/sci/ - Science & Math » Thread #12393869

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PUZZLES MEGATHREAD

Anonymous Fri 27 Nov 23:55:21 2020 No.12393869 View Reply Original Report

Quoted By: >>12394148 >>12394379 >>12394438 >>12394572 >>12394598 >>12394625

Anyone else feeling like flexing their brains in a bit of fun?
Take a puzzle leave a puzzle!

>You drop a pin of length 'a' on the floor
>Floor is made from parallel wooden planks of width 'b'. (just think parallel lines)
What are the odds the pin will lie on two planks (cross a line)?

Anonymous

Anonymous Sat 28 Nov 2020 00:17:02 No.12393959 Report

Quoted By: >>12393971

>puzzle thread
>ask a question that everyone's already heard of

Anonymous

Anonymous Sat 28 Nov 2020 00:22:18 No.12393971 Report

Quoted By: >>12394006 >>12394016 >>12394031 >>12394058 >>12394609 >>12402881 >>12403916

>>12393959
>>12393959
I haven't heard of this puzzle.
I'll give it a go, and add one I've seen.

List out the numbers from 1 to 100.
Randomly select ten of those numbers.
Have a "bucket" on your left and a "bucket" on your right.
You can put each number from those ten into either the left bucket, the right bucket, or neither bucket.
If you can make the sum of the numbers in the left bucket equal the sum of the numbers in the right bucket, then you have achieved the goal for that group of random ten.

Can you achieve the goal for ALL unique groups of ten.

Anonymous

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Anonymous Sat 28 Nov 2020 00:26:32 No.12393989 Report

Quoted By: >>12394013 >>12394016 >>12394109 >>12394135 >>12394468 >>12394652 >>12395823 >>12397417 >>12403792 >>12403880

Two envelopes are offered to you. It is known that one envelope contains twice the amount of money as the other one.
You pick an envelope at random, open it and see that there is $10 in it. You are offered to keep the money or switch to the other envelope.
Is it in your interest to switch?

Anonymous

Anonymous Sat 28 Nov 2020 00:32:22 No.12394006 Report

Quoted By: >>12394019 >>12394058

>>12393971
no...
if the sum of those 10 numbers is an odd number, then you can't divide it into 2 equal groups so it would be impossible. Unless I'm misunderstanding the problem and you don't need to use ALL of the numbers when putting them in the buckets

Anonymous

Anonymous Sat 28 Nov 2020 00:34:03 No.12394013 Report

Quoted By: >>12394020 >>12395823

>>12393989
yes because if you switch correctly you will gain 10 dollars but if you are incorrect you only lose 5.

Anonymous

Anonymous Sat 28 Nov 2020 00:34:40 No.12394016 Report

Quoted By: >>12394020

>>12393971
Yes

>>12393989
Yes

Anonymous

Anonymous Sat 28 Nov 2020 00:36:32 No.12394019 Report

Quoted By:

>>12394006
You dont need to use all the numbers
>You can put each number from those ten into either the left bucket, the right bucket, or neither bucket.
The neither bucket option means that number goes unused :)

Anonymous

Anonymous Sat 28 Nov 2020 00:36:43 No.12394020 Report

Quoted By: >>12394416 >>12394468

>>12394013
>>12394016
WRONG

Anonymous

Anonymous Sat 28 Nov 2020 00:38:43 No.12394031 Report

Quoted By: >>12394036

>>12393971
Put none of the numbers in the buckets. 0=0

Anonymous

Anonymous Sat 28 Nov 2020 00:40:11 No.12394036 Report

Quoted By: >>12394055

>>12394031
yeah I realised after I wrote it that i fucked up.
Obviously I meant that you have to have atleast one number in each bucket

Anonymous

Anonymous Sat 28 Nov 2020 00:45:48 No.12394055 Report

Quoted By:

>>12394036
SPOILERS!! SOLUTION AHEAD!
.
.
.
.
.

There are 2^10 -1 = 1023 nonempty subsets. Each subset has a sum. The largest sum possible is 91+92+...+100 = 5050 - 90*(91)/2 = 955.
1023 > 955
By pigeonhole, two different subsets give the same sum. Remove the common elements from the two subsets to give two nonempty disjoint subsets with a common sum.

Anonymous

Anonymous Sat 28 Nov 2020 00:46:28 No.12394058 Report

Quoted By:

>>12394006
nevermind I'm an idiot

>>12393971
I think I understand the basic solution now, but having a hard time explaining it in words. Basically, has to do with the fact that you will always have either the same/opposite ones digit or same/opposite tens digit (guaranteed to happen with 10 numbers) and from there you can build anything

Anonymous

Anonymous Sat 28 Nov 2020 01:03:31 No.12394109 Report

Quoted By: >>12394112

>>12393989
There's not enough information to complete the EV calculation.

Anonymous

Anonymous Sat 28 Nov 2020 01:04:16 No.12394112 Report

Quoted By: >>12394116 >>12394138

>>12394109
What information is lacking?

Anonymous

Anonymous Sat 28 Nov 2020 01:06:01 No.12394116 Report

Quoted By: >>12395823

>>12394112
not that anon, but I think the marginal utility information of $10 versus $5 versus $20.
Say im going to die without my $9 insulin, it would never make sense to gamble that money away.

Anonymous

Anonymous Sat 28 Nov 2020 01:10:29 No.12394135 Report

Quoted By: >>12395823

>>12393989
It's in my interest to switch, because I can't stand not knowing what's in the other envelope. I'd risk losing a mere $5 to satisfy my curiosity.

Anonymous

Anonymous Sat 28 Nov 2020 01:11:34 No.12394138 Report

Quoted By: >>12395823

>>12394112
We don't know the method used to select the amounts in each envelope so we can't say anything about the probabilities. The bait is designed to make brainlets think switching has a 50% chance of double and 50% of half and calculate the expected value using those probs, but there's no reason to think that.

Anonymous

Anonymous Sat 28 Nov 2020 01:12:03 No.12394139 Report

Quoted By: >>12394147 >>12402863

Thank you anons, for helping me thread my thread.

Have another one:
>You're on a raft, middle of the ocean
>Have a stopwatch in your pocket
What's the circumference of the Earth?

Anonymous

Anonymous Sat 28 Nov 2020 01:14:16 No.12394147 Report

Quoted By: >>12394271

>>12394139
Lmao

Anonymous

Anonymous Sat 28 Nov 2020 01:14:19 No.12394148 Report

Quoted By:

>>12393869
50%
Either it does or doesn't

Anonymous

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Anonymous Sat 28 Nov 2020 01:44:02 No.12394234 Report

Quoted By: >>12394360 >>12394630 >>12402881

One places nine beetles on a circular track, where the nine distances, measured in meters, between two consecutive beetles are the first nine prime numbers, 2,3,5,7,11,13,17,19 and 23. The order is arbitrary, and each number appears exactly once as a distance. At starting time, each beetle decides randomly whether she would go, traveling at a speed of 1 meter per minute, clockwise or counter-clockwise. When two beetles bump into each other, they immediately do "U-turn", i.e. reverse direction (like billiard balls). We assume that the size of the beetles is negligible. At the end of 50 minutes, after many collisions, one notices the distances between the new positions of the beetles. The nine distances are exactly as before, the first nine primes numbers! How to explain this miracle?

Anonymous

Anonymous Sat 28 Nov 2020 01:58:05 No.12394271 Report

Quoted By: >>12394279

>>12394147
It's doable, I swear.

Anonymous

Anonymous Sat 28 Nov 2020 01:59:59 No.12394279 Report

Quoted By:

>>12394271
is the answer 24,901 mi?

Anonymous

Anonymous Sat 28 Nov 2020 02:24:08 No.12394360 Report

Quoted By:

>>12394234
Primes are multi dimensional 1st order identities

Anonymous

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Anonymous Sat 28 Nov 2020 02:36:33 No.12394379 Report

Quoted By: >>12394401 >>12394536 >>12394633

>>12393869
Well right now I only manage to make it work in cases where we know b>=a, if that is the case then what I did was, first parametrize the angle of the stick such that:
$-\frac{\pi}{2}\leq \theta\leq\frac{\pi}{2}$
Notice now that the "length" of our stick can be calculated as follows:
$a\cos\theta$
Let's look at it from the perspective of the leftmost plank on which the stick lands. Then, the probability that our stick will ONLY be on that plank is:
$\frac{b-a\cos\theta}{b}$
We can make a little graph of this the x axis is gonna be our angle, refer to the first part of pic related. We are interested in the area outside the curve and such that y is less than 1, since we're looking for the probability the stick lands in two or more planks. We then divide this area by the total and we get our answer:
$\frac{2a}{\pi b}$

Now, if you look at the right part of pic related, this is what happens when b < a. As you can see, just calculating the integral raw as before is not gonna be enough, we're gonna have to start integrating from the point on which the curve starts being positive:
$\int _{\arccos \left(\frac{b}{a}\right)}^{\frac{\pi }{2}}\:\frac{b-acosx}{b}dx$
Notice that I'm only doing it in one half since the other is symmetrical. So if you calculate that and then from there calculate the shaded area, I think the answer is:
$\frac{2b\arccos \frac{b}{a} + 2a - 2\sqrt{a^{2}-b^{2}}}{\pi b}$
Tho I'm not sure since it looks pretty ugly and generally that's not a good sign, so hopefully some anon can give me a hand for the second half.

Anonymous

Anonymous Sat 28 Nov 2020 02:44:57 No.12394401 Report

Quoted By: >>12394536

>>12394379
I'm not sure what happened there, I swear the preview of the Latex looks fine. Anyways, the part that got spoiled says:

$\frac{b-a\cos\theta}{b}$

We can make a little graph of this the x axis is gonna be our angle, refer to the first part of pic related. We are interested in the area outside the curve and such that y is less than 1, since we're looking for the probability the stick lands in two or more planks. We then divide this area by the total and we get our answer:
$\frac{2a}{\pi b}$

Anonymous

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Anonymous Sat 28 Nov 2020 02:49:59 No.12394416 Report

Quoted By: >>12394423 >>12394425 >>12394617 >>12395483 >>12395520 >>12395823 >>12398047 >>12398108 >>12403667

>>12394020
You are the personification of the Dunning-Kruger Syndrome. You are so incompetent that you cannot even state the problem correctly, and you still have the courage to say others are wrong, in all caps no less.

Let the amounts in the envelopes be Y and 2Y, and the amount in the first envelope be X.

In your question, you told us that X = 10. We can use the expectation formula to conclude we expect to gain $2.5 by switching.

$\frac{1}{2} 20 + \frac{1}{2} 5 = 12.5$ , which is 2.5 more than 10.

In the standard version of the problem, you don't get to know how much money is inside the first envelope, that is, you do not know X. X must be treated as a random variable (as opposed to a constant!), and the expected value of X is $\frac{3Y}{2}$ . Then if you calculate how much money you would have by switching, you still get $\frac{3Y}{2}$ - you can't distinguish the two envelopes and they have the same expected amount of money. I assume this is the problem you were trying to trick us with.

But since you told us X, it's always better to switch. X is no longer random and we don't need to rely on its expectation to estimate it.

Anonymous

Anonymous Sat 28 Nov 2020 02:52:53 No.12394423 Report

Quoted By: >>12394436 >>12394454 >>12395823

>>12394416
Do you really think that without seeing what's in the envelope there's no difference but after seeing what's in the envelope it's better to switch? Think about it carefully.

Anonymous

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Anonymous Sat 28 Nov 2020 02:53:55 No.12394425 Report

Quoted By: >>12394436 >>12403476

>>12394416
>You are so incompetent that you cannot even state the problem correctly
What did you mean by this?

Anonymous

Anonymous Sat 28 Nov 2020 02:56:35 No.12394436 Report

Quoted By: >>12394452 >>12395823

>>12394423
Did you even read my post? In one case X is a constant. In another case X is a random variable. Do you really think that the two cases are the same? Think about it carefully.

>>12394425
This is the version he meant to post (I assume): https://en.wikipedia.org/wiki/Two_envelopes_problem, it's a well known paradox in probability. It explicitly states "before inspecting it" in the problem statement.

Anonymous

Anonymous Sat 28 Nov 2020 02:57:17 No.12394438 Report

Quoted By:

>>12393869
https://en.wikipedia.org/wiki/Buffon%27s_needle_problem

Anonymous

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Anonymous Sat 28 Nov 2020 03:01:47 No.12394452 Report

Quoted By: >>12394460 >>12394468 >>12394476

>>12394436
I posted the problem that I wanted to post. You accuse me of being incompetent for not posting a similar problem that you've seen before and yet you have the gall to accuse others of having Dunning-Kruger syndrome. lol.
>Did you even read my post? In one case X is a constant. In another case X is a random variable. Do you really think that the two cases are the same? Think about it carefully.
Yes. The point here is that observing the amount of money inside the envelope doesn't actually give you any useful information and thus should not change what is beneficial for you.
There is actually quite a simple answer to this problem that I'm rather surprised nobody has brought up.

Anonymous

Anonymous Sat 28 Nov 2020 03:01:54 No.12394454 Report

Quoted By: >>12394495

>>12394423
changing the information you are allowed to know changes the probability. That's how all probability problems work. In one example the sample space is a known constant range, in the other example the range of values for x is infinite.

Anonymous

Anonymous Sat 28 Nov 2020 03:03:53 No.12394460 Report

Quoted By:

>>12394452
>The point here is that observing the amount of money inside the envelope doesn't actually give you any useful information
it does give you useful information. If you were given a negative value you wouldn't switch

Anonymous

Anonymous Sat 28 Nov 2020 03:06:24 No.12394468 Report

Quoted By: >>12394477 >>12395823

>>12394452
Well if you posted the problem you wanted, that is, >>12393989 has no mistakes in it, then the correct answer is yes, you should switch. Why did you make this post >>12394020 then? Since you loudly claimed it was wrong, I made a justified assumption that you posted the wrong problem.

>Yes. The point here is that observing the amount of money inside the envelope doesn't actually give you any useful information and thus should not change what is beneficial for you.
Knowing the amount of money changes everything, because it breaks the symmetry of the problem.

Anonymous

Anonymous Sat 28 Nov 2020 03:08:38 No.12394476 Report

Quoted By: >>12394495 >>12395823 >>12403667

>>12394452
https://plus.maths.org/content/two-envelopes-problem-resolution
Read the section "What if you open envelope A?"

Anonymous

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Anonymous Sat 28 Nov 2020 03:09:18 No.12394477 Report

Quoted By: >>12394491

>>12394468
>Since you loudly claimed it was wrong, I made a justified assumption that you posted the wrong problem.
Maybe consider for a moment that you might actually be wrong instead, my dear Dunning-Kruger.
>Knowing the amount of money changes everything, because it breaks the symmetry of the problem.
Think about it carefully.

Anonymous

Anonymous Sat 28 Nov 2020 03:10:14 No.12394482 Report

Quoted By: >>12394495

>oh shit I realized I made a fool of myself on /sci/ let me just post anime pictures and pretend to be retarded

Anonymous

Anonymous Sat 28 Nov 2020 03:14:45 No.12394491 Report

Quoted By:

>>12394477
Well can you explain to us why he's wrong then? Why should I keep my $10 rather than the expected $12.5?

Anonymous

Anonymous Sat 28 Nov 2020 03:14:46 No.12394492 Report

Quoted By: >>12394502

why doesn't someone just write a computer program where it always switches and see how much money it makes on average? We can end this meme right away

Anonymous

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Anonymous Sat 28 Nov 2020 03:16:02 No.12394495 Report

Quoted By:

>>12394482
>>12394476
>>12394454
Still wrong.

Anonymous

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Anonymous Sat 28 Nov 2020 03:18:07 No.12394502 Report

Quoted By: >>12394512

>>12394492
You are very close to the actual answer but your suggestion is a bit off. We know the average money will be about 12.5. But is that the point? Think about it carefully.

Anonymous

Anonymous Sat 28 Nov 2020 03:22:09 No.12394512 Report

Quoted By: >>12394516

>>12394502
>We know the average money will be about 12.5.

Then why do you keep posting this nonsense? I don't know if you're just trolling or not but pretty much everyone ITT understands the reasoning why you claim there's a paradox, it's just that everyone except you doesn't understand why you're wrong. Unless there's some sort of secret reasoning you're not sharing I'm going to assume you're trolling

Anonymous

Anonymous Sat 28 Nov 2020 03:23:18 No.12394516 Report

Quoted By: >>12394519

>>12394512
It's a puzzle. That's why I posted it. Think about it carefully for yourself.

Anonymous

Anonymous Sat 28 Nov 2020 03:25:26 No.12394519 Report

Quoted By: >>12394522

>>12394516
how about you just explain yourself instead of posting "think about it carefully". Think about what? the fact that you're retarded?

Anonymous

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Anonymous Sat 28 Nov 2020 03:26:29 No.12394522 Report

Quoted By: >>12394530

>>12394519
>Think about what?
About the puzzle. Think about it carefully.

Anonymous

Anonymous Sat 28 Nov 2020 03:29:03 No.12394530 Report

Quoted By: >>12394534

>>12394522
I have, and I don't know what realization you want me to make, other than the fact that revealing the information in the envelope influencing your decision to switch is counter-intuitive and somewhat irrational, which literally everyone already knows.

Anonymous

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Anonymous Sat 28 Nov 2020 03:30:43 No.12394534 Report

Quoted By: >>12394562

>>12394530
>other than the fact that revealing the information in the envelope influencing your decision to switch is counter-intuitive and somewhat irrational, which literally everyone already knows.
I can assure you that that's NOT the realization I want you to make. Why do you want to give up on the problem so quickly? What's the rush?

Anonymous

Anonymous Sat 28 Nov 2020 03:31:02 No.12394536 Report

Quoted By:

>>12394379
>>12394401
Well I just confirmed from Wikipedia that both of my answers are correct.

https://en.wikipedia.org/wiki/Buffon's_needle_problem

Thanks for nothing /sci/

Anonymous

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Anonymous Sat 28 Nov 2020 03:31:53 No.12394540 Report

Quoted By: >>12394559

>having a good and productive puzzle thread
>anon posts his puzzle but messes it up
>tells everyone who solved it they're WRONG
>gets proven wrong himself
>derails the entire thread with trolling, anime posting, and by pretending to be obtuse
>can dish it out but can't take it himself
Let's just move on and post other problems in an attempt to save the thread.

Anonymous

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Anonymous Sat 28 Nov 2020 03:38:37 No.12394559 Report

Quoted By:

>>12394540
I already said that I didn't mess up the problem. There's no point in lying about that. I posted the version that I wanted to post. It's there in one of the links anon posted. You still insisting that I messed it up makes you the troll, not me.
I can assure you that I'm not trolling, there is an actual solution to this and that everyone ITT so far has gotten it wrong. If you weren't so quick to pass it off and instead spent a little bit of time thinking about it you would realize I'm right.
Also
>>>can dish it out but can't take it himself
Lol what? At this point it's just random accusations with no basis in reality.

Anonymous

Anonymous Sat 28 Nov 2020 03:41:02 No.12394562 Report

Quoted By: >>12394563

>>12394534
>Why do you want to give up on the problem so quickly? What's the rush?

because a) I don't give a shit, and b) I suspect you're trolling. Why don't you just tell us what you think the answer is and we can all move on :)

Anonymous

Anonymous Sat 28 Nov 2020 03:42:09 No.12394563 Report

Quoted By:

>>12394562
The answer is that it doesn't matter. The puzzle is figuring out what's wrong with the explanations that say it does.

Anonymous

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Anonymous Sat 28 Nov 2020 03:46:32 No.12394572 Report

Quoted By: >>12394592 >>12394603

>>12393869
you all need to have a look on 'baba is you'.

Anonymous

Anonymous Sat 28 Nov 2020 03:58:10 No.12394592 Report

Quoted By: >>12394603 >>12397513

>>12394572
https://www.youtube.com/watch?v=E24WWi9X8u0

this video gives a good impression.

the game will send you so deep down the rabbit hole like no other game.

Anonymous

Anonymous Sat 28 Nov 2020 04:00:43 No.12394598 Report

Quoted By:

>>12393869
We would need to know the width of the pin and the size of the gaps between the boards.
>inb4 both are 0
then the pin doesn't exist

Anonymous

Anonymous Sat 28 Nov 2020 04:03:37 No.12394603 Report

Quoted By: >>12394607

>>12394572
>>12394592
this game was all over the internet a couple years ago, where you living under a rock? also it's pretty fun

Anonymous

Anonymous Sat 28 Nov 2020 04:06:48 No.12394607 Report

Quoted By:

>>12394603
i played around last xmas.
i guess there are enough puzzle fans who do not know about it.

Anonymous

Anonymous Sat 28 Nov 2020 04:07:20 No.12394609 Report

Quoted By: >>12395655

>>12393971
If we take the numbers 1,3,4,5,6,7,8,9,10, their sum would be 53, which is a number that you can't divide by 2, proving that you can't do it.

Anonymous

Anonymous Sat 28 Nov 2020 04:10:22 No.12394617 Report

Quoted By:

>>12394416
Based ichihime
Still got 4x4 in my todays games. If only mahjong could be solved mathematically in my mind

Anonymous

Anonymous Sat 28 Nov 2020 04:14:09 No.12394625 Report

Quoted By: >>12394633

>>12393869
Ignoring width of pin and width of gaps between floor:
If pin lands perpendicular to floor boards then the chance is a/b
Assume floor boards lie parallel to y-axis
Since pin does not necessarily land perpendicular to floor boards, the chance is (a/b)(average difference of x-value of one end of pin and x-value of other end of pin)
We can consider just one quadrant of unit circle
So we need to find the average value of cos(?) on the interval (0,?/2)
Average value of f(x) is ?f(x)dx/(b?a)
So we need ?cos(?)d?/(?/2)
which is sin(?)/(?/2)
so sin(?/2)/(?/2)-sin(0)/(?/2)
which is 2/?
Therefore the average value of |x| on the unit circle is 2/?
So the answer is 2a/?b

Anonymous

Anonymous Sat 28 Nov 2020 04:15:35 No.12394630 Report

Quoted By: >>12395625

>>12394234
After colliding the 2 beetles start moving apart towards other beetles that start a prime away. In the time it takes to cover that distance the same distance has opened up between the original 2 beetles so on average the gaps stay prime sized.
It would be the same with any size gap though, primes are a red herring.

Anonymous

Anonymous Sat 28 Nov 2020 04:16:30 No.12394633 Report

Quoted By: >>12403675

>>12394379
I quit school after graduating high school and school doesn't teach fuckall and I just used cold hard logic to calculate the problem and I can confirm that you are indeed correct. My method is here: >>12394625

Anonymous

Anonymous Sat 28 Nov 2020 04:22:08 No.12394652 Report

Quoted By:

>>12393989
no because I am 10 dollars short of paying rent and idgaf if I get the other 10

Anonymous

Anonymous Sat 28 Nov 2020 12:35:13 No.12395483 Report

Quoted By: >>12395530

>>12394416
Why have you assumed the probabilities of it being 20 or 5 are both 50%? Thats not stated anywhere in the problem.

Anonymous

Anonymous Sat 28 Nov 2020 12:54:05 No.12395520 Report

Quoted By: >>12395584

>>12394416
My dude, I've read your explanation and it seems reasonable but it leads to this problem:

Instead of looking at the inside of the envelope, to fix X (for example, X=10), the person imagines there's 10 inside. If X=10, it would make sense for him to switch.
Now, he similarly imagines an alternative scenario, where X=11. It would again make sense for him to switch.
By induction, it follows that whatever (positive) number he imagines is inside the envelope, it will make sense for him to switch.

Since the contents of the envelope must match one of his (infinitely many) imaginary scenarios and in that scenario it makes sense for him to switch, this means it makes sense for him to switch.

Is there an error in my reasoning? Where?

Anonymous

Anonymous Sat 28 Nov 2020 12:56:15 No.12395530 Report

Quoted By: >>12395545

>>12395483
How could they be anything else?

Anonymous

Anonymous Sat 28 Nov 2020 13:01:27 No.12395545 Report

Quoted By: >>12395566

>>12395530
Imagine the experimental protocol is this: before you are given the envelopes, $10 is put in one envelope, and then $5 is put in the other with probability 0.9, or $20 with probability 0.1. This still satisfies all the information we are given in the puzzle, but the EV is very different to assuming there is a 0.5 chance of either.

The probabilities are completely unknown to us.

Anonymous

Anonymous Sat 28 Nov 2020 13:10:14 No.12395566 Report

Quoted By: >>12395596

>>12395545
But if you averaged all the possible probability choices for $5 and $20 ( (0.9,0.1) in your case, or (1.0,0.0) or (0.3,0.7) ) wouldn't you come up with 0.5,0.5?
Thus, with no knowledge of this probability choice, 0.5,0.5 is the only sensible assumption?

Anonymous

Anonymous Sat 28 Nov 2020 13:17:18 No.12395584 Report

Quoted By: >>12395591

>>12395520
That's an interesting argument, consider this.

He imagines that X = 10, and imagines that the other envelope (let's call it Z) has Z = 12.5 in it.

By your very argument, since he's imagining that Z = 12.5, he should switch again and imagine that X = 15.625. But then he should imagine that Z = 19.5, and so on, so he should switch forever, never picking an envelope and eternally upping the ante in his mind.

The problem here is that X itself is a random variable, specifically it is an expectation of a random variable. So when you imagine that X = 10, what you are really saying is, "let me imagine that the smaller envelope has 20/3 and the larger has 40/3, which makes X = 10". Then when you think about switching, you will also imagine that Z = 10, so there's no benefit to switching.

Anonymous

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Anonymous Sat 28 Nov 2020 13:19:53 No.12395591 Report

Quoted By: >>12395593

>>12395584
>By your very argument, since he's imagining that Z = 12.5, he should switch again and imagine that X = 15.625. But then he should imagine that Z = 19.5, and so on, so he should switch forever, never picking an envelope and eternally upping the ante in his mind.
We have reached peak brainrot. It doesn't get stupider than this.

Anonymous

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Anonymous Sat 28 Nov 2020 13:21:23 No.12395593 Report

Quoted By:

>>12395591
>retard calls others retarded

Anonymous

Anonymous Sat 28 Nov 2020 13:22:12 No.12395596 Report

Quoted By: >>12395615 >>12402870

>>12395566
If you did that yeah, but why would you do that? By averaging like that you're assuming again that the meta-probability of the experimenters choosing a particularly probability is uniform, when in fact its unknown. What if they are more likely the pick a probability that favours $5?

All we know is that there is some probability P that the other envelope has double and (1 - P) that it has half. We can't calculate expected value from that without knowing P.

Anonymous

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Anonymous Sat 28 Nov 2020 13:34:45 No.12395615 Report

Quoted By: >>12395639

>>12395596
You are closer to the true answer than others but still not quite there.

Anonymous

Anonymous Sat 28 Nov 2020 13:38:37 No.12395625 Report

Quoted By: >>12397400 >>12397741 >>12397751

>>12394630
>In the time it takes to cover that distance the same distance has opened up between the original 2 beetles
Why would that be the case?
>It would be the same with any size gap though
Are you sure about that?

Anonymous

Anonymous Sat 28 Nov 2020 13:47:24 No.12395639 Report

Quoted By: >>12395647

>>12395615
Not sure that I'm missing anything desu. The answer to the stated question "Is it in your interest to switch?" is "it doesn't matter", not because its EV neutral, but because its impossible to say anything about the value of the other envelope other than it has either a 5 or a 20 in it.

Hope you're gonna post the frog one after this thats always good for a laugh

Anonymous

Anonymous Sat 28 Nov 2020 13:52:34 No.12395647 Report

Quoted By: >>12395661

>>12395639
>the frog one
?

Anonymous

Anonymous Sat 28 Nov 2020 13:56:07 No.12395655 Report

Quoted By:

>>12394609
>>12394609
If we used your numbers. 1 and 2 in the left bucket, 3 in the right bucket. They sum to the same number

Anonymous

Anonymous Sat 28 Nov 2020 13:57:55 No.12395661 Report

Quoted By:

>>12395647
Sorry I assumed you were the anon who always posts the frog poison/antidote one in these sorts of threads as you post very much like him.

Anonymous

Anonymous Sat 28 Nov 2020 14:06:28 No.12395686 Report

Quoted By: >>12395710 >>12397403

This isn’t exactly a logic puzzle but a Lateral Logic Puzzle
It is solvable but the answer is not what you’d think
Apologies if it doesn’t fit with the spirit of the thread
> A man pushing his car stops in front of a hotel
> As soon as he stops, he knows he is bankrupt
> Why?

Anonymous

Anonymous Sat 28 Nov 2020 14:10:37 No.12395693 Report

Quoted By:

The amount of money makes a big difference.

10 dollars. I would say screw it and switch for a chance of 20 dollars.

If the envelope had 1million dollars. I wouldn't risk settling for 500k.

But back to the 10 dollars.
The probability seems 50/50 that you got the high payout. Which indeed it is. But the issue at hand isn't the 10dollars. It is the person handing you the envelope. If they are a person who just likes to troll. Then if you switch you will probably get the 5 dollars.

If the person is nice and likes to see people be happy. Then it will be doubled to 20.

So you gotta ask yourself... do you trust a gamble from another person.
Screw that. The world is filled with douche bags. So keep the ten and go.

Another perspective. The guy wasn't willing to put 100 dollars in. So he is cheap. And most likely try to get you to take the 5. Since there is only one switch. He probably gave you the ten. So you will take the 5. Since it is human nature to try and be greedy. Example above. Where I would risk for 20. The envelope guy is betting you will switch. While you're betting you will double your money.

It's the same for any amount. Even with the million. If you are willing to risk the million for 2million. If they guy thinks you're greedy. He betting you will switch.

In either case the chances are better for the dealer to give you the high amount first. Since the dealer will lose less if you switch.
And since the only thing we know is the 10 dollars. The poster can say 5 or 20. But we will never know since it wasn't really done.

So here is puzzle/riddle.

You never know the day you will die. But do you know the day you will never die?

Anonymous

Anonymous Sat 28 Nov 2020 14:18:40 No.12395710 Report

Quoted By: >>12395726

>>12395686

Then man just realised he was bankrupt. Because he couldn't afford the 80 dollars for a tow truck. When he just got paid/had a savings above 80 dollars. He just didn't realise until he stopped.
(Wrong answer most likely)
But it would be nice to know the logic behind how he knows

Anonymous

Anonymous Sat 28 Nov 2020 14:22:22 No.12395726 Report

Quoted By: >>12395860

>>12395710
It’s not the correct answer according to the book I got it out of
It’s deceptive, the idea is it makes perfect logical sense in a different frame of reference - the trick to the puzzle is to find that frame of reference

Anonymous

Anonymous Sat 28 Nov 2020 15:05:02 No.12395823 Report

Quoted By: >>12395885 >>12395980 >>12397093

>>12393989
>>12394013
>>12394116
>>12394135
>>12394138
>>12394416
>>12394423
>>12394436
>>12394468
>>12394476

Google is your friend, chodes.

https://brilliant.org/wiki/two-envelope-paradox/

Anonymous

Anonymous Sat 28 Nov 2020 15:24:26 No.12395860 Report

Quoted By: >>12395878

>>12395726
>It’s not the correct answer according to the book I got it out of
First thing I thought of was that he's playing a game of Monopoly, but I can't see how there can be a "correct" answer for this sort of thing.

Anonymous

Anonymous Sat 28 Nov 2020 15:33:24 No.12395878 Report

Quoted By: >>12395955

>>12395860
That is the right answer - it may seem dumb but it’s a different frame of reference where pushing a car, stopping in front of a hotel, and bankruptcy all make logical sense

Anonymous

Anonymous Sat 28 Nov 2020 15:40:21 No.12395885 Report

Quoted By:

>>12395823
The explanation in that website is wrong.

Anonymous

Anonymous Sat 28 Nov 2020 16:07:01 No.12395955 Report

Quoted By: >>12395965

>>12395878
Haha. That's great. I love monopoly. I am surprised I didn't think that. But I often don't pushed pieces across the board.. except the car.

But can you figure out the

"You never know the day you die.
But do you know the day you will never die?"

Anonymous

Anonymous Sat 28 Nov 2020 16:11:14 No.12395965 Report

Quoted By: >>12396083

>>12395955
My first thought is yesterday but that’s probably incorrect

Anonymous

Anonymous Sat 28 Nov 2020 16:18:27 No.12395980 Report

Quoted By:

>>12395823
They're right that you'll win more than $X by using an auxiliary distribution... but you'll also win more than $X by just always switching, as everyone has already explained, by blindly switching you expect 1.25X. Using the auxiliary distribution is actually worse because you might only expect to win 1.1X or 1.2X.

Anonymous

Anonymous Sat 28 Nov 2020 16:46:07 No.12396083 Report

Quoted By:

>>12395965
It is correct. There is today, tomorrow, Monday, tuesday.... Sunday.

The only day you can never die is yesterday. Since you are alive today. The soonest anyone can die is today. Though you might not be found until another point in time. And it's about the individuals perspective.

Anonymous

Anonymous Sat 28 Nov 2020 23:35:03 No.12397093 Report

Quoted By:

>>12395823
>brilliant.org
Fuck off.

Anonymous

Anonymous Sun 29 Nov 2020 01:51:01 No.12397400 Report

Quoted By:

>>12395625
>Why
They all move at the same speed.
Beetles 1&2 are going the same way, distance d apart.
Beetles 2&3 collide and reverse directions.
Beetles 1&2 now approach while 2&3 move apart. After d/2 minutes 1&2 collide and reverse, 2&3 end up d apart and going the same way.
>sure
yes. Nothing special about d except before looking you need to wait at least long enough for it to close and open back up.

Anonymous

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Anonymous Sun 29 Nov 2020 01:52:12 No.12397403 Report

Quoted By:

>>12395686

Anonymous

Anonymous Sun 29 Nov 2020 01:57:45 No.12397417 Report

Quoted By:

>>12393989
Yes.

Expected value problem. $5 loss vs $10 gain.

Anonymous

Anonymous Sun 29 Nov 2020 02:37:23 No.12397513 Report

Quoted By:

>>12394592
>posting egg's series
the memories of frustration...

Anonymous

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Anonymous Sun 29 Nov 2020 04:10:24 No.12397741 Report

Quoted By:

>>12395625
Since the identity of the individual beetles is irrelevant, the situation is unchanged if we pretend they are "phantasmagoric", such that two colliding beetles simply phase through each other and continue moving in their original direction, instead of reversing course.
In this scenario, after 50 = (2+3+...+23)/2 minutes each ghost-beetle will have travelled exactly half of the track distance, i.e. it will end up at the antipode of where it had originally started. Mathematically, this amounts to rotating the track by an angle of $\pi/2$ , which preserves the distances between the pairs of beetles.

Anonymous

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Anonymous Sun 29 Nov 2020 04:14:10 No.12397751 Report

Quoted By:

>>12395625
Since the identity of the individual beetles is irrelevant, the situation is unchanged if we pretend that they are "phantasmagoric", such that when any two beetles meet, rather than colliding and reversing course, they simply phase through each other and continue moving in their original direction.
In this scenario, after 50 = (2+3+...+23)/2 minutes each ghost-beetle will have travelled exactly half of the track distance, i.e. it will end up at the antipode of where it had originally started. Mathematically, this amounts to rotating the track by an angle of $\pi$ , which preserves the distances between the beetles.

Anonymous

Anonymous Sun 29 Nov 2020 05:43:37 No.12398047 Report

Quoted By:

>>12394416
God you are dunning-Kruger
It’s ironic.

Anonymous

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Anonymous Sun 29 Nov 2020 05:48:32 No.12398061 Report

Quoted By: >>12398158 >>12398392

fun math problem time!

consider a ketchup cup with radius R and sides with length L. what is the optimal angle you should bend the sides out to maximize the volume that the cup can hold?

Anonymous

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Anonymous Sun 29 Nov 2020 06:08:03 No.12398108 Report

Quoted By: >>12401112 >>12403557 >>12403885

>>12394416
probability is a junk science,
I will explain why below but only 1/20 will understand the answer

The question is stupid.
Very stupid.
Your answer relys on averages. On average, I.e. if you did the experiment many times, on average your strategy would gain more times then you would fail. But there is nothing that says you will get to do this many times.
> conclude we expect to gain $2.5 by switching.
Your donning Kruger is showing. This is a nonsense statement in a one off event. With a known $10 you either gain $10 (other envelope has 2x=$20) or you lose $5 (other envelope has x/2=5$) there is zero situation where as you said
>we expect to gain $2.5.
Does not exist. Unless you get to do this many many many times with some crazy that just keeps handing you envelope pairs.
An illrational brain puts more weight on current winning rather than future losses.
What if you need 10$ to pay your rent? You might lose $5 lose your roof.
This is clearer with bigger numbers.
What if you saw $1000?
Would you gamble and get either $2000 or $500? The odds of these incomes are EXACTLY the same. It’s 2 envelopes, with no other info (it’s also not like money hall where things change) I can not stress that part enough.
Is the bet rational? This question probability can not answer.

Anonymous

Anonymous Sun 29 Nov 2020 06:24:18 No.12398158 Report

Quoted By: >>12403515

>>12398061
bend it out into a flat sheet and make a mountain of ketchup on top

Anonymous

Anonymous Sun 29 Nov 2020 08:19:31 No.12398392 Report

Quoted By: >>12399821

>>12398061
Is there a formula for the area of that shape? Or do you have to use a cone and then subtract the part of cone that extends below table?

Anonymous

Anonymous Sun 29 Nov 2020 17:38:37 No.12399821 Report

Quoted By:

>>12398392
personally i would compute the radius as a function of height, integrate the volume using discs, differentiate the volume, and then set the derivative to zero

Anonymous

Anonymous Sun 29 Nov 2020 23:27:28 No.12401112 Report

Quoted By:

>>12398108
Consider a fee. After choosing your final envelope with you have to pay back half of the total, 1/2Y
a) Y=15, you lose $2.5 by switching, gain $2.5 by not.
b) Y=30, you gain $5 by switching, lose $5 by not.

Anonymous

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Anonymous Mon 30 Nov 2020 05:41:11 No.12402297 Report

Quoted By: >>12402611 >>12402881

Two circles (of different sizes) overlap, with common chord XY.
At point X there is a lighthouse, which emits a single beam of light (straight line, bidirectional) that intersects each of the circles at points A and B respectively.
Prove that the length of the lightbeam AXB is maximized when it is angled to be perpendicular to the chord XY.

(You could do this with calculus, but the puzzle is to find an argument that works using only elementary geometry.)

Anonymous

Anonymous Mon 30 Nov 2020 08:37:37 No.12402611 Report

Quoted By: >>12402946

>>12402297
I wish I was Smart.
something with the 90degree angle being the biggest, any tilt would shorten it in off sized circles. Even if that’s the right track I have no idea how to prove it.
If you fold it in half along xy, the lines overlap when perpendicular unfolded, any tilt, when folded would make them form an angle with x at vertex. And you could see intuitively if you moved axb off perpinculer, ax would get shorter faster than bx would get longer.

Anonymous

Anonymous Mon 30 Nov 2020 11:20:17 No.12402863 Report

Quoted By: >>12402946

>>12394139
Is anyone interested in the solution to this problem?

Anonymous

Anonymous Mon 30 Nov 2020 11:27:06 No.12402870 Report

Quoted By: >>12403551

>>12395596
I was thinking about it some more, you've lead me down the wrong path. See, the envelopes were filled with money before you chose one. So the probability literally can't be anything other than 50%.

Anonymous

Anonymous Mon 30 Nov 2020 11:31:22 No.12402881 Report

Quoted By:

>>12393971
>>12394234
>>12402297
Thank you anons.

Anonymous

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Anonymous Mon 30 Nov 2020 12:10:26 No.12402946 Report

Quoted By: >>12402972 >>12403272

>>12402863
Hmm. Do you need to be able to see the Sun for this?
>>12402611
You really only need the inscribed angle theorem (pic related) and the fact that the diameter is the longest possible chord.

Anonymous

Anonymous Mon 30 Nov 2020 12:23:57 No.12402972 Report

Quoted By: >>12403049

>>12402946
Oh I thought the cord had to go through x.

Anonymous

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Anonymous Mon 30 Nov 2020 13:04:59 No.12403049 Report

Quoted By: >>12403096

>>12402972
>the cord had to go through x.
It does, I wasn't referring to the chord XY there. It's a hint for you that diameters are involved somehow.

Leaving another puzzle for the rest:

A cable containing 10 identical wires is laid from a remote military outpost to the HQ where you are stationed. Unfortunately, no one knows which wire is which, as they forgot to label the individual endpoints, and you are tasked to do so. All you have are:
>tape and a marker for labeling the wires
>a battery and lightbulb to test if a circuit is closed or not
The outpost is very far away, so you want to minimize the number of round-trips you make there. What is the smallest number of trips needed?

Anonymous

Anonymous Mon 30 Nov 2020 13:25:21 No.12403096 Report

Quoted By: >>12403115

>>12403049
3

Anonymous

Anonymous Mon 30 Nov 2020 13:35:19 No.12403115 Report

Quoted By:

>>12403096
Actually, you only need to visit the outpost once.

Anonymous

Anonymous Mon 30 Nov 2020 14:44:21 No.12403272 Report

Quoted By: >>12403404

>>12402946
>do you need to see the Sun
Yes, yes you do, good thinking! That's why I put you in the middle of the ocean - guaranteed unimpeded view of the horizon.

Anonymous

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Anonymous Mon 30 Nov 2020 15:47:44 No.12403404 Report

Quoted By:

>>12403272
Well, my original idea was to time the angular velocity of the Sun (from sunrise to sunset), but then I realized that to get anything like the Earth's circumference out of it, you'd need to know your latitude and probably the current season as well, which is not given.

Still, prompted by your mentioning of viewing the horizon, I've thought of another method which I think is probably the intended solution, though it assumes you know the height of your eyes above sea-level when standing (H) and squatting (H'). (Depending on what else you have with you, you might be able to calculate those with shadow sticks or something, but I don't think that's relevant here.)

The method is to wait for the sunrise while standing facing that direction. At the instant the Sun appears on the horizon, start the timer and squat down, whence you will see the Sun disappear back beneath the horizon. Stop the timer when it reappears, and divide the recorded time by 1 day to get the angle $a$ through which the Earth rotates in the meantime.
Together with H and H', you can use the not-to-scale diagram in picrelated, plus a bit of trigonometry, to calculate the Earth's radius R and hence circumference.

Anonymous

Anonymous Mon 30 Nov 2020 16:05:56 No.12403476 Report

Quoted By: >>12403532

>>12394425
The original 2 envelope problem you cannot inspect the contents of your first choice

Anonymous

Anonymous Mon 30 Nov 2020 16:13:32 No.12403506 Report

Quoted By:

For those with a 3d modeling program, I riddle you this.
Show how a cube could be passed through a hole made in another cube of the same size without splitting the cube into two pieces

Anonymous

Anonymous Mon 30 Nov 2020 16:16:38 No.12403515 Report

Quoted By: >>12403572

>>12398158
Is this correct please respond... I don't know enough about ketchup fluid dynamics to know

Anonymous

Anonymous Mon 30 Nov 2020 16:22:05 No.12403532 Report

Quoted By: >>12403557

>>12403476
So? This is a different problem.

Anonymous

Anonymous Mon 30 Nov 2020 16:30:18 No.12403551 Report

Quoted By:

>>12402870
Of course it can, for all you know there was never a chance of getting $20. An experimental protocol with just $5 and $10 fits all the information we are given.

Anonymous

Anonymous Mon 30 Nov 2020 16:32:46 No.12403557 Report

Quoted By: >>12403581 >>12403583

>>12398108
You somehow managed to both overthink the question and be retarded at the same time.

>>12403532
From the poster's responses, it was very clear he posted the wrong problem and had no idea what he was talking about.

Anonymous

Anonymous Mon 30 Nov 2020 16:41:19 No.12403572 Report

Quoted By:

>>12403515
no, in that case the cup has zero volume. it's one of the two worst solutions

Anonymous

Anonymous Mon 30 Nov 2020 16:42:50 No.12403581 Report

Quoted By: >>12403596

>>12403557
You're retarded. I posted the problem that I wanted to post. It's not my fault that you didn't solve it correctly.
You're literally like the most autistic person ever. I post a problem and just because it's similar to a problem you've seen before your autistic brain goes WOWZEEE HE POSTED THE WRONG PROBLEM. Just because the two problems are similar doesn't mean one of them is wrong. I am genuinely amazed that someone of your intelligence is still capable enough to post on 4chan.

Anonymous

Anonymous Mon 30 Nov 2020 16:44:09 No.12403583 Report

Quoted By: >>12403596

>>12403557
Also the phrase "wrong problem" doesn't even make any sense. There are wrong answers to problems, but you cannot say a problem is wrong. That doesn't even make any sense. God you're retarded.

Anonymous

Anonymous Mon 30 Nov 2020 16:47:05 No.12403596 Report

Quoted By: >>12403642

>>12403581
You're retarded. You posted the wrong problem, everyone solved it, you called them wrong when they were right, then when people proved you wrong you started posting anime pictures and >hehe think about it carefully xD!!

Your absolutely idiotic responses such as
>Do you really think that without seeing what's in the envelope there's no difference but after seeing what's in the envelope it's better to switch? Think about it carefully.
>The point here is that observing the amount of money inside the envelope doesn't actually give you any useful information and thus should not change what is beneficial for you.
prove without a shadow of doubt that you are retarded. Even if you posted the problem you meant to post, you are still retarded for typing the two sentences above. Fuck off retard.

>>12403583
It seems reading comprehension is also very difficult for you.

Anonymous

Anonymous Mon 30 Nov 2020 16:57:59 No.12403642 Report

Quoted By: >>12403648

>>12403596
Not that anon but what you've quoted there is entirely accurate

Anonymous

Anonymous Mon 30 Nov 2020 16:59:04 No.12403648 Report

Quoted By: >>12403661

>>12403642
retard

Anonymous

Anonymous Mon 30 Nov 2020 17:02:13 No.12403661 Report

Quoted By: >>12403667 >>12403707

>>12403648
Prior to opening the envelope, you know there is money inside, given that you can't put a negative amount of money into an envelope, you know the amount is positive. Is there any positive value X that upon opening the envelope, you wouldn't conclude that switching has a greater EV? If no, opening the envelope gave you no information you didn't already have.

Anonymous

Anonymous Mon 30 Nov 2020 17:04:03 No.12403667 Report

Quoted By: >>12403689

>>12403661
Read the explanation >>12394416 or >>12394476
As has been explained a dozen times ITT, knowing the amount in the envelope (X) changes the problem because X is now a constant as opposed to a random variable.

Anonymous

Anonymous Mon 30 Nov 2020 17:06:40 No.12403675 Report

Quoted By:

>>12394633
actual thought pollution

Anonymous

Anonymous Mon 30 Nov 2020 17:09:22 No.12403689 Report

Quoted By: >>12403696

>>12403667
I'll read that link if you answer the question: is there a value that upon opening, you wouldn't decide to switch?

Anonymous

Anonymous Mon 30 Nov 2020 17:10:35 No.12403696 Report

Quoted By: >>12403701

>>12403689
You don't fucking get it, do you? The point of the question is that you don't open the envelope in the first place.

But if I were to open the envelope, then you should always switch, as everyone already answered.

Anonymous

Anonymous Mon 30 Nov 2020 17:12:28 No.12403701 Report

Quoted By: >>12403763 >>12403776

>>12403696
So therefore you aren't getting any information, which is what the other anon said. If you would ALWAYS switch after opening the envelope, then opening it clearly is not influencing your decision, because you know you will switch whatever happens after opening it.

Anonymous

Anonymous Mon 30 Nov 2020 17:14:41 No.12403707 Report

Quoted By: >>12403710

>>12403661
Puts in envelope
"You owe me 10 dollars"

Anonymous

Anonymous Mon 30 Nov 2020 17:15:47 No.12403710 Report

Quoted By:

>>12403707
Thats not money

Anonymous

View Same Google iqdb SauceNAO Untitled.png, 5KiB, 800x600

Anonymous Mon 30 Nov 2020 17:18:34 No.12403719 Report

Quoted By: >>12403721

Hey, I created a "puzzle" this night. I can't make a new thread, because some monkey decided to get my IPrange banned, so this thread is perfect.

(1/4)
Splinter Paradox

This is a nice example of how the Real numbers can be counter-intuitive, that came in my mind during a long night of insomnia. You can present this to highschoolers you might be tutoring and college freshmen, you can present it as a really interesting puzzle, none of that boring game theory shit like “in a mexican standoff there’s Frank, Kennedy and Gary...”.

Now, imagine we are working with wood. We want to make a wood panel by connecting two pieces of wood, birch (a fair wood) and walnut (a dark wood). Gluing two squares is boring, so we want to turn the connection point into something interesting. We decide to make the junction as triangles, as the image below (Please ignore the scale. Orient yourself by the values written).

Anonymous

View Same Google iqdb SauceNAO Untitled2.png, 5KiB, 800x600

Anonymous Mon 30 Nov 2020 17:19:40 No.12403721 Report

Quoted By: >>12403728

>>12403719
(2/4)
As anyone can see, we have a perfect walnut triangle, and for those who know a bit of geometry, an identical birch triangle, but split in half. If you work out the areas, you will find that in the middle, 50% of the wood is birch, and 50% is walnut, and in total the piece is 2 units2 walnut and 2 units2 birch. Imagine now that we decide to have more than one walnut triangle, but keeping the width and height of the panel.

Anonymous

Anonymous Mon 30 Nov 2020 17:20:48 No.12403728 Report

Quoted By: >>12403731

>>12403721
(3/4)
Nice. We notice that the areas haven’t changed: the middle section still is 50-50, and each wood type is exactly 2 units2 in area. We notice that this change in the junction can be interpreted as a function of the angle alpha in the previous image. We can progressively reduce this angle, and more triangles will fit in that 1 unit width panel. But we are not savages, we will choose angles that allow for the whole walnut triangles to fit in, but it is intuitive that in these intermediary angles, our 50-50 property and 2 units2 area will be preserved.“oh, but what if instead make the triangle very obtuse?” Well, then we will have to reduce the height of the triangle as we grow the angle, and when we hit 180°, there will be just two rectangular shapes, simple planks, with 2 units2 each, but the central area will vanish, as it will be the border line, so talking about the distribution of the area is nonsense.

The really interesting stuff is when we make the angle more acute, turning the triangles into these 2 units long splinters, birch, walnut, birch, walnut, etc. The fun in this is that the areas are preserved, no matter how thin the splinters are. And the middle section will still be 50% walnut and 50% birch. So in our journey we make the angle as small as we can imagine. 0.1°, 0.001°, 1/TREE(3)°, etc. Always 50-50, and 2 units2 for each wood. Then we give up in our (literally) never ending quest of finding the lowest number above zero... and we turn the angle into zero itself.

Anonymous

View Same Google iqdb SauceNAO D8CRtMS.jpg, 41KiB, 374x374

Anonymous Mon 30 Nov 2020 17:21:54 No.12403731 Report

Quoted By: >>12403853

>>12403728
(4/4)
This might sound absurd, because it is, this is completely unrelated to our mundane reality, but thankfully our workshop is in the realm of imagination, so we can do that. In fact, that’s how we solve integrals. Well, if the angle is zero, then each splinter is infinitesimally thin, and we know that every splinter is sandwiched between two splinters of the opposing wood (except the right and left borders, but we can ignore that). And since they are infinitesimally thin, it means that infinite splinters of each wood can be found in there.

So here I present you the first question:

What is the proportion of each wood in the middle section, and what is the total area of each wood?

With your answer in mind, I present you the second question:

If we pull each piece of wood apart, what is the total area of the birch piece and the total area of the walnut piece?

Anonymous

Anonymous Mon 30 Nov 2020 17:28:15 No.12403763 Report

Quoted By: >>12403779 >>12403974

>>12403701
>you aren't getting any information
Holy shit, are you retarded? You open the envelope and get to see what X is, how can you say "you aren't getting any information"?

If you open the envelope, you put an actual value to X. Like X = 10, for example. You know without a doubt that X = 10. You also know that the other envelope has an expected value of 1.25X in it. Therefore you should switch. I think you get this part.
Suppose you don't open the envelope. You don't know what X is. 50% of the time, it's Y. 50% of the time, it's 2Y. But you don't know which, so there's no point in switching. Both envelopes have expected value 3Y/2.

Anonymous

Anonymous Mon 30 Nov 2020 17:31:32 No.12403776 Report

Quoted By:

>>12403701
You are completely correct. The other anon is retarded.

Anonymous

Anonymous Mon 30 Nov 2020 17:32:25 No.12403779 Report

Quoted By: >>12403789 >>12403896

>>12403763
There are two envelopes, one containing X$, one containing 2X$. Your probability of choosing either of them is 50%. Your probability of increasing your money by switching envelopes is also 50%.

Knowing that the envelope you chose contains 10$ gives you no information regarding whether your envelope is the X$ or 2X$ envelope, hence it changes nothing. Absolute morons on this board I swear.

Anonymous

Anonymous Mon 30 Nov 2020 17:34:02 No.12403789 Report

Quoted By: >>12403798

>>12403779
Glad to finally see someone understand what's going on. I am currently writing up my own discussion of the puzzle. Will post it soon.

Anonymous

Anonymous Mon 30 Nov 2020 17:34:19 No.12403792 Report

Quoted By:

>>12393989
Switch because I'm addicted to gambling
Also potential gain is more than the potential loss?
In any case you get at least 5 fucking dollars man that's something I can buy a PS2 memory card for that which I really need right now tbf

Anonymous

Anonymous Mon 30 Nov 2020 17:35:20 No.12403798 Report

Quoted By:

>>12403789
Please don't.

Anonymous

Anonymous Mon 30 Nov 2020 17:48:10 No.12403853 Report

Quoted By: >>12403888 >>12403911

>>12403731
are you arguing that each splinter's area becomes 0? it feels like the counterintuitiveness of that is just a consequence of imposing an impossible limit to a physical scenario, and doesn't have a lot of mathematical nuance

Anonymous

Anonymous Mon 30 Nov 2020 17:53:38 No.12403880 Report

Quoted By:

>>12393989
all of you brainlets forget that the reaction to loss vs gain is asymmetric as well. it is human nature to suffer from losses a lot more than gaining from wins. so no, keep the money and run with it.

Anonymous

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Anonymous Mon 30 Nov 2020 17:54:06 No.12403885 Report

Quoted By: >>12403890 >>12403902 >>12403926 >>12406320

>>12398108
I am the person who posted the problem. So far you're the closest one to the actual solution.
The most important point here is that this is a PHILOSOPHICAL problem, not a mathematical problem. The answer to this problem lies almost completely in the way you translate it into an actual mathematical problem.
>On average, I.e. if you did the experiment many times
This is very important. We don't know what happens if we repeat this experiment many times.
It's a one off event, and therefore the expectation is much less relevant than if we were allowed to repeat this experiment infinitely many times.
However, even to define the probability and expectations, some assumptions need to be made, we need to mathematically model the situation.
There's a problem in that there are many different models. It could be that one envelope contains $10 and the other contains $20, and we condition on the even that you chose the $10 envelope. Alternatively you could have $10 and $5. Another alternative is that you have $10 and half of the time the other envelope has $5 and the other half it has $20. It's not at all clear which algorithm of the three you would write to simulate the probabilities, since each algorithm satisfies equally well the problem statement, and hence the description of the situation is too vague to talk about the probabilities involved, without additional assumptions. In the wikipedia explanation for the version of the problem where you do not see what's in the envelope, their model assumes that the total sum of the envelopes is always the same, hence the expectation comes out to be 0.
This is a general problem in probability: a description of a one-off event is rarely enough to form a probabilistic model of what's happening.

Anonymous

Anonymous Mon 30 Nov 2020 17:54:42 No.12403888 Report

Quoted By:

>>12403853
It is not a physical scenario. It is a mathematical scenario, where the Reals are the norm. Just like when you apply calculus to find the area below a curve.

Anonymous

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Anonymous Mon 30 Nov 2020 17:55:09 No.12403890 Report

Quoted By: >>12403894 >>12403902 >>12403926

>>12403885
(2/?)
Another example of this is the Monty Hall. There are at least two possible models of the situation. One is where the presenter actually picks one of the other doors at random, and you just discard the cases where he opened the car, i.e. he just so happens to open a door with a goat. In that case, the probability of winning comes out to be 1/2 if you switch. Another case, the standard assumption is that the presenter knows which doors are hiding the goats, so always opens the door with the goat (note here the word ALWAYS: the explanation depends on what happens if you repeat this experiment many times). In that case the probability of winning when you switch is 2/3, as in the standard answer. The commonly made assumption that the presenter knows behind which doors the goats are can be made less obvious if we let the person opening the door be just a person on the stage. We don't know what we knows, so the probability of such a situation is not well defined.
Let us explore the concept of expectation a bit more. People ITT take the model where each time you pick $10 and half of the time the other envelope has $20 and the other half it has $5. They calculate the expectation if you switch to be $12.5. They infer that it's more beneficial to switch because if you switch the expectation is higher. But instead of looking at the money, let's "change the coordinates" and look at log_2(money) instead. Then with this model the expectation if you switch comes out to be 1/2(log_2(10)+1)+ 1/2(log_2(10) - 1) = log_2(10). The same as the expectation if you stay with your chosen envelope. So the expectation of log(money) is the same whether we switch or not. Curious! If we were to apply the same faulty logic to this, we would say we expect log_2(money) to be the same in either case, and hence we expect money to be the same in either choice.

Anonymous

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Anonymous Mon 30 Nov 2020 17:56:10 No.12403894 Report

Quoted By: >>12403898 >>12403902 >>12403926 >>12404005 >>12406337

>>12403890
(3/?)
Alternatively, we could replace log_2 with arctan(x) and then you will actually see that the expectation of arctan(x) is higher when you stay than when you switch, even though arctan is an increasing function. You can equally say that when you stay you expect arctan(money) to be higher, so since arctan(money) is an increasing function you expect money to be higher. The trouble here, of course, lies in the fact that we're trying to use expectation to tell us something about one-off event. Expectation is only relevant in the (infinitely) long run, when we apply the central limit theorem.
Coming back to calculating probabilities, you could say that even though you don't know what the experiment looks like when you repeat it infinitely many times you can still calculate the probabilities based on your degrees of knowledge. I.e. since for all you know the other envelope can have either $20 or $5, and these options are epistemologically symmetric, you assign probability 1/2 to both. This is a valid move but it still doesn't make the expectation relevant since it's a one-off event.
The most obvious rebuttal to all this is to say that sure, this is a one-off event, but in life there are many such one-off events. Isn't it true, that if we always chose the option with the higher expected value, even though we can't repeat each individual even infinitely many times, the distinct events add up and because we've chosen the better of two options in the long run we end up better off? The answer is that this doesn't work. Consider the following situation. At each step n, you have to make a decision, whether to get $1 or play a game. The game consists of flipping a coin. In the k'th turn of the game, if you get heads, you lose and the game ends, you move on to step n+1 without any new money. If you get tails, you have a choice to either end the game and take $2^k with you, or throw the coin again for the opportunity to win $2^(k+1).

Anonymous

Anonymous Mon 30 Nov 2020 17:56:31 No.12403896 Report

Quoted By:

>>12403779
Holy shit you have Dunning-Kruger. Or you're the anime retard samefagging and continuing his quest to shit up the thread. Maybe both.

>Knowing that the envelope you chose contains 10$ gives you no information regarding whether your envelope is the X$ or 2X$ envelope
This is right. You can never know if you have X or 2X.
>hence it changes nothing.
This is wrong.

If you don't open the envelope, then to you, both envelopes are identical. You had 50% chance of picking 2X, 50% chance of picking X. If you switch, you still have 50% chance of picking 2X, and 50% chance of picking X. Hence, there is no reason to switch.

If you do open the envelopes, you know one has 10. You stand to gain $2.5 on average by switching.

This is the paradox, and you fell for hook, line, and sinker.

Anonymous

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Anonymous Mon 30 Nov 2020 17:57:13 No.12403898 Report

Quoted By: >>12403902 >>12403926 >>12406337

>>12403894
(4/4)
Suppose you were to follow the strategy which maximizes your expectation at each point that you have to make a choice. In that case, what you would do is to at each step n, you play the game, and at each turn of the game you flip a coin, never taking out the profit. If you continue this, you will always eventually lose the game and NEVER get any money. To get any money whatsoever from the game, you NEED to make choices for which the expectation of money gained is smaller. This conclusively debunks the rebuttal.
Another thing to realize here is that the solution to this problem is NO DIFFERENT than the solution where you don't know what's inside either envelope, at least assuming that money is infinitely divisible (for example, if you got 1 cent in one envelope, you would know that the other envelope has 2 cents if we can't divide cents into smaller currency). Whatever your solution is to the first problem where you know the value, by symmetry it will not depend on the particular value you find in the envelope, so your solution will not use in any way the value you find in the envelope, and that means the problem is equivalent to one where you don't know the value. From this we can see that it actually doesn't matter whether or not you switch. Anons ITT who suggest that "the money inside the envelope is a random variable if you don't open it" are wrong because they already presuppose the probabilistic model they want to apply to the problem when this is the question in the first place. And as I already explained, the only natural probabilistic model to apply here is the one based on epistemological symmetry and that even using that model, there is nothing that actually suggests it's better to switch (blindly applying expectation doesn't work).

In conclusion, the answer to this problem is that it doesn't matter if you switch or not, neither choice is better because the problem is actually symmetric.

Anonymous

Anonymous Mon 30 Nov 2020 17:57:52 No.12403902 Report

Quoted By:

>>12403885
>>12403890
>>12403894
>>12403898
retard

Anonymous

Anonymous Mon 30 Nov 2020 18:00:22 No.12403911 Report

Quoted By:

>>12403853
Complementing, the "paradox" is that when you pull those pieces away from eachother, you will end up in one hand with a birch rectangle that is 3 units2, and a walnut rectangle that is 3 units2. I got inspired by the fuckery in te Banach-Tarski "paradox".

Anonymous

Anonymous Mon 30 Nov 2020 18:01:22 No.12403916 Report

Quoted By:

>>12393971
You can put them in neither bucket as well? Wtf kinda problem is that.
8, 16, 32, 64, 65, 66, 69, 95, 99, 100..

Anonymous

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Anonymous Mon 30 Nov 2020 18:02:49 No.12403926 Report

Quoted By: >>12403953

>>12403885
>>12403890
>>12403894
>>12403898
>muh "it was philosophical all along"
what a backtracking retard

Anonymous

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Anonymous Mon 30 Nov 2020 18:07:38 No.12403953 Report

Quoted By: >>12403961

>>12403926
>backtracking
When did I imply it wasn't a philosophical problem? The problem statement itself is philosophical, i.e. it's not a question that's formalizable in ZFC. The formalizability of the question is part of the question, just like with so many problems in probability and physics.
You sound like a really dumb person.

Anonymous

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Anonymous Mon 30 Nov 2020 18:09:31 No.12403961 Report

Quoted By: >>12403998

>>12403953
imagine being assblasted so hard you rack your brain for 3 days and the best you can come up with is that it was philosophical all along

Anonymous

Anonymous Mon 30 Nov 2020 18:11:54 No.12403974 Report

Quoted By: >>12403986 >>12404023

>>12403763
You, and the link you posted are making the same mistake I've pointed out multiple times iit. You are making the completely unsupported assumption that the distribution of $5 and $20 is uniform, there's even a comment on your link that points this out.

The idea that your EV is 1.25x is so utterly wrongheaded that sometimes its difficult to know how best to approach it. Think about the implications of what you are saying:

>original choice is a random binary selection
>switching is the opposite of a random binary selection
>the opposite of a random binary selection is just as random as the original selection
>but somehow the first random selection is systematically worse

Anonymous

Anonymous Mon 30 Nov 2020 18:14:15 No.12403986 Report

Quoted By: >>12404053

>>12403974
True no one said what the distribution was, 50/50 was an assumption, perhaps unsupported, but I think it's reasonable. All that it would change is the EV, not the overall answer.

Even if it was 10/90, or even 0/100, it doesn't change this fact: If you don't see the envelope content, there's no point in switching. That's all I've been trying to say this entire time.

Anonymous

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Anonymous Mon 30 Nov 2020 18:17:05 No.12403998 Report

Quoted By: >>12404002 >>12404013

>>12403961
Holy shit. Philosophical as opposed to what? Mathematical? Do you see any mathematics in the question? Obviously a big part of the solution is how you use reason and potentially mathematics to figure it out. It's very sad that it took me pointing it out for you to realize it.
Also just because the question is philosophical in nature, doesn't mean it doesn't have an answer. I explained the answer in my post. I suggest you actually read the posts you reply to before getting so pissy about them.

Anonymous

Anonymous Mon 30 Nov 2020 18:17:22 No.12404002 Report

Quoted By: >>12404010

>>12403998
retard

Anonymous

Anonymous Mon 30 Nov 2020 18:18:06 No.12404005 Report

Quoted By: >>12404200 >>12406337

>>12403894
>The most obvious rebuttal to all this is to say that sure, this is a one-off event, but in life there are many such one-off events. Isn't it true, that if we always chose the option with the higher expected value, even though we can't repeat each individual even infinitely many times, the distinct events add up and because we've chosen the better of two options in the long run we end up better off? The answer is that this doesn't work
Kinda regret defending you now, this clearly does work, its what I do for a living.

Anonymous

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Anonymous Mon 30 Nov 2020 18:18:40 No.12404010 Report

Quoted By: >>12404036

>>12404002
Why are you so angry lmao

Anonymous

Anonymous Mon 30 Nov 2020 18:19:01 No.12404013 Report

Quoted By: >>12404349

>>12403998
>dude I'm not wrong if I just interpret the question differently!

Anonymous

Anonymous Mon 30 Nov 2020 18:19:39 No.12404018 Report

Quoted By: >>12404217

Here's a puzzle for all you smoothbrains.

You are given two envelopes, one containing X$, one containing 2X$, where X > 0. After choosing an envelope with an unknown amount Z$ inside it, and without opening it, you may switch envelopes any number of times.
Since you are a mathematical genius you know that switching envelopes has an EV of (5/4)*Z which is greater than Z, so switching is a net positive choice. How often do you have to switch envelopes to get a cool EV of at least 100Z$?

Anonymous

Anonymous Mon 30 Nov 2020 18:20:49 No.12404023 Report

Quoted By:

>>12403974
It isn't always worse. Suppose you open the envelope and the value is 7. If you feel coinage in the other envelope, then you don't switch.
I get it, you don't like money. Doesn't mean the rest of us have to be communists. I'll take my 1.25 ev upfront and compounded infinite times thank you very much.

Anonymous

Anonymous Mon 30 Nov 2020 18:22:15 No.12404036 Report

Quoted By: >>12404040

>>12404010
retard

Anonymous

Anonymous Mon 30 Nov 2020 18:23:28 No.12404040 Report

Quoted By: >>12404043

>>12404036
SEETHING brainlet

Anonymous

Anonymous Mon 30 Nov 2020 18:23:45 No.12404043 Report

Quoted By: >>12404051

>>12404040
retard animeposter

Anonymous

Anonymous Mon 30 Nov 2020 18:24:59 No.12404051 Report

Quoted By: >>12404068

>>12404043
Go read a textbook on rigorous probability. You're clearly way in over your head.

Anonymous

Anonymous Mon 30 Nov 2020 18:25:35 No.12404053 Report

Quoted By: >>12404059 >>12404167

>>12403986
The EV IS the overall answer. If the EV is less than 10 (or X if unopened) you don't switch, if it is, you do, but the EV is can't be calculated without knowing the probability distribution, therefore there is never any reason to switch or not switch.

I know what you've been trying to say, its wrong. You don't get any meaningful information from opening the envelope.

Anonymous

Anonymous Mon 30 Nov 2020 18:26:28 No.12404059 Report

Quoted By: >>12404122

>>12404053
Wrong. If you open the envelope the entire question changes.

Anonymous

Anonymous Mon 30 Nov 2020 18:28:01 No.12404068 Report

Quoted By:

>>12404051
Go read a textbook on philosophy. You clearly know nothing about it.

Anonymous

Anonymous Mon 30 Nov 2020 18:30:25 No.12404078 Report

Quoted By: >>12404135

>google "two envelope paradox"
>5000000 results saying there's no point in switching when you don't open the envelope
Holy shit can all you envelope retards fuck off?

Anonymous

Anonymous Mon 30 Nov 2020 18:39:53 No.12404122 Report

Quoted By: >>12404133

>>12404059
It doesn't, which you'd understand if you'd taken on board what I said about the unknown probabilitiy distribution. Perhaps you should explain why you wouldn't switch an unopened envelope to clarify your position.

Anonymous

Anonymous Mon 30 Nov 2020 18:42:01 No.12404133 Report

Quoted By: >>12404154

>>12404122
It does. When you don't open the envelopes, the two sealed envelopes are the same. You have no advantage when picking either one. The distribution makes no difference once the money is in the envelope - it's all just X or 2X.

Anonymous

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Anonymous Mon 30 Nov 2020 18:42:38 No.12404135 Report

Quoted By: >>12404199

>>12404078
>consensus is truth
>science over facts
>wikipedia deboonked this

Anonymous

Anonymous Mon 30 Nov 2020 18:46:21 No.12404154 Report

Quoted By: >>12404170

>>12404133
Its still X and 2X after opening though, you just now know the value of either X or 2X, but not which it is.

Anonymous

Anonymous Mon 30 Nov 2020 18:50:20 No.12404167 Report

Quoted By: >>12404218

>>12404053
It's always 50/50 because that's the chance you pick either envelope. Doesn't matter what the distribution was.

Anonymous

Anonymous Mon 30 Nov 2020 18:50:59 No.12404170 Report

Quoted By:

>>12404154
You have a choice between a hundred envelopes. Each one contains x^n dollars, where n is from 0-100. You pick an envelope and open it, inside is one dollar. Do you switch envelopes?

Anonymous

Anonymous Mon 30 Nov 2020 18:56:07 No.12404199 Report

Quoted By:

>>12404135
I don't get what you're trying to say. Is he wrong just because everyone else agrees?
Do you also think that 1 + 1 = 3?

Anonymous

Anonymous Mon 30 Nov 2020 18:56:12 No.12404200 Report

Quoted By: >>12404242

>>12404005
It doesn't work. Here's another example (I realized my previous example was a bit flawed). You start with 1$. If your balance is at least 1$ I offer you a choice of either 1$ more or flipping a coin. Tails means you give me your whole balance and heads means you quadruple your balance. If you always choose the option with the higher expected value, you will always flip the coin and eventually run out of money. Always choosing the option with higher expected value leads you to go bankrupt while always choosing the option with lesser expected value leads you to as big of an amount of money as you want.
This clearly demonstrates that always choosing the option with higher expected value is not an an optimal strategy (in this case, it's the WORST strategy).

Anonymous

Anonymous Mon 30 Nov 2020 18:59:32 No.12404217 Report

Quoted By:

>>12404018
Basically this is the same ligic as red and black on a fair roulette table. Do you let it ride?
Fuck it, do it! Vagas wouldnt give you those odds

Anonymous

Anonymous Mon 30 Nov 2020 18:59:46 No.12404218 Report

Quoted By: >>12404239

>>12404167
Prior to opening you think there is X and 2X, after opening therefore you now hold either X or 2X. The implication that switching is better suggests the other envelope has a greater chance of being the 2X envelope. Does this make sense?

Anonymous

Anonymous Mon 30 Nov 2020 19:05:32 No.12404239 Report

Quoted By: >>12404263 >>12404268

>>12404218
But the chance is 50/50 that the other envelope has 2X.

Anonymous

Anonymous Mon 30 Nov 2020 19:06:05 No.12404242 Report

Quoted By: >>12404748 >>12406372

>>12404200
Mate I bet on sports for a living, one off events can have their EV calculated in such a way that one can profit long term, thats all I was saying.

What you are talking about there is more akin to expected growth, which is different, and requires an optimal staking strategy, going all in is the exact opposite of an optimal staking strategy. Making the bet there isn't the mistake, the amount staked is.

Anonymous

Anonymous Mon 30 Nov 2020 19:09:55 No.12404263 Report

Quoted By: >>12404312

>>12404239
So first pick: 50% X, 50% 2X
After opening, switching is now: 50% X, 50% 2X

Where is the difference?

Anonymous

Anonymous Mon 30 Nov 2020 19:11:09 No.12404268 Report

Quoted By: >>12404273 >>12404312

>>12404239
So if it contained a value of 10, in 50 swaps you may have lost a max of approx 10, but you could earn 10^50
Why would you not now swap 50 times
Sound much better than a casino to me

Anonymous

Anonymous Mon 30 Nov 2020 19:12:37 No.12404273 Report

Quoted By:

>>12404268
Oops, 10*2^50

Anonymous

Anonymous Mon 30 Nov 2020 19:20:58 No.12404312 Report

Quoted By: >>12404333 >>12404335

>>12404263
I've been saying this the whole time.

To be clear, my argument is
>open the envelope, always switch
>don't open the envelope, no point in switching

>>12404268
That's exactly the fallacy in the "switch when you don't open the envelope" argument. The money cannot increase forever.

Anonymous

Anonymous Mon 30 Nov 2020 19:26:00 No.12404333 Report

Quoted By: >>12404343

>>12404312
Forever? It worth doing a few hunfmdred times.

Its true the law of large numbers means given enough swaps you will end up even

But, over a small sample, with it be exponential decay/growth in value, its makes more sense to swap

Anonymous

Anonymous Mon 30 Nov 2020 19:26:55 No.12404335 Report

Quoted By: >>12404352 >>12404368

>>12404312
But the logic I posted shows there is no difference. Prior to opening, the chance the 2X is in the other envelope is 50%, after opening, it is still 50%. You have gained zero information because you don't know if the $10 you've seen is the X or the 2X.

Anonymous

Anonymous Mon 30 Nov 2020 19:27:14 No.12404340 Report

Quoted By:

>only one anon gave a fuck about my poor's man sphere duplication
>everyone else is fighting instead of posting nice puzzles

:(

Anonymous

Anonymous Mon 30 Nov 2020 19:27:51 No.12404343 Report

Quoted By:

>>12404333
Being the potention returns are greater than the potentisl loses... Significantly

Anonymous

Anonymous Mon 30 Nov 2020 19:29:41 No.12404349 Report

Quoted By:

>>12404013
How am I interpreting it differently? Differently from what? How is your interpretation different from mine?

Anonymous

Anonymous Mon 30 Nov 2020 19:30:30 No.12404352 Report

Quoted By: >>12404365

>>12404335
It doesnt matter the actual value, just that the next swap is based on your current value of x

Anonymous

Anonymous Mon 30 Nov 2020 19:33:35 No.12404365 Report

Quoted By: >>12404373 >>12404402

>>12404352
You don't know if you have X, there's a 50% chance you're holding 2X. This is true whether or not you open the envelope.

Anonymous

Anonymous Mon 30 Nov 2020 19:34:23 No.12404368 Report

Quoted By: >>12404385

>>12404335
You gained information because now you know either X or 2X is 10. You also know the other thing is 5 or 20, so you can compute the EV.

If you don't know it's 10 (ie you don't open the envelope), both EVs are 3X/2.

Anonymous

Anonymous Mon 30 Nov 2020 19:35:22 No.12404373 Report

Quoted By: >>12404380 >>12404385

>>12404365
Upon opening the envelope, there is a 50% chance you double your money by switching.

Anonymous

Anonymous Mon 30 Nov 2020 19:36:45 No.12404380 Report

Quoted By: >>12404410

>>12404373
No, the chance is only 1/4. You have to condition on actually picking the smaller envelope.

Anonymous

Anonymous Mon 30 Nov 2020 19:37:50 No.12404385 Report

Quoted By: >>12404391

>>12404368
You can't compute the EV after opening, this is madness, a few posts ago you were agreeing with me that the probability distribution is unknown.

>>12404373
No there isn't, there is an entirely unknown probability of doubling it.

Anonymous

Anonymous Mon 30 Nov 2020 19:39:05 No.12404391 Report

Quoted By: >>12404431

>>12404385
No, I said the probability distribution doesn't matter because it's always 50/50. Of course you can compute the EV after you open.

Anonymous

Anonymous Mon 30 Nov 2020 19:41:56 No.12404402 Report

Quoted By:

>>12404365
Oh so we are swapping the same 2 evelopes over and over? My bad.

I thought once you swap and you have x or 2x, then the next time would be x/2 or 4x, and the x/2 or 8x
Sorry i was making this more interesting than it is

Anonymous

Anonymous Mon 30 Nov 2020 19:44:12 No.12404410 Report

Quoted By: >>12404425 >>12404434 >>12404447

>>12404380
Casino gives you odds, 2:1 on a coin flip. Do you play that game? What is the house edge? How many other games are more profitable?

Anonymous

Anonymous Mon 30 Nov 2020 19:47:55 No.12404425 Report

Quoted By:

>>12404410
I would if they gave me half the money i bet back if i lost

Anonymous

Anonymous Mon 30 Nov 2020 19:48:25 No.12404431 Report

Quoted By: >>12404476

>>12404391
You are confusing two entirely different probability distributions. There is no information in the puzzle that tells you the methodology in which the money was put into the envelopes.

Imagine prior to you entering, the experimenters put $10 in one envelope, and then $20 in the other with probability 0.1 and $5 with probability 0.9. This satisfies the information within the puzzle but will not produce an EV of 12.50. We can't just assume, upon seeing a $10, that $5 and $20 are equally likely, their probabilities are unknown. This is why switching makes no difference regardless of opening or not.

The probability of choosing X or X2 (the higher or lower amount) is 50/50, but that doesn't give us any meaningful information, because we don't know if the value we have is the high or low value, and we knew that prior to opening.

Anonymous

Anonymous Mon 30 Nov 2020 19:49:07 No.12404434 Report

Quoted By: >>12404476

>>12404410
How do you get zero in this game?

Anonymous

Anonymous Mon 30 Nov 2020 19:50:59 No.12404447 Report

Quoted By: >>12404476

>>12404410
How do you get 0 unless all evelopes are empty? It not a valid comparison

Anonymous

Anonymous Mon 30 Nov 2020 19:54:52 No.12404476 Report

Quoted By: >>12404558 >>12404897

>>12404431
Wrong. The envelopes in front of you is either X or 2X. It does not matter about how money was allocated on other envelopes. Your excuse to throwaway money is fine, I'll bag it.
>>12404434
>>12404447
You have ten dollars in the envelope. Do you risk $5 to make $10?

Anonymous

Anonymous Mon 30 Nov 2020 20:11:47 No.12404558 Report

Quoted By:

>>12404476
>two envelopes in front of you
>one has X, one has 2X
>you want the 2X
>you pick one at random, 50% chance you have X, 50% chance you have 2X
>therefore 50% the other envelope has 2X
>you look in your envelope and see a value
>still don't know if its X or 2X
>but now the other envelope magically has a greater chance to contain the 2X
Genius

Anonymous

Anonymous Mon 30 Nov 2020 21:06:12 No.12404748 Report

Quoted By:

>>12404242
Please explain, point by point, the thought process that leads you to think it's better to switch the envelope.

Anonymous

Anonymous Mon 30 Nov 2020 21:39:07 No.12404897 Report

Quoted By:

>>12404476
Oh i see, thanks, its still a bit grey...

So if i was presented this option 6 times
Option 1: just take the 10 = 60
Option 2: swap 50% @ 5, 50% @20 = 75

But this assumes each option is completely independent from each other and these numbers just used to demostrate the logic

So i would always swap. What have i misunderstood?

Anonymous

Anonymous Mon 30 Nov 2020 23:42:54 No.12405386 Report

Quoted By: >>12406219

>There's an irregular cake
>a person's enjoyment value of a slice of this cake is subjective, depending not only on its size but on the distribution of ingredients contained in it or whatever
>say the enjoyment provided by a full cake is C
>the method for distributing this cake between two people so each get a slice worth at least a subjective C/2 is the following: person 1 cuts it into two slices he considers equal, and person 2 chooses the slice he likes the best
>extrapolate this method for 3 people, so each of them gets a slice he considers at least worth C/3
I have no idea what the answer is, I wanted to try solving it and the page where I read it later died. If you solve it, please share.

Anonymous

Anonymous Tue 01 Dec 2020 04:12:05 No.12406219 Report

Quoted By:

>>12405386
Person A cuts the cake into 3 pieces he thinks are fair, person B picks his favorite 2, and trims a bit off of his 1st favorite such that his 1st favorite becomes equal to his second. Person C gets his favorite, person B one of his remaining favorites, the one that has been trimmed if person C hasn't taken it, and person A gets the other one. The trimming is cut in three pieces that person C considers fair, person B takes his favorite, person C takes one and person A takes the other. Person A isn't jealous because he got a fair third of the cake plus some more cake, person B isn't jealous because he got his favorite slice and favorite part of the trimming, person C isn't jealous because he got his favorite slice and a fair bit of the trimming. This can be generalized for n people but the number of cuts grows very quickly.

Anonymous

Anonymous Tue 01 Dec 2020 04:51:22 No.12406320 Report

Quoted By:

>>12403885
Based

Anonymous

View Same Google iqdb SauceNAO 874644E9-5C70-4812-8F0A-99504603 (...).png, 455KiB, 1336x538

Anonymous Tue 01 Dec 2020 05:01:53 No.12406337 Report

Quoted By:

>>12403898
>>12403894
Based
>>12404005
> its what I do for a living.
I bet you think you can scientifically prove causation by isolating variables without a ransomed control group, like in economics.
Here is what will happen if you do this for a living, you will make reasonable returns for several years and then get wiped out

Anonymous

View Same Google iqdb SauceNAO F9D73A77-09C4-48A9-A82A-309A2BDA (...).png, 406KiB, 1170x666

Anonymous Tue 01 Dec 2020 05:20:28 No.12406372 Report

Quoted By:

>>12404242
If it’s so sure then why are you not retired as a 9 figure millionaire after 5 years or so if you have a guaranteed average positive return over the long run ?

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