It's just extremely basic probability.

Anonymous

Anonymous Sun 27 Oct 2019 01:49:39 No.11094649 Report

Quoted By: >>11094749

>>11094480
There isn't enough information to answer.
You would need to know the probability distribution that was used to select the dollar amounts in the envelopes.

The expected profit from switching is [P(5,10)*(-5) + P(10,20)*(10)] / [P(5,10) + P(10,20)]

Anonymous

Anonymous Sun 27 Oct 2019 02:24:39 No.11094749 Report

Quoted By: >>11094788 >>11095063

>>11094649
bruh, there's two envelopes, one has the normal amount, one has the double amount, its 50/50 which one you get

Anonymous

Anonymous Sun 27 Oct 2019 02:41:42 No.11094788 Report

Quoted By: >>11095000 >>11097900 >>11105019

>>11094749
ITT idiots who don't understand The Bayes Theorem.

You cannot have a flat prior from 0-infinity in the real world.

Anonymous

Anonymous Sun 27 Oct 2019 04:15:35 No.11094959 Report

Quoted By:

>>11094480
The average would be a gain of $2.5 over many trials.

Anonymous

Anonymous Sun 27 Oct 2019 04:41:57 No.11095000 Report

Quoted By: >>11095138

>>11094788
>ITT idiots who don't understand The Bayes Theorem.
All reputable statisticians reject Bayes "theorem".

Anonymous

Anonymous Sun 27 Oct 2019 05:03:23 No.11095038 Report

Quoted By:

>>11094480
This is kind of retarded but let me try:
>You technically never lose money, only get less than you would have otherwise.
>Other envelope could be the larger one ($20) or the smaller one ($5)
>You could gain $10 or "lose" $5
>So on average you would expect to gain $5 by switching to the other envelope.

Anonymous

Anonymous Sun 27 Oct 2019 05:10:33 No.11095049 Report

Quoted By:

>>11094504
>the expectation of gain in 0
Am I not getting the the money in either of the envelopes? How can I even lose money? Someone gave me 2 envelopes, I didn’t have those before.

Anonymous

Anonymous Sun 27 Oct 2019 05:18:23 No.11095063 Report

Quoted By: >>11097621

>>11094749
>its 50/50 which one you get
Suppose I prepared 100 pairs of envelopes with one pair being (10,20) and the other 99 pairs being (5,10) then I pick a pair at random and present OP's game to you.
Obviously the chance of the pair being (5,10) is way higher than the chance of the pair being (10,20).

Anonymous

Anonymous Sun 27 Oct 2019 05:18:56 No.11095065 Report

Quoted By:

>>11094534
You don't understand probability.

Anonymous

Anonymous Sun 27 Oct 2019 05:21:18 No.11095074 Report

Quoted By: >>11098534

>>11094534
the value of "x" in the first example is smaller than the value of "x" in the second example, though. They are not equivalent.

Anonymous

Anonymous Sun 27 Oct 2019 05:32:52 No.11095093 Report

Quoted By:

>>11094480
The paradox is in the labeling. To correct the phrasing of the problem, one envelope has x dollars, and the other envelope has 2*x dollars. You can choose to pay the cost of one of the envelopes to gain what is in the other envelope (the problem says you can lose money, but it doesn’t explain how, so i’m filling in the gaps). So, you can either gain x dollars, or lose x dollars, depending on your choice. However, you don’t have enough information to judge if the $10 is a 2x, or an x, since you don’t know what the distribution of x is, you don’t have enough information to compute the odds ratio of x = 5 vs x = 10. The problem tries to suggest that the ratio is 50 - 50, but that is misleading, instead, you need to compute the probability of x = 5 given x = 5 or x = 10, and x = 10 given x = 5 or x = 10.

Anonymous

Anonymous Sun 27 Oct 2019 05:36:31 No.11095096 Report

Quoted By: >>11097903

>>11094504
According to the information given, you cant even compute expected value. That is the solution of the paradox.

Anonymous

Anonymous Sun 27 Oct 2019 05:42:38 No.11095107 Report

Quoted By: >>11095117

This is just pot odds. You have a 50% to gain 10 dollars if you switch and a 50% to lose 5 dollars. That's a bet you take every time.

Anonymous

Anonymous Sun 27 Oct 2019 05:49:27 No.11095117 Report

Quoted By: >>11095120

>>11095107
Just because you had a 50 - 50 chance of picking the smaller envelope at first, doesn’t mean the probabilities are still the same after you learn what was in your envelope. After you learn that your envelope has $10 in it, that could be enough to tell you that the other envelope is guaranteed to contain $5, because your distribution may not include $20.

Anonymous

Anonymous Sun 27 Oct 2019 05:50:29 No.11095120 Report

Quoted By: >>11095129

>>11095117
lol

Anonymous

Anonymous Sun 27 Oct 2019 05:52:39 No.11095129 Report

Quoted By:

>>11095120
I can tell you haven’t seen 21

Anonymous

Anonymous Sun 27 Oct 2019 05:58:27 No.11095138 Report

Quoted By:

>>11095000
Based frquentist.

Anonymous

Anonymous Sun 27 Oct 2019 20:46:46 No.11096837 Report

Quoted By:

>>11094480
People mix up two scenarios:
There is $10 and $20 in the envelopes, so you gain $10 for switching
There is $10 and $5 in the envelopes, so you lose $5 by switching.
Apply the same problem after every switch. It of course doesn't really work, since the only option once you switch is either you have now $20, so you lose $10 by switching back, or you now have $5 and you gain $5 by switching back, so there is really no paradox.

Anonymous

Anonymous Sun 27 Oct 2019 20:47:52 No.11096843 Report

Quoted By: >>11098530

>>11094480
People mix up two scenarios:
There is $10 and $20 in the envelopes, so you gain $10 for switching
There is $10 and $5 in the envelopes, so you lose $5 by switching.
Apply the same reasoning after every switch. It of course doesn't really work, since the only option once you switch is either you have now $20, so you lose $10 by switching back, or you now have $5 and you gain $5 by switching back, so there is really no paradox.

Anonymous

Anonymous Mon 28 Oct 2019 00:04:39 No.11097489 Report

Quoted By: >>11097915

>>11094480
How much money can you expect to win or lose?

You will either get 5 or 20
in case of 5 you lose 5
in case of 20 you gain 10

The average change in amount gained can be expressed as
(-5 + 10) / 2 = 2.5

The average change in difference from your previous amount, in effect absolute amount gained can be expressed as
(|-5| + |10|) / 2 = 7.5

There is no paradox. The phrasing is simply unclear, could be either of those.

The amount won or lost is 5 or 10
The difference from prior prize is -5 or (+)10
Logically we expect the latter, but consider what is meant by:
How much money can you expect to lose?
If you answer -5, what does that mean; losing a negative amount is winning.

If only asked how much can be expected to win, then -5 and 10 is correct.
But when asked how much is expected to win (+10) or lose (+5) is correct, since you lose a positive amount of money.

The amount of 7.5 is the amount won or lost, on average.
The problem (or paradox if you like) is that this number makes no sense unless you realize that it is the average of the absolute value of delta.

Anonymous

Anonymous Mon 28 Oct 2019 00:33:21 No.11097569 Report

Quoted By: >>11100324

>>11094494
>Get 11 bucks
>Switch
>Now you have 5 bucks and 50 cents
Fucking genius

Anonymous

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Anonymous Mon 28 Oct 2019 00:49:14 No.11097612 Report

Quoted By: >>11098505 >>11100256

AAAAAAAA
THIS IS TWISTING MY BRAIN
ITS LIKE YOURE TRAPPED FOR ETERNITY SWITCHING BETWEEN ENVELOPES BECAUSE THE GRASS IS ALWAYS GREENER ON THE OTHER SIDE
WHAT KIND OF CYCLICAL HELL IS THIS

Anonymous

Anonymous Mon 28 Oct 2019 00:53:13 No.11097621 Report

Quoted By:

>>11095063
>b-but, you're ignoring base rates!
Fuck figuring that shit out, it's 50/50. Either it happens or it don't

Anonymous

Anonymous Mon 28 Oct 2019 01:06:32 No.11097661 Report

Quoted By: >>11097803 >>11097865 >>11104736 >>11107446

>>11094490
>but it is better to switch since you'd be making $10 as opposed to losing $5 (the expected value of the switch is positive)
How can this explanation be squared with this one:
>one envelope contains N, the other 2N
>switching can result in either gaining N or losing N

Anonymous

Anonymous Mon 28 Oct 2019 01:19:12 No.11097703 Report

Quoted By:

About $7. Did it in my head though

Anonymous

Anonymous Mon 28 Oct 2019 01:59:21 No.11097803 Report

Quoted By:

>>11097661
>How can this explanation be squared with this one:
>>one envelope contains N, the other 2N
>>switching can result in either gaining N or losing N
What needs squaring?

Anonymous

Anonymous Mon 28 Oct 2019 02:17:35 No.11097865 Report

Quoted By: >>11097885

>>11097661
Neither explanation contradicts the other, one says:
>you either gain $10 by switching from $10 to $20, or lose $5 by switching from $10 to $5
the other says:
>you either gain $10 by switching from $10 to $20, or lose $10 by switching from $20 to $10

Anonymous

Anonymous Mon 28 Oct 2019 02:24:09 No.11097885 Report

Quoted By: >>11098538

>>11097865
the other says*:
>(if N=10 then) you either gain $10 by switching from $10 to $20, or lose $10 by switching from $20 to $10
>(if N=5 then) you either gain $5 by switching from $5 to $10, or lose $5 by switching from $10 to $5

But you've already found that your envelop contains $10, hence we eliminate two possibilities:
>(if N=10 then) you gain $10 by switching from $10 to $20
>(if N=5 then) you lose $5 by switching from $10 to $5

So the expected gain is ($10 - $5)/2 = $2.50, so switching is the optimal strategy.

Anonymous

Anonymous Mon 28 Oct 2019 02:24:59 No.11097889 Report

Quoted By:

>>11094480
uh
2.5
.5 prob of -5 + .5 prob of +10 = .5*(-5+10) = .5(5) = 2.5

Anonymous

Anonymous Mon 28 Oct 2019 02:26:16 No.11097897 Report

Quoted By:

>>11094480
If I switch and it has 5, I lost 5 in the switch.
If I switch and it has 20, I gained 10 in the switch.
It makes sense to switch since the potential gain is greater than the potential loss.

Anonymous

Anonymous Mon 28 Oct 2019 02:26:42 No.11097900 Report

Quoted By:

>>11094788
>he's never heard of improper priors

Anonymous

Anonymous Mon 28 Oct 2019 02:28:11 No.11097903 Report

Quoted By:

>>11095096
was gonna say this
but it seems implied that there's no logical explanation that they would use 5's or 20's since its a fake ass scenario so the assumption is it isn't based on anything and thus the chance is the same

Anonymous

Anonymous Mon 28 Oct 2019 02:31:42 No.11097911 Report

Quoted By:

>>11094480
The "paradox" is just that the wording is ambiguous.

Anonymous

Anonymous Mon 28 Oct 2019 02:32:00 No.11097913 Report

Quoted By:

>>11094480
(20+5)/2

its not a paradox really

Anonymous

Anonymous Mon 28 Oct 2019 02:33:13 No.11097915 Report

Quoted By: >>11098477

>>11097489
>average of the absolute value of delta.
so you're saying you can expect to lose -2.5 on average or expect to gain 2.5 on average (means the same thing)
the averages are based on a data set consisting of both gains and losses that aren't separated in any way by the language of the word problem. Your interpretation has added complexity IMO

Anonymous

Anonymous Mon 28 Oct 2019 03:01:08 No.11097965 Report

Quoted By: >>11098715

>>11094490
How is basic expected value math a paradox though?

Anonymous

Anonymous Mon 28 Oct 2019 09:38:07 No.11098477 Report

Quoted By:

>>11097915
Yes, but the wording of the problem does not ask for average loss or average gain, it asks how much do you expect to [gain or lose] on average, it asks about the change in amount, not the direction of change.

An average taking in account the direction of change asks what the average of the difference in end results would be. Here they did not ask that. They asked about the average volatility of the system. (it is 7.5) By what amount can you expect your result to change.

Anonymous

Anonymous Mon 28 Oct 2019 09:59:06 No.11098505 Report

Quoted By: >>11098530 >>11098536 >>11105046

>>11097612
that's the paradox.
as long as you don't open one envelope expected value leads to continued switching never settling on any envelope. the moment you open one you're expected to switch one last time. i don't know how people here don't get it, dumb highschoolers

Anonymous

Anonymous Mon 28 Oct 2019 10:19:54 No.11098530 Report

Quoted By:

>>11098505
No.>>11096843

Anonymous

Anonymous Mon 28 Oct 2019 10:23:21 No.11098532 Report

Quoted By: >>11100357 >>11100484 >>11100512 >>11100513 >>11100547

>>11094480
half a chance you randomly selected the envelope which if switched would win x and half a chance you randomly picked the envelope if switched would lose x
(1/2*x)+(1/2*(-x))=0
switching is of no benefit

Anonymous

Anonymous Mon 28 Oct 2019 10:24:24 No.11098534 Report

Quoted By: >>11100539

>>11095074
they are

Anonymous

Anonymous Mon 28 Oct 2019 10:26:02 No.11098536 Report

Quoted By:

>>11098505
But there are only two envelopes. You have to settle for the second envelope.

Anonymous

Anonymous Mon 28 Oct 2019 10:28:02 No.11098538 Report

Quoted By: >>11098542

>>11097885
why would switching be an optimal strategy, you randomly selected one envelope and the values inside are fixed, how is that logical in any ways

Anonymous

Anonymous Mon 28 Oct 2019 10:29:55 No.11098542 Report

Quoted By: >>11098543 >>11098546 >>11098547

>>11098538
Your potential gain is greater (in dollar amount) than your potential loss.

Anonymous

Anonymous Mon 28 Oct 2019 10:32:45 No.11098543 Report

Quoted By: >>11098552

>>11098542
if the values inside the envelope would magically change when you swap them.
just think about it in real terms, what in the hell is the point in keeping going back and forth between the two envelopes

Anonymous

Anonymous Mon 28 Oct 2019 10:36:14 No.11098546 Report

Quoted By: >>11098558 >>11098565

>>11098542
But only the first time.

You see X dollars in one envelope.

So, either there is 2X in the second envelope, and you gain X for switching to the second envelope, and you lose X by switching back.

Or, there is 1/2X in the second envelope, and you lose 1/2X by switching and you gain 1/2X by switching back.

So it's X gain vs 1/2X loss for the first switch and X loss vs 1/2x gain for the second switch, that is you lose by switching back.

Anonymous

Anonymous Mon 28 Oct 2019 10:36:21 No.11098547 Report

Quoted By: >>11098555 >>11098571

>>11098542
one envelopes contains 10 the other 20
you randomly selected the 10 one and switch. great you won 10
you randomly selected the 20 one and switch. darn it you lost 10

Anonymous

Anonymous Mon 28 Oct 2019 10:36:41 No.11098549 Report

Quoted By:

>>11094480
You stand to gain $10 or lose $5 and as far as you know it's 50/50 odds. There's nowhere to get confused. The number in the first envelope you choose tells you the baseline for the game. By switching you stand to gain 1.5x of that baseline on average.

Of course, I doubt some impoverished graduate student doing the study is going to be shoving $20's in there, so I'll happily take my lunch money.

Anonymous

Anonymous Mon 28 Oct 2019 10:37:46 No.11098552 Report

Quoted By: >>11098565

>>11098543
You only switch once. Your first envelope has a set amount. great. You could be not greedy and settle for that. On average you would lose 2.5 dollars. If you switch once, you are in effect gambling for a coin toss where you either lose 5 bucks or win 10. The average of that bet is a 2.5 dollar win. You should take that bet. You are not guaranteed to win, it is still 50/50, but since the prize money is not proportional to the risk, you should gamble.

Anonymous

Anonymous Mon 28 Oct 2019 10:39:31 No.11098555 Report

Quoted By: >>11098561 >>11098588 >>11098595

>>11098547
You could just as easily be playing a game with $20 and $40. If the denominations are arbitrary, then it's a winning proposition to switch every time.

Anonymous

Anonymous Mon 28 Oct 2019 10:40:30 No.11098558 Report

Quoted By: >>11098565 >>11098588 >>11098595

>>11098546
Yes it is a 50/50 chance of losing or winning, but the prize of winning is greater than the penalty for losing so on average you win.

Anonymous

Anonymous Mon 28 Oct 2019 10:41:04 No.11098561 Report

Quoted By:

>>11098555
he money in the envelope doesn't magically increase dumbass

Anonymous

Anonymous Mon 28 Oct 2019 10:42:06 No.11098565 Report

Quoted By: >>11098574

>>11098552
>>11098546
>>11098558
WRONG
JUST KYS

Anonymous

Anonymous Mon 28 Oct 2019 10:44:09 No.11098571 Report

Quoted By: >>11098583

>>11098547
You do not know the amount of money in the second envelope. You know that it is either half or double the amount of the envelope you just opened. You gain more by doubling than you lose by halving. Your potential gain is greater than your potential loss.

Anonymous

Anonymous Mon 28 Oct 2019 10:45:19 No.11098574 Report

Quoted By:

>>11098565
Nice argument friend. You are a true scholar.

Anonymous

Anonymous Mon 28 Oct 2019 10:50:36 No.11098583 Report

Quoted By: >>11098692

>>11098571
You not knowing doesn't mean the money in reality magically increases. the money is fixed, someone either put 5 and 10 or 10 and 20, switching doesn't cause you to teleport into a parallel universe where 5 turns into 20. try it yourself get two envelope put money in them.

Anonymous

Anonymous Mon 28 Oct 2019 10:52:17 No.11098588 Report

Quoted By:

>>11098555
There can't be $40 in the first envelope when you know it's $10 there, dumbass.
>>11098558
same, only with 4X and X instead.

Anonymous

Anonymous Mon 28 Oct 2019 10:54:47 No.11098595 Report

Quoted By: >>11100345

>>11098555
>>11098558
Imagine the reverse scenario. You knwo there is twice as much money in one envelope as in the other one. You are given one envelope and you are shown there is $10 in the other. Will you take the certain $10, or will you keep the closed envelope?

Anonymous

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Anonymous Mon 28 Oct 2019 11:04:36 No.11098610 Report

Quoted By:

>>11094480
Ashe's conformity experiment. OP posts a question, to which the answer is obvious. Confederates in the thread then post the wrong answer.
The paradox is that the social pressure to maintain conformity outstrips hard logic.

The correct answer is C.

Anonymous

Anonymous Mon 28 Oct 2019 11:07:33 No.11098612 Report

Quoted By: >>11098708 >>11098755 >>11100341 >>11100789 >>11105092

Why are brainlets so attracted to this? Why do people still argue that there is a gain by switching? You can literally just google the two envelopes problem and discover you’re retarded

Anonymous

Anonymous Mon 28 Oct 2019 11:42:03 No.11098669 Report

Quoted By: >>11098708 >>11098755 >>11100348 >>11100402

Always switching should have the same outcome as never switching, this isn't like monty hall

Anonymous

Anonymous Mon 28 Oct 2019 11:54:26 No.11098692 Report

Quoted By: >>11098707 >>11098755

>>11098583
>either put 5 and 10 or 10 and 20
Exactly. That is why switching once will make you a profit on average. You either gain -5 or gain 10. Your average gain is 2.5 you should never keep the first envelope.

Anonymous

Anonymous Mon 28 Oct 2019 12:03:40 No.11098707 Report

Quoted By: >>11098710 >>11098755

>>11098692
never reply to me again ugly nigger

Anonymous

Anonymous Mon 28 Oct 2019 12:04:02 No.11098708 Report

Quoted By: >>11105014

>>11098612
>>11098669

How can you not get it. Once you have one open envelope with 10 bucks in it, you KNOW the other envelope has an equal chance to either contain 5 or 20.
Expected average value when switching:
0.5 * -5 + 0.5 * 20 = 12.5
That is more than ten. You should pick the second envelope.

You can only do this once. The amounts in the envelopes do not magically change. No-one is saying that. When you have one open envelope the correct action is always to switch it, (you can only do this once since there is only two envelopes.)

Anonymous

Anonymous Mon 28 Oct 2019 12:05:39 No.11098710 Report

Quoted By:

>>11098707
Nice argument friend.

Anonymous

Anonymous Mon 28 Oct 2019 12:08:18 No.11098715 Report

Quoted By: >>11098720 >>11098731

>>11097965
Imagine instead of one person playing, its two, they get an envelope each. Should they both switch? Should they both switch if they'd chosen the opposite envelope in the first place? How can both envelopes have a higher EV than the other?

Anonymous

Anonymous Mon 28 Oct 2019 12:12:47 No.11098720 Report

Quoted By: >>11098735

>>11098715
Yes, in any instance when this game is played the expected value of the second envelope is higher.
The probability to lose is 0.5 the effect of losing is x/2 The probability to win is 0.5 the effect of winning is 2x
0.5 * x/2 + 0.5 * 2x =
.25x + x = 1.25x

You always gain 25% of switching ON AVERAGE.

Anonymous

Anonymous Mon 28 Oct 2019 12:15:15 No.11098723 Report

Quoted By:

>>11094480
7.5 money

Anonymous

Anonymous Mon 28 Oct 2019 12:25:57 No.11098731 Report

Quoted By:

>>11098715
Either you have an envelope with $x, and you gain $x by switching. Or you have an envelope with $2x, and you lose $x by switching.

Anonymous

Anonymous Mon 28 Oct 2019 12:30:13 No.11098735 Report

Quoted By: >>11098794 >>11100338

>>11098720
Yes I understand how EV works. You aren't getting why its a paradox. You open an envelope, you see it has 'a value' in it, therefore you should switch. In that case, considering you know both envelopes have 'a value' in them prior to opening, the same logic should apply prior to opening, making infinitely switching the best strategy.

Anonymous

Anonymous Mon 28 Oct 2019 12:30:25 No.11098737 Report

Quoted By: >>11098739

POLL TIME GUYS
LET'S SETTLE THIS ONCE AND FOR ALL
http://www.strawpoll.me/18857505

Anonymous

Anonymous Mon 28 Oct 2019 12:31:46 No.11098739 Report

Quoted By: >>11098741

>>11098737
Scientific fact can not be settled by popular vote.

Anonymous

Anonymous Mon 28 Oct 2019 12:36:18 No.11098741 Report

Quoted By:

>>11098739
But let's try anyway:
https://www.strawpoll.me/18857522

Anonymous

Anonymous Mon 28 Oct 2019 12:42:37 No.11098746 Report

Quoted By: >>11098751

>>11094480
The "paradox" is that "on average" is meaningless here.

Anonymous

Anonymous Mon 28 Oct 2019 12:46:56 No.11098751 Report

Quoted By:

>>11098746
I'm curious, the average of what numbers are you talking about, [-5, 10] or [5, 10]?

Anonymous

View Same Google iqdb SauceNAO 1565668053161.gif, 2MiB, 420x420

Anonymous Mon 28 Oct 2019 12:49:35 No.11098755 Report

Quoted By:

>>11098612
>>11098669
>>11098692
>>11098707
>>11098707
And as soon as the right answer is posted, it gets slid off the front page

Anonymous

Anonymous Mon 28 Oct 2019 13:08:07 No.11098794 Report

Quoted By: >>11098821

>>11098735
You only switch one time. After you have switched the situation is different because you already know the amount in each envelope. If you get to switch again, you just pick the the envelope with more money. It is not a paradox that the same probabilities do not apply when you have a differing amount of information. Once you have switched and know the amount in each envelope, the game becomes meaningless.

Anonymous

Anonymous Mon 28 Oct 2019 13:21:26 No.11098821 Report

Quoted By: >>11098833 >>11098853 >>11098951 >>11100332

>>11098794
You aren't getting any information from opening the first envelope, you've implicitly shown this yourself by using X rather than 10 in your EV calculation, the value itself is irrelevant. You already know the envelope has X in it before opening.

If offered the chance of switching prior to opening the envelope would you take it? If not, why not? After all X*2 is greater than X/2.

Anonymous

Anonymous Mon 28 Oct 2019 13:25:33 No.11098833 Report

Quoted By: >>11100504 >>11100520 >>11100532

>>11098821
Or to put it another way, is there a positive value for X where upon opening, you wouldn't switch? If no, opening doesn't matter.

Anonymous

Anonymous Mon 28 Oct 2019 13:33:57 No.11098853 Report

Quoted By: >>11098866

>>11098821
>If offered the chance of switching prior to opening the envelope would you take it?

That is not really switching, in effect I would be selecting one envelope.

Anonymous

Anonymous Mon 28 Oct 2019 13:40:15 No.11098866 Report

Quoted By: >>11099230 >>11100236

>>11098853
You aren't gaining any information from opening it though. Its exactly the same scenario just with X instead of a known value.

Anonymous

Anonymous Mon 28 Oct 2019 14:11:23 No.11098951 Report

Quoted By: >>11098958

>>11098821
>If not, why not? After all X*2 is greater than X/2.
Because that isn't what's going on. Either there is X in your envelope, and you gain X from switching. Or there is 2X in your envelope and you lose X from switching. You cannot write is as 2X vs 1/2X because the X would have a different value in each of the two cases.

e.g. if there is $10 and $20 in the envelopes, then in first case, X would equal $10 and in the second case X would equal $20.

Anonymous

Anonymous Mon 28 Oct 2019 14:16:13 No.11098958 Report

Quoted By:

>>11098951
I know that, I'm trying to explain why the simple EV calculation resulting in a 1.25x gain from switching is incoherent. I don't think switching gains you anything.

Anonymous

Anonymous Mon 28 Oct 2019 14:18:59 No.11098962 Report

Quoted By:

>>11094480
you can expect to loose 5$ or gain 10$

Anonymous

Anonymous Mon 28 Oct 2019 15:47:50 No.11099159 Report

Quoted By: >>11099180 >>11099217

>>11094480
gain $10 lose $5, what kind of dumbass wouldn't just pick up the one that feels heaviest though

Anonymous

Anonymous Mon 28 Oct 2019 15:55:18 No.11099180 Report

Quoted By: >>11099182

>>11099159
>A $20 bill weighs twice as much as a $10 bill
This is bait isn’t it

Anonymous

Anonymous Mon 28 Oct 2019 15:56:09 No.11099182 Report

Quoted By: >>11099217

>>11099180
>2 $10 bills vs 1 $10 bill

Anonymous

Anonymous Mon 28 Oct 2019 16:02:26 No.11099195 Report

Quoted By:

People who say you should switch must be shell shook by monty hall.

Anonymous

Anonymous Mon 28 Oct 2019 16:14:33 No.11099217 Report

Quoted By:

>>11099182
>>11099159
>He's able to tell the difference in weight between one bill and two bills.
Wew, we can't compete against you, you balance machine.

Anonymous

Anonymous Mon 28 Oct 2019 16:21:53 No.11099230 Report

Quoted By: >>11099810

>>11098866
This is the crux of the "paradox".
Even if the value isn't important, knowing the value is. The system only changes at the point of knowing, regardless of the value.

Anonymous

Anonymous Mon 28 Oct 2019 20:32:45 No.11099810 Report

Quoted By:

>>11099230
no it doesn't

Anonymous

View Same Google iqdb SauceNAO anon_wallpaper.jpg, 237KiB, 2306x913

Anonymous Mon 28 Oct 2019 20:35:52 No.11099816 Report

Quoted By:

>>11094480
twice as much or half as much, stupid question

Anonymous

Anonymous Mon 28 Oct 2019 22:28:22 No.11100236 Report

Quoted By: >>11100504

>>11098866
>You aren't gaining any information from opening it though.
How is the amount of money in the envelope not information?

Anonymous

Anonymous Mon 28 Oct 2019 22:32:39 No.11100255 Report

Quoted By:

If the outcomes are equally as likely, you're expected to gain 1/4 of the known value by switching, on average

Anonymous

Anonymous Mon 28 Oct 2019 22:32:52 No.11100256 Report

Quoted By:

>>11097612
What does this mean for life choices

Anonymous

Anonymous Mon 28 Oct 2019 22:41:11 No.11100292 Report

Quoted By: >>11100297 >>11100310 >>11100316

This is now a brainlet containment thread.

You may only leave once you understand that the probability of the other envelope containing $20 is not equal to the probability that it contains $5.

Anonymous

Anonymous Mon 28 Oct 2019 22:42:51 No.11100297 Report

Quoted By:

>>11100292
>You may only leave once you understand that the probability of the other envelope containing $20 is not equal to the probability that it contains $5.
Why's that?

Anonymous

Anonymous Mon 28 Oct 2019 22:44:57 No.11100308 Report

Quoted By:

It isn´t a paradox, it just an experiment that measures risk-aversion

Anonymous

Anonymous Mon 28 Oct 2019 22:45:30 No.11100310 Report

Quoted By: >>11101604

>>11100292
Obviously. The $20 envelope would have a higher than that of the $5 envelope because it’s desired more and that desire increases expectation of the $20 there by increasing the probability of the envelope containing $20.

Anonymous

Anonymous Mon 28 Oct 2019 22:46:56 No.11100316 Report

Quoted By:

>>11100292
Well, what's P($20) then?
If they're not equal, there must still be a relation between the two probabilities

Anonymous

Anonymous Mon 28 Oct 2019 22:50:15 No.11100324 Report

Quoted By:

>>11097569
>1,100 units of currency is odd
XD

Anonymous

Anonymous Mon 28 Oct 2019 22:54:12 No.11100332 Report

Quoted By:

>>11098821
>You aren't getting any information from opening the first envelope, you've implicitly shown this yourself by using X rather than 10 in your EV calculation, the value itself is irrelevant.
How is the amount of money in the envelope not information?

Anonymous

Anonymous Mon 28 Oct 2019 22:55:49 No.11100338 Report

Quoted By: >>11100504

>>11098735
>In that case, considering you know both envelopes have 'a value' in them prior to opening, the same logic should apply prior to opening, making infinitely switching the best strategy.
Wrong, because each envelope has different 'values'.

Anonymous

Anonymous Mon 28 Oct 2019 22:57:11 No.11100341 Report

Quoted By:

>>11098612
>You can literally just google the two envelopes problem and discover you’re retarded
What makes you think the no-switch proponents on google are any less wrong than the no-switch proponents in this thread?

Anonymous

Anonymous Mon 28 Oct 2019 22:57:41 No.11100342 Report

Quoted By: >>11100504

The information given by opening one envelope doesn't matter because regardless of what the original envelope is, you'll always get an average increase in value by 25% by switching once

Anonymous

Anonymous Mon 28 Oct 2019 22:58:16 No.11100345 Report

Quoted By:

>>11098595
>Will you take the certain $10, or will you keep the closed envelope?
Keep the closed envelope, since you expect to gain by switching.

Anonymous

Anonymous Mon 28 Oct 2019 22:59:17 No.11100348 Report

Quoted By:

>>11098669
>Always switching should have the same outcome as never switching, this isn't like monty hall
Never switching always has the outcome of $10, switching has the outcome of either $5 or $20.

Anonymous

Anonymous Mon 28 Oct 2019 23:02:48 No.11100357 Report

Quoted By: >>11100484

>>11098532
>half a chance you randomly selected the envelope which if switched would win x and half a chance you randomly picked the envelope if switched would lose x
Wrong: half a chance you randomly selected the envelope which if switched would win $10 and half a chance you randomly picked the envelope if switched would lose $5

Anonymous

View Same Google iqdb SauceNAO .png, 14KiB, 304x133

Anonymous Mon 28 Oct 2019 23:14:01 No.11100392 Report

Quoted By:

Anonymous

Anonymous Mon 28 Oct 2019 23:16:02 No.11100402 Report

Quoted By: >>11100432

>>11098669
>this isn't like monty hall
Let's fix that.

You have THREE envelopes containing $X, $2X, $4X, and a host who knows everything.
You pick and open one envelope, revealing the amount.
If you picked $X or $4X, the host opens the $2X envelope.
If you picked $2X, the host opens a random envelope.
You're then offered a choice to switch to the unopened envelope.
Do you?

Anonymous

View Same Google iqdb SauceNAO file.png, 8KiB, 405x133

Anonymous Mon 28 Oct 2019 23:29:43 No.11100432 Report

Quoted By:

>>11100402
That's not akin to the original problem as the average delta is 0 in your case, so switching has the exact same outcomes, mirrored

Anonymous

Anonymous Mon 28 Oct 2019 23:48:47 No.11100484 Report

Quoted By: >>11100512

>>11100357
wrong refer to>>11098532

Anonymous

Anonymous Mon 28 Oct 2019 23:53:54 No.11100504 Report

Quoted By: >>11100520 >>11100532

>>11100236
Because it doesn't in any way help you decide your strategy. You know from the beginning that one envelope contains double the value of the other, your goal is to maximise the probability of choosing the larger value. After opening an envelope, does knowing that value give you any information you can use to inform your choice? No.

If you believe that switching provides you with on average a 1.25x gain, then it does that without ever needing to see the initial amount. See >>11098833. Therefore given the opportunity to switch prior to opening, you should switch, and given that the situation is symmetric, you should switch again, and again, infinitely.

>>11100338
See >>11098833

>>11100342
>you'll always get an average increase in value by 25% by switching once
No you don't.

Anonymous

Anonymous Mon 28 Oct 2019 23:57:28 No.11100512 Report

Quoted By: >>11100513

>>11100484
>wrong refer to>>11098532
Hint: After opening the envelope, you eliminate one the two possibilities you listed (winning x or losing x).

Anonymous

Anonymous Mon 28 Oct 2019 23:58:45 No.11100513 Report

Quoted By:

>>11100512
wrong refer to>>11098532 (You)

Anonymous

Anonymous Mon 28 Oct 2019 23:59:16 No.11100520 Report

Quoted By: >>11100528 >>11100530

>>11100504
>See >>11098833
>is there a positive value for X where upon opening, you wouldn't switch?
No.

>If no, opening doesn't matter.
Correct, in the sense that I would switch to the other envelope once irregardless of whether I opened the first envelop.

Anonymous

Anonymous Tue 29 Oct 2019 00:01:08 No.11100528 Report

Quoted By: >>11100537

>>11100520
Would you switch prior to opening?

Anonymous

Anonymous Tue 29 Oct 2019 00:01:16 No.11100530 Report

Quoted By:

>>11100520
This anon is right

Anonymous

Anonymous Tue 29 Oct 2019 00:01:49 No.11100532 Report

Quoted By: >>11100542

>>11100504
>If you believe that switching provides you with on average a 1.25x gain, then it does that without ever needing to see the initial amount. See >>11098833. Therefore given the opportunity to switch prior to opening, you should switch, and given that the situation is symmetric, you should switch again, and again, infinitely.
I don't believe that switching provides you with on average a 1.25x gain, only that switching once provides you with on average a 1.25x gain. Switching twice obviously provides you no gain on average. Hence the situation is completely asymmetric.

Anonymous

Anonymous Tue 29 Oct 2019 00:02:52 No.11100537 Report

Quoted By: >>11100586

>>11100528
>Would you switch prior to opening?
As I said, I would switch to the other envelope once irregardless of whether I opened the first envelop.

Anonymous

Anonymous Tue 29 Oct 2019 00:03:09 No.11100539 Report

Quoted By: >>11100547

>>11098534
2 Possibilities:
1) You have x:
x = 10

2) You have 2x:
x = 10/2
= 5

x is not equivalent between scenarios.

Anonymous

Anonymous Tue 29 Oct 2019 00:03:46 No.11100542 Report

Quoted By: >>11100549

>>11100532
>imagine unironically believing picking an envelope and then passing it back magically makes you money
JUST

Anonymous

Anonymous Tue 29 Oct 2019 00:04:48 No.11100547 Report

Quoted By:

>>11100539
wrong refer to>>11098532 (You)

Anonymous

Anonymous Tue 29 Oct 2019 00:05:34 No.11100549 Report

Quoted By: >>11100552

>>11100542
>>imagine unironically believing picking an envelope and then passing it back magically makes you money
If by magically you mean by containing more money on average, then yes.

Anonymous

Anonymous Tue 29 Oct 2019 00:06:30 No.11100552 Report

Quoted By:

>>11100549
LMAO

Anonymous

Anonymous Tue 29 Oct 2019 00:14:39 No.11100586 Report

Quoted By: >>11100615 >>11100633

>>11100537
If you switch prior to opening, you are in exactly the same situation you were before switching. There's no reason not to switch again if you believe it was advantageous the first time.

Imagine you switch once before opening, and then open, you see $10. Do you switch? If no, how does the first switch make it more likely the original envelope contained $5? Or imagine you have two people who believe this 1.25x nonsense, they switch once, then open, do they both have 1.25x EV? If you repeated indefinitely would both players end up with 1.25x more money that was in their original envelopes?

Anonymous

Anonymous Tue 29 Oct 2019 00:16:50 No.11100593 Report

Quoted By:

>134 posts
>49 posters
holy shit this brainlet won't let go

Anonymous

Anonymous Tue 29 Oct 2019 00:22:07 No.11100615 Report

Quoted By: >>11100634

>>11100586
>If you switch prior to opening, you are in exactly the same situation you were before switching.
Why are the envelopes exactly the same when they contain different amounts of money?

Anonymous

Anonymous Tue 29 Oct 2019 00:30:28 No.11100633 Report

Quoted By: >>11100644 >>11100650

>>11100586
>Imagine you switch once before opening, and then open, you see $10. Do you switch?
Yes.

>Or imagine you have two people who believe this 1.25x nonsense, they switch once, then open, do they both have 1.25x EV?
After they open, one will have 2x and the other will have 0.5x.

>If you repeated indefinitely would both players end up with 1.25x more money that was in their original envelopes?
No, there are only two envelopes.

Anonymous

Anonymous Tue 29 Oct 2019 00:30:34 No.11100634 Report

Quoted By: >>11100636

>>11100615
Because you have no idea which is which. There is a value in envelope A that if you saw, you would switch, there is also a value in envelope B that if you saw, you would switch. Given that you know prior to opening that are you going to switch whichever envelope you currently hold, you should switch indefinitely.

Anonymous

Anonymous Tue 29 Oct 2019 00:33:13 No.11100636 Report

Quoted By:

>>11100634
>Given that you know prior to opening that are you going to switch whichever envelope you currently hold, you should switch indefinitely.
Switching twice is the same as not switching, so there's no reason to consider any case of switching more than once.

Anonymous

Anonymous Tue 29 Oct 2019 00:37:48 No.11100644 Report

Quoted By:

>>11100633
>Yes.
No*

Anonymous

Anonymous Tue 29 Oct 2019 00:41:09 No.11100650 Report

Quoted By: >>11100671

>>11100633
>After they open, one will have 2x and the other will have 0.5x.
Well no, one will have X and the other will have 2X, or X and 0.5X if you prefer, but regardless. I'm asking about EV, if you repeated the process would both players bankroll converge on 1.25x of the amounts in their pre-switch envelopes? If not, then one of them doesn't have 1.25x EV.

Anonymous

Anonymous Tue 29 Oct 2019 00:48:00 No.11100671 Report

Quoted By: >>11100712

>>11100650
>I'm asking about EV, if you repeated the process would both players bankroll converge on 1.25x of the amounts in their pre-switch envelope?
If you repeat the process, then '1.25x of the amounts in their pre-switch envelope' can change each trial, so their bankroll doesn't necessarily converge to it.

Anonymous

Anonymous Tue 29 Oct 2019 01:10:46 No.11100712 Report

Quoted By: >>11100718

>>11100671
Ok lets simplify the question, would both players end with a greater amount than they would have had if they hadn't switched?

Anonymous

Anonymous Tue 29 Oct 2019 01:13:39 No.11100718 Report

Quoted By: >>11100726

>>11100712
>would both players end with a greater amount than they would have had if they hadn't switched?
Only one player can end up with a great amount, since there are only two envelopes.

Anonymous

Anonymous Tue 29 Oct 2019 01:15:12 No.11100726 Report

Quoted By: >>11100730

>>11100718
Over arbitrarily many trials

Anonymous

Anonymous Tue 29 Oct 2019 01:15:50 No.11100730 Report

Quoted By: >>11100757

>>11100726
>Over arbitrarily many trials
What is kept fixed in each trial?

Anonymous

Anonymous Tue 29 Oct 2019 01:25:31 No.11100757 Report

Quoted By: >>11100776 >>11100916

>>11100730
You're hitting on here the actual fundamental problem with this question which is that without knowing anything about the methodology or probability distribution of how the values are put in the envelopes, there is not enough information to give any answer at all. I wanted to avoid that because it doesn't help demonstrate why the 1.25x EV argument in particular is mental.

Assume for the sake of argument the values are always either 5 & 10 or 10 & 20 and there is an even distribution of both. Clearly both players can't have a total higher than they would have had without the switch because that amount is just the other player's total.

Anonymous

Anonymous Tue 29 Oct 2019 01:29:48 No.11100776 Report

Quoted By: >>11100781

>>11100757
still the average of those 4 values is $11.25 so you should still always switch if your envelope is 10

Anonymous

Anonymous Tue 29 Oct 2019 01:32:00 No.11100781 Report

Quoted By:

>>11100776
But in this scenario they are switching prior to opening.

Anonymous

Anonymous Tue 29 Oct 2019 01:35:42 No.11100789 Report

Quoted By:

>>11098612
>two envelopes problem
this is worded as the Alibaba problem and you are the retard.

Anonymous

Anonymous Tue 29 Oct 2019 02:28:42 No.11100916 Report

Quoted By:

>>11100757
>Assume for the sake of argument the values are always either 5 & 10 or 10 & 20 and there is an even distribution of both. Clearly both players can't have a total higher than they would have had without the switch because that amount is just the other player's total.
In this case the players have EV of either +2.5 (if they start with a 10) or -2.5 (if they don't start with 10).

Anonymous

Anonymous Tue 29 Oct 2019 05:05:49 No.11101396 Report

Quoted By: >>11101488

To you retards who still discuss this. The whole point of the monty hall problem was that someone knows where the prize is.

Here it doesn't matter which one you pick.

Anonymous

View Same Google iqdb SauceNAO 00_44_o94sg.jpg, 171KiB, 450x600

Anonymous Tue 29 Oct 2019 05:52:17 No.11101488 Report

Quoted By:

>>11101396

Anonymous

Anonymous Tue 29 Oct 2019 06:53:31 No.11101591 Report

Quoted By:

Anyone noticed a slight uptick in board quality in the last day?
BASED BRAINLET CONTAINMENT THREAD

Anonymous

Anonymous Tue 29 Oct 2019 06:58:21 No.11101603 Report

Quoted By: >>11101647

Guys I have a question that I made in the shower. Suppose there are three boxes named Red, Blue and Green. There is a gift inside one of these boxes. What is the probability that the gift is inside Red ? Now Green is opened but there is no gift inside it. What is the now the probability that Red has the gift inside it?
PS : Is this similar to the Monty Hall problem?

Anonymous

Anonymous Tue 29 Oct 2019 06:58:27 No.11101604 Report

Quoted By:

>>11100310
your free to go

Anonymous

Anonymous Tue 29 Oct 2019 07:14:27 No.11101647 Report

Quoted By:

>>11101603
If Green is opened every time without knowledge of the gift, so 1/3 of the time the gift really is in Green, then it is in Red 1/2 the time after.

The Monty Hall problem relies on someone picking either Green or Blue such that the opened box never has a gift in it.

Anonymous

Anonymous Tue 29 Oct 2019 07:53:48 No.11101707 Report

Quoted By: >>11101753

>>11094480
Better to switch, you'd gain $10 or lose $5, expected value of switching is positive.

Anonymous

Anonymous Tue 29 Oct 2019 08:23:10 No.11101753 Report

Quoted By: >>11101778

>>11101707
Better to switch back after you have switched. You'd gain $10 or lose $5, expected value of switching is positive.

Anonymous

Anonymous Tue 29 Oct 2019 08:34:59 No.11101771 Report

Quoted By: >>11101781 >>11101790 >>11102588

I tried to resolve it by programming it and let it run X many trials, the problem is, how do I choose the values randomly? How exactly? Do I pick one value, then use twice as much for the other one? From a certain range? Then there will be a problem when switching from above the range always loses, and it loses more than you win on average. Or 0 to inf with decreasing probability? Then there is non uniform probability and you are more likely to lose the higher value there is.

In order to solve the problem, the exact mechanism for generating the values must be decided.

Anonymous

Anonymous Tue 29 Oct 2019 08:41:53 No.11101778 Report

Quoted By:

>>11101753
If you were given the problem anew, with a fresh pair of envelopes with the new value you switched to appearing in the new chosen envelope, and still (importantly, as this but really makes the answer "not enough information") no information about which is a more likely maximum value to be in the envelopes, then yes, switch again.

If you have information about the likely maximum (ie, $20 is less likely than $10 to be the maximum value in any envelope) that changes things. If you have already seen both envelopes, obviously that changes things.

Anonymous

Anonymous Tue 29 Oct 2019 08:44:11 No.11101781 Report

Quoted By:

>>11101771
Hmm, you've actually convinced me there is no benefit to switching

Anonymous

Anonymous Tue 29 Oct 2019 08:49:59 No.11101790 Report

Quoted By: >>11101795 >>11101830

>>11101771
It's given that there is $10 in the envelope you picked so you only have two consider two cases ($10,$5) and ($10,$20).

Anonymous

Anonymous Tue 29 Oct 2019 08:54:53 No.11101795 Report

Quoted By: >>11101807

>>11101790
So then if it is equally likely to be the two cases, then everyone is right and you should switch, but if it will be ($10,$20) only a third of the time or less you should stay.

It's not so simple

Anonymous

Anonymous Tue 29 Oct 2019 09:01:42 No.11101807 Report

Quoted By:

>>11101795
Now, you have figured out that just because there are two possible outcomes, that the probability of those outcomes occurring isn’t necessarily 50 - 50. You may leave the brainlet containment thread.

Anonymous

Anonymous Tue 29 Oct 2019 09:20:24 No.11101830 Report

Quoted By:

>>11101790
It is also given that the money was in the envelopes before you opened one.

Anonymous

Anonymous Tue 29 Oct 2019 17:33:48 No.11102588 Report

Quoted By: >>11102626

>>11101771
Here's how I did it:
http://wikisend.com/download/356552/enve.c

Anonymous

Anonymous Tue 29 Oct 2019 17:53:12 No.11102626 Report

Quoted By: >>11102645 >>11103005

>>11102588
So you have
25% chance of having $10, with $5 in the other envelope.
25% chance of having $10, with $20 in the other envelope.
25% chance of having $20, with $10 in the other envelope
25% chance of having $5, with $10 in the other envelope.
Right?

So you there is, on average, $45/4 in your envelope and $45/4 in the other envelope, i.e. you gain nothing from switching.

Anonymous

Anonymous Tue 29 Oct 2019 18:03:39 No.11102645 Report

Quoted By:

>>11102626
Yes, exactly, just like in the problem description.
I randomly select which envelope gets the $ 10
Then I randomly select whether the other gets half that ($5) or double that ($20).
Then I randomly select one envelope.
Then I keep a tally of what the result is when keeping and switching the envelope.

Anonymous

Anonymous Tue 29 Oct 2019 20:28:30 No.11103005 Report

Quoted By: >>11103055 >>11104589 >>11104591

>>11102626
No, the 25% figures are completely baseless

Trump 2020

Trump 2020 Tue 29 Oct 2019 20:34:58 No.11103015 Report

Quoted By:

It’s because the probability is 50/50 and not about the gain or loss of money

Anonymous

Anonymous Tue 29 Oct 2019 21:05:27 No.11103055 Report

Quoted By: >>11103400

>>11103005
Can you elaborate on what you claim to be baseless? What do you not understand?

Anonymous

Anonymous Tue 29 Oct 2019 23:28:51 No.11103400 Report

Quoted By: >>11103430

>>11103055
What do you not understand, the numbers are not substantiated, are not generally true, and don’t follow from the statement of the problem.

Anonymous

Anonymous Tue 29 Oct 2019 23:43:06 No.11103430 Report

Quoted By: >>11103725

>>11103400
You don't know which envelope you chose.

You know that one envelope contained $10, because you chose it and opened it. You do not know which one it was.

Envelope A Envelope B
Either could have contained $10
The other one will be either half or double, so

10 5
10 25
5 10
20 10

These are the options for what either envelope will contain. There are 4 possible combination, that is where the 25% comes from.

Anonymous

Anonymous Wed 30 Oct 2019 01:21:03 No.11103725 Report

Quoted By: >>11103859

>>11103430
I mean, if you think about it, its a 50 - 50 chance you have downs syndrome, either you have it, or you don’t
So it should be 50-50

Anonymous

Anonymous Wed 30 Oct 2019 01:55:44 No.11103859 Report

Quoted By: >>11103958

>>11103725
Even if that were true, how is that applicable to the problem at hand?

Anonymous

Anonymous Wed 30 Oct 2019 02:34:06 No.11103958 Report

Quoted By:

>>11103859
i think he just mustered the courage to come out as down syndromed

Anonymous

Anonymous Wed 30 Oct 2019 07:24:27 No.11104589 Report

Quoted By:

>>11103005
What about it is baseless?

How do you interpret his code, if not the way i described?

How else would you choose the values, so that both
1. one envelope contains twice as much money as the other
2. once you choose one of them, the other envelope has an equal chance of having one half or twice the amount as the envelope you chose?

Anonymous

Anonymous Wed 30 Oct 2019 07:25:37 No.11104591 Report

Quoted By: >>11104961

>>11103005
What about it is baseless?
How do you interpret anon's code, if not the way I described?

How else would you choose the values, so that both
1. one envelope contains twice as much money as the other
2. once you choose one of them, the other envelope has an equal chance of having one half or twice the amount as the envelope you chose?

Anonymous

Anonymous Wed 30 Oct 2019 08:08:02 No.11104640 Report

Quoted By:

>>11094480
Doesn't seem like a paradox to me

The first selection basically doesn't even matter

it's a double or half proposition

double or nothing with free money is neutral ev

of course double or half with free money would be profitable

Anonymous

Anonymous Wed 30 Oct 2019 09:26:17 No.11104736 Report

Quoted By:

>>11097661
old post but one contains N
the other contains 2N OR .5N

Anonymous

Anonymous Wed 30 Oct 2019 11:21:45 No.11104894 Report

Quoted By:

It is highly interesting that it matters that either envelope may have contained the $10.

When calculating probabilities, you need to take into account all possibilities. If you had initially gotten the envelope with $5 or $20 you would still not know if it was the consolation prize or the jackpot. Were you to play the game over and over, (and the amounts changed (but ratios constant)) you would never know if you had the larger or smaller amount.

There are 4 ways to have one envelope contain $10 and the other either half or double that.
10 5
10 20
5 10
20 10

In any instance there are just two envelopes, no matter what dollar amount it contains, you do not know if it is the larger or smaller one. There are never more than two envelopes in the game.

The problem with assigning dollar values is that we do not know if we have the higher or lower value.
We could open the envelope and get $20. If we get the higher value, there is no way to get more. There never was $40 in any envelope, we can not know this for this game, (the $10 we got might be the x/2 of the game) but we can construct our own game with the same rules and see that it would be impossible under the rules, if not in this particular game.
if 10 is the low there can only be 20
if 10 is the high there can only be 5
But we don't know what envelope we got. The amount does not matter.
In these instances we might have gotten
20 and the other contains 10 or
5 and the other contains 10.

This is why there are four options. each with a probability of .25

Anonymous

Anonymous Wed 30 Oct 2019 12:06:03 No.11104950 Report

Quoted By:

>>11094480
What have you done

Anonymous

Anonymous Wed 30 Oct 2019 12:17:52 No.11104961 Report

Quoted By: >>11104984 >>11105000 >>11105035

>>11104591
>What about it is baseless?
It implies that the distribution of envelopes was designed such that its likely you would find one with $10. A proper prior distribution unbiased by that information would not lead to an equal chance of the other envelope containing $5 or $10. There are proper distributions that do in fact lead to the expected value of switching always being higher than the amount in the first envelope, however such distributions have an infinite mean and are therefore unrealistic.

Anonymous

Anonymous Wed 30 Oct 2019 12:21:06 No.11104968 Report

Quoted By:

What is the probability of getting a set of envelopes with a total of $15 and the probability of getting a set of envelopes with a total of $30 and why?

Anonymous

Anonymous Wed 30 Oct 2019 12:25:32 No.11104976 Report

Quoted By: >>11104995

>>11094480
I flip a coin. Heads I get $10, Tails I lose $5.
How's this different?

Anonymous

Anonymous Wed 30 Oct 2019 12:30:47 No.11104984 Report

Quoted By: >>11105081 >>11105135

>>11104961
>It implies that the distribution of envelopes was designed such that its likely you would find one with $10.
I disagree. It only works back from the fact that you did indeed find $10 and constructs all scenarios that would have been possible for this to happen.

Anonymous

Anonymous Wed 30 Oct 2019 12:36:03 No.11104994 Report

Quoted By: >>11105004

Why are people assuming that the $10 envelope gets chosen first in all scenarios?

Anonymous

Anonymous Wed 30 Oct 2019 12:36:25 No.11104995 Report

Quoted By:

>>11104976
The chosen envelope is not present in the coin flip. It is relevant because its content is tied to the prize.

Anonymous

Anonymous Wed 30 Oct 2019 12:37:33 No.11105000 Report

Quoted By: >>11105035 >>11105067

>>11104961
1. I was commenting on the computer code posted by another anon, where the value are generated such a way.

2. While not directly stated in the problem, the only possible way to create such a scenario is to pick a value (randomly or not), then randomly decide if it is the higher or lower value. No other way will fulfil both conditions that a) one envelope countains double the value of the other one b) no matter which envelope you pick, the other one is equally likely to contain 1/2 or 2x the amount of the envelope you hold.

Anonymous

Anonymous Wed 30 Oct 2019 12:38:47 No.11105004 Report

Quoted By:

>>11104994
Assigning a set dollar amount confuses people by seemingly making a third middle option that in reality never exists in the game.

Anonymous

Anonymous Wed 30 Oct 2019 12:43:55 No.11105014 Report

Quoted By:

>>11098708
>you KNOW the other envelope has an equal chance to either contain 5 or 20
how

Anonymous

Anonymous Wed 30 Oct 2019 12:45:18 No.11105019 Report

Quoted By: >>11105204

>>11094788
this is not the real world and we have no other information given, so we have to assume it’s 50/50

Anonymous

Anonymous Wed 30 Oct 2019 12:52:17 No.11105035 Report

Quoted By: >>11105041

>>11104961
>>11105000
Consider the following: You pick a random banknote for one envelope and put two of the same in the other envelope.

Then you pick:
$1 envelope. You always win $1
$2 envelope. 50% you win $2 or lose $1
$4 envelope. 100% you lose $2 by switching.
$5 envelope 100% you win $5 by switching.
$10 envelope 50% winning $10 or losing $5
$20 envelope 50% of winning $20 or losing $10
$40 envelope 100% of losing $20
$50 envelope 100% chance of winning $50
$100 envelope 50% chance of winning $100 or losing $50
$200 envelope 100% chance of losing $100

So on average you gain $1+2+5+10+20+50+100 and lose $1+2+5+10+20+50+100 by switching. Any possible scenario you create where the values are placed inside the envelopes beforehand is going to have this property.

Anonymous

Anonymous Wed 30 Oct 2019 12:55:12 No.11105041 Report

Quoted By:

>>11105035
(there are two possible envelopes where the probability is 50%, so they count twice)

Anonymous

Anonymous Wed 30 Oct 2019 12:57:31 No.11105046 Report

Quoted By: >>11105074

>>11098505
>as long as you don't open one envelope
that is not the problem stated.
One is already open and shows you 10 bucks. Do you switch with no additional information besides the other one having either 20 or 5?

Anonymous

Anonymous Wed 30 Oct 2019 13:06:31 No.11105067 Report

Quoted By:

>>11105000
Here is a revised version of that code, you will see that the results are still the same, and switching does not matter in the long run.

http://wikisend.com/download/101712/enve2.c

Anonymous

View Same Google iqdb SauceNAO 1375549615821.png, 107KiB, 400x400

Anonymous Wed 30 Oct 2019 13:07:35 No.11105070 Report

Quoted By: >>11105079 >>11105088

>There's two envelops
>One has 1000000000 times more money inside it
>You open one at random
>It's $1
Do you switch?

Anonymous

Anonymous Wed 30 Oct 2019 13:09:16 No.11105074 Report

Quoted By: >>11108639

>>11105046
The problem also doesn't state that you RANDOMLY pick the envelope with $10 EVERYTIME

Anonymous

Anonymous Wed 30 Oct 2019 13:11:56 No.11105079 Report

Quoted By: >>11105095

>>11105070
Yes, but not because of math. I did not pay to participate and losing a buck does not matter, but winning a trillion dollars does.

Anonymous

Anonymous Wed 30 Oct 2019 13:12:07 No.11105081 Report

Quoted By: >>11105101

>>11104984
>I disagree. It only works back from the fact that you did indeed find $10 and constructs all scenarios that would have been possible for this to happen.
That's incorrect in two ways.

1. A prior distribution can't be constructed from the conditional information. It has to represent the subject's knowledge before receiving that information.

2. It doesn't construct any realistic scenario, let alone all possible scenarios. Please show me a proper prior distribution of envelopes in which $10 is as likely to be found in the first envelope as any other value in the distribution yet the chance of $5 or $20 in the other is equal after having found the $10 envelope. It can't be done. You have to bias the distribution towards $10, which means you're just begging the question to get the answer you want.

Here let's try to construct it: We know (5,10) and (10,20) exist. But then we need envelopes with 5 and 20 to get their probabilities the same as 10. So we add (2.5,5) and (20,40) but now we have the same problem with 2.5 and 40. So it's easy to see we can't both have all values being equally probable while having both (5,10) and (10,20) being equally likely. You can try this with any proper distribution of envelopes and you'll get the same result.

Anonymous

Anonymous Wed 30 Oct 2019 13:15:21 No.11105088 Report

Quoted By:

>>11105070
>There's two envelops
>One has 1/1000000000 times more money inside it
>You open one at random
>It's $1000000000
Do you switch?

Anonymous

Anonymous Wed 30 Oct 2019 13:16:57 No.11105092 Report

Quoted By: >>11106137

>>11098612
>just google a different problem, dumbass
when you don’t know what is in the first envelope, it doesn’t matter. But the problem CLEARLY states that one envelope contains 10 dollars. You wanna tell me you wouldn’t pick the other one?

Anonymous

View Same Google iqdb SauceNAO 1382762764241.png, 42KiB, 184x237

Anonymous Wed 30 Oct 2019 13:18:04 No.11105095 Report

Quoted By: >>11105216

>>11105079
Okay, here's your $0.000000001.

Anonymous

Anonymous Wed 30 Oct 2019 13:20:50 No.11105101 Report

Quoted By:

>>11105081
If you were to take into account all those values, constantly halving and doubling the smallest and biggest value, you would at some point have enough distinct possible values that any one of them could be disregarded and the result would still be accurate within a margin of error proportional to the group of values included in that group. The result would approach the result from the model where either envelope could have contained the $10. It is a simpler way to do the same thing more accurately.

Anonymous

Anonymous Wed 30 Oct 2019 13:36:59 No.11105135 Report

Quoted By: >>11105150

>>11104984
>I disagree. It only works back from the fact that you did indeed find $10
That's a posterior distribution, not a prior distribution.

>and constructs all scenarios that would have been possible for this to happen.
It only represents scenarios in which (5,10) and (10,20) were equally likely in the prior distribution. This doesn't represent any proper prior distribution: https://www.tandfonline.com/doi/abs/10.1080/00031305.1996.10473551

Anonymous

Anonymous Wed 30 Oct 2019 13:44:00 No.11105150 Report

Quoted By: >>11105555

>>11105135
>(5,10) and (10,20) were equally likely in the prior distribution

What other number could that 10 be paired with other than 5 and 20?

Anonymous

Anonymous Wed 30 Oct 2019 13:47:06 No.11105156 Report

Quoted By: >>11105301

>First envelope has 10 dollars
>Second envelope has 5 dollars or 20 dollars
>Average money of second envelope is 10 dollars
>Switching has no effect

PROBLEM SOLVED

Anonymous

Anonymous Wed 30 Oct 2019 14:16:40 No.11105204 Report

Quoted By:

>>11105019
You assume a flat prior, not a flat posterior.

Anonymous

Anonymous Wed 30 Oct 2019 14:21:14 No.11105216 Report

Quoted By:

>>11105095
LOL

Anonymous

Anonymous Wed 30 Oct 2019 14:50:25 No.11105301 Report

Quoted By:

>>11105156
>Average money of second envelope is 10 dollars

You calculated wrong, but that is not the only error.
The instances are not related. You would need to calculate averages from all possible scenarios, even those where you first chose the other envelope.

Anonymous

Anonymous Wed 30 Oct 2019 15:31:45 No.11105438 Report

Quoted By: >>11105441

Choosing one object at random from a pair has an expected value of 1.5M, with an actual value of either 2M or M.

Let M = x , a discrete random variable of integers ( 1 , 2 , 3 , ... , n ) , with a probability distribution of ( 1 / N ) for each variable. You can think of these as pennies, or the smallest possible discrete values of M.

For example, let N = 50 . The maximum value is [2(50)] - [1(50)] with an expected value of 75 and probability of 1/50, the minimum value is [2(1)] - [1(1)] with an expected value of 0.75 and probability of 1/50. There are also 48 other possible values with a probability of 1/50 each and expected value equal to 2(n)/2 + 1(n)/2 or 1.5(n).

Let us select a pair at random (1/50 possibilities), then an object at random (1/2 possibilities), and the object contains 1 penny. There is only one pair that contains this object, namely [2(1)] - [1(1)]. The expected value and actual value of switching to the paired object is +1. Similarly, if you open the object that contains 100 pennies, the expected and actual value of switching to the paired object is -50.

If we select a pair and object that contains 10 pennies, there are now only 2 possibilities:

>1. We selected the pair [2(5)] - [1(5)] and the object [2(5)].
>2. We selected the pair [2(10)] - [1(10)] and the object [1(10)].

The paired object is equally likely to be [1(5)] or [2(10)] and the expected value of switching to one of them is +2.5 pennies.

Anonymous

Anonymous Wed 30 Oct 2019 15:32:50 No.11105441 Report

Quoted By:

>>11105438
But why does switching BEFORE observing the contents have an expected value of 0 and AFTER observing increase or decrease the expected value?

First, let us take a look at the expected value of any object:

The expected value for an object is:

>1/100 + 2/100 + 2/100 + 4/100 + 3/100 + 6/100 + ... + 100/100 = 38.25

The expected value for an object from a pair is:

>1.5/50 + 3/50 + 6/50 + 7.5/50 + ... + 75/50 = 38.25

Because all objects and paired objects have equal expected value, the expected value of switching is +/-0. However, what if we remove some of these objects or pairs? It is obvious that removing objects or pairs containing less than 38.25 will increase their expected value and vice versa.

If we take the pair [2(1)] - [1(1)] and remove the object [1(1)], then the expected value of selecting an object increases from +1.5 to +2. This is exactly what happens when we observe the contents of an object; namely the expected value of the paired object changes because we are removing 98 or 99 of the possibilities for that object that could have existed before we determined that it was paired to a certain object.

Anonymous

Anonymous Wed 30 Oct 2019 15:56:27 No.11105511 Report

Quoted By:

>>11094480
ITT: Reading comprehension and assumptions

Anonymous

Anonymous Wed 30 Oct 2019 16:21:15 No.11105555 Report

Quoted By:

>>11105150
The envelopes don't need to contain 10.
Do you understand what a prior distribution is?

Anonymous

Anonymous Wed 30 Oct 2019 20:54:12 No.11106137 Report

Quoted By: >>11108129

>>11105092
lmao tard

Anonymous

Anonymous Thu 31 Oct 2019 08:09:17 No.11107391 Report

Quoted By:

SAGE

Anonymous

Anonymous Thu 31 Oct 2019 09:02:00 No.11107446 Report

Quoted By:

>>11097661
You're not losing N though, you're losing .5N

Anonymous

Anonymous Thu 31 Oct 2019 17:31:55 No.11108129 Report

Quoted By:

>>11106137
nice argument

Anonymous

Anonymous Thu 31 Oct 2019 22:29:09 No.11108639 Report

Quoted By:

>>11105074
no, it states that one envelope contains 10 with certainty.

Anonymous

Anonymous Fri 01 Nov 2019 11:36:35 No.11109788 Report

Quoted By:

Not enough information, we aren't told the distribution of the pairs of (5,10) and (10,20), which the answer depends on.

If you assume 50/50, you switch, because you gain more on average

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