/sci/ - Science & Math

>>9953207
if you say 50% of the time the other envelope has $20, and 50% of the time it has $5, the E.V. of switching is $25/2=$12.50 > $10.00 so it appears you should switch.
However: the situation is symmetrical. You chose an envelope at random, and nothing changed afterwards. So the idea that switching gives you a benefit is absurd.

>>9953217
How is it symmetrical if 1 envelope is opened and the other is unopened?

>>9953207

The trick is realizing that you have an improper prior. There are two variables; the total sum in the envelopes, and whether you got higher or lower.

Use Bayes rule and you'll see why assuming that the total sum is a uniformly distributed across all amounts of money leads to this result.

https://math.stackexchange.com/questions/964381/if-you-have-two-envelopes-and

>>9953226
it's symmetrical because if you had opened the other envelope, the same logic would tell you to switch.

It's not a paradox. Just insufficient information to answer.

>>9953217
>You chose an envelope at random, and nothing changed afterwards.
But if you switch envelopes you no longer have the one with $10, so it's wrong to say nothing changed.

>>9953252
But if you opened the second envelope and find $10 instead, then you wouldn't know what's in the first envelope

>>9953252
>it's symmetrical because if you had opened the other envelope, the same logic would tell you to switch.
Yes, because the optimal strategy is to switch.

>>9953256
>>9953258
The $10 doesn't matter to that logic, EV is 5/4 of the opened envelope
>>9953259
as detailed at the link here >>9953236 switching and not switching have equivalent E.V.s

>>9953270
What is the EV is not switching and what is the EV of switching?

>>9953272
>What is the EV is not switching and what is the EV of switching?
10$ and 12.5$ respectfully

>>9953207
>I don't understand how this is a paradox.
It's only a paradox in the same way that the Fermi "paradox" is a paradox (it's not because the absence of evidence is not evidence of absence).

I.e. it's only a paradox for brainlets with a popsci understanding of probability

>>9953276
So it actually is better to switch

>>9953289
>So it actually is better to switch
Of course, because you either get half or double what's in the envelope you opened. Unless you're a dumb poorfag who would be happy with $10, any intelligent gambler would take the risk for the doubling.

>>9953287
What's the "paradox" then?

>>9953294
>What's the "paradox" then?
There isn't one, just like how there isn't one for the Fermi "paradox" either. It's a common theme in popsci to label things as paradoxes even when they're not, i.e. Fermi's, Zeno's, Theseus, etc.

>>9953236
There isn't even a paradox here. Just look at the EV. Others already worked it out.

>https://math.stackexchange.com/questions/964381/if-you-have-two-envelopes-and
>The assumption that is unrealistic is that there is a [math]\frac{1}{2}[/math] chance that the other envelope contains twice the money. >Realistically, there is an underlying distribution of values and that distribution dictates the probability that a given amount is the smaller.
This "logic" contradicts the fact that we had a 50% chance of picking the higher valued envelope in the first place.
If the first pick somehow wasn't 50/50, it should be stated in the problem.

>>9953294
The paradix is that intuitively, it seems like you should switch but also intuitively it seems like shouldn't matter if you switch. The paradox is resolved when you do math and realize it doesn't matter if you switch.

>>9953320
>but also intuitively it seems like shouldn't matter if you switch
Your intuition is faulty if yours is telling you this.

>>9953305
so there actually is a paradox here: switching with a probability depending on the amount you open is a better strategy than always switching and always staying. it's not really symmetrical.

>>9953325
switching doesn't matter so idk

>>9953329
>switching doesn't matter so idk
The only way switching wouldn't matter is if the envelopes contained the same amount of money, which is not the case.

>>9953331
to be precise: on average, any stategy that involves switching is no better than any strategy that doesn't involve switching

thanks for injecting this autism into my life, I was missing it

>>9953338
>to be precise: on average, any stategy that involves switching is no better than any strategy that doesn't involve switching
In what sense?

>>9953326
No, there's no paradox, at least not as far as OP's puzzle is concerned. It was stated that $10 is in the envelope. "The amount you open" is fixed. You're either losing $5 or gaining $10, and the fact that $10 was in your envelope has no bearing on what the other envelope contains barring some outside information about your adversary's finances.
But in the fantastic "all real values" problem you're referencing, yes, intuitively, we can see an advantage by realizing that as our number gets lower, it becomes likelier that it's the lower of the two.

https://blog.xkcd.com/2010/02/09/math-puzzle/comment-page-1/#comment-17804

>>9953331
the absolute state of /sci/
>durr, no matter which envelope I chose, I should still switch, making the whole act of switching completely pointless

>>9953345
the absolute state of /sci/
>durr, no matter which envelope I chose, I shouldn't switch, making the whole point of there being a second envelope completely pointless

>>9953345
>implying you know beforehand which envelope you picked
Oh look, schizo-kun is back.

>>9953207
What's the point of switching? You have a 50% chance of getting the larger sum, 50% chance of getting the smaller sum.
Knowing how much money is in one envelope is meaningless.

>>9953349
>What's the point of switching? You have a 50% chance of getting the larger sum, 50% chance of getting the smaller sum.
The gain from the larger sum is larger than the loss from the smaller sum.

>>9953343
it applies to this puzzle too, since you know a positive amount of money is divided between the two envelopes, switching with some probability gives you a slightly higher EV
In fact, if the experiment is posed where you ALWAYS know the first envelope has $10, you should always switch.

>>9953352
brainlet who fell for the illusion detected.

>>9953357
>brainlet who fell for the illusion detected.
brainlet who fell for the x = x/2 illusion (i.e. x=0, which is not the case since the envelope is given to contain 10$) detected

>>9953352
you don't get any greater information from knowing what you got, that's just to trick retards into thinking switching matters.
>The gain from the larger sum is larger than the loss from the smaller sum.
The total amount in both envelopes is a constant {\displaystyle c=3x} {\displaystyle c=3x}, with {\displaystyle x} x in one envelope and {\displaystyle 2x} 2x in the other.
If you select the envelope with {\displaystyle x} x first you gain the amount {\displaystyle x} x by swapping. If you select the envelope with {\displaystyle 2x} 2x first you lose the amount {\displaystyle x} x by swapping. So you gain on average {\displaystyle G={1 \over 2}(x)+{1 \over 2}(-x)={1 \over 2}(x-x)=0} {\displaystyle G={1 \over 2}(x)+{1 \over 2}(-x)={1 \over 2}(x-x)=0} by swapping.
Swapping is not better than keeping. The expected value {\displaystyle \operatorname {E} ={\frac {1}{2}}2x+{\frac {1}{2}}x={\frac {3}{2}}x} {\displaystyle \operatorname {E} ={\frac {1}{2}}2x+{\frac {1}{2}}x={\frac {3}{2}}x} is the same for both the envelopes.

>>9953361
>you don't get any greater information from knowing what you got
What definitions of 'information' and 'knowledge' are you using that allow for non-informational knowledge?

>>9953371
useful information

>>9953207
>>9953217
The beauty of these types of threads is I can code up the solution in 5 minutes.
If 100,000,000 trials doesn't convince you to switch then all I have to say is enjoy being poor.

If you would switch when the first envelope contains $10, you would switch no matter how much money was in the first envelope, right? So if you predetermine that no matter what you open first, you'll switch, why don't you just pick the other envelope first and stick with it? And if that's the case, how can switching ever provide any benefit?

>>9953429
you ran the experiment wrong, you are begging the question. you need to put x into one envelope 2x into the other envelope and see how the strategy of always switching affects your EV.

>>9953468
>using the problem's given conditions is begging the question

>>9953429
You are so dumb lol

>>9953451
>So if you predetermine that no matter what you open first, you'll switch, why don't you just pick the other envelope first and stick with it?
Why would you stick with the first envelope if you predetermine that you'll switch?

>>9953361
you have to use [math] [/ math] for in-line LaTeX rendering or [eqn] [/ eqn] for blocked line (remove the space after the slash)

>>9953481
The problem gives a scenario in which, on choosing an envelope randomly, you happen to have chosen a 10. This is different from scenarios in which you are actually guaranteed to always have a 10 and there is actually a 50% chance that the other envelope will be higher, which will only happen in particular distributions, which you don't know a priori from the problem scenario.

A 'correct' simulation would place money from a particular distribution, then filter out everything except the cases where you happen you chose a 10 initially. In which case the expected value will depend on the distribution chosen. There is actually no 'neutral' or mathematically pure way of simulating the problem, and that's what makes it interesting.

>>9953429
>coding in Basic
You might as well end your suffering.

Did a simulation of this to confirm that switching is the better strategy. Reminds me of the Monty Hall "paradox"

>>9953914
> Reminds me of the Monty Hall "paradox"

Yes, in that simulations usually fail to capture the actual nuances of the problem and reduce it to something simpler instead.

>>9953429
>Visual Basis LUL
Autism/10

>>9953914
Nevermind, my simulation was faulty. Here is the proper explanation of a simulation.

$10 in lesser envelope
$20 in greater envelope

We would initially randomly pick the lower envelope half the time, then per our strategy, switch to the greater envelope. Meaning we get $20 in this case.

Then we would initially pick the higher envelope half the time, then switch to the lower envelope. Meaning we get $10 in this case. Expected value is therefore 20*.5 + 10*.5 = 15.

If we don't switch, then when we pick the greater envelope, we get $20. And when we pick the lesser envelope, we get $10. Expected value is still $15.

Kind of embarrassed it took me this long desu

>>9953969
That is not exactly right. We don't know if the 10 dollars is the low amount or the high amount. The other envelope could be 5 or 20.
2x, or 0.5x, and the average of the two is 1.25x . +25% should be expected gain.

>>9953972
>+25% should be expected gain.
no
https://en.wikipedia.org/wiki/Two_envelopes_problem

>>9953972
It doesn't matter what the low or high amount may be. There are only two envelopes, each with a fixed amount in them during every iteration of a simulation, one envelope having x and the other having 2x. Your reasoning implies that even if we happen to choose the higher envelope, there is still a 50% probability of there being twice as much money in the other envelope, which is wrong. Read my post again.

>>9953207
50% chance for 5$, 50% chance for 20$ on switching which yields a higher return on average. Where's the paradox?

>>9953976
>>9953976
Not in this case because know that it has 10 dollars in it. That value isn't going to change because I switch envelopes.

>>9953217
>You choose an envelope at random
No you don't, you know you got the envelope with 10$ and that's why you want to switch.
View it like this: There are 3 envelopes: 5, 10 and 20. The gamemaster chooses 5 + 10 or 10 + 20 with equal chance and hands them to you. If you get 5 or 10 you switch if you get 20 you stay.

>>9953996
If it is supposed to be completely random what values the envelopes contain then the initial distribution of one of the values matters. you cant have a uniform distribution over all real numbers

>>9953207
Is this really as fucking stupid as "gaining $10 is better than losing $5 when the odds are equal"? Fuck this gay earth

>>9953993
There is a 50% probability you chose x, the smaller envelope. Let's say it has $10. There is a 50% chance you chose 2x, the bigger envelope, let's say it has $20.

If you initially choose the bigger envelope, and you switch, you end up with $10. If you initially choose the smaller envelope and you switch, you end up with $20. Expected value is still $15.

If you initially choose the bigger envelope and stay, you get $20. If you initially choose the smaller envelope and stay, you get $10. Expected value is $15.

The logic still works whether we're talking about a simulation of 10000 trials, or one individual offer.

>>9953996
>There are 3 envelopes
This is where you're screwing things up. You're giving the 'game' a built in advantage that IS NOT THERE, by assuming that you have three envelopes, which is obviously false.

>>9954003
The person who designed this game put $10 in one envelope. Then he threw a dice. If the result was 1 or 2 he put $5 in the other envelope else he put $20 in it. You don't know the result of the die throw and happen to have gotten the $10 envelope by random choice. Given that wouldn't it be rational to switch?

OP's problem is the same exact it doesn't say how the envelopes were filled.

>>9954003
>by assuming that you have three envelopes, which is obviously false
If you reject this then there has to be a distribution of envelope contents, otherwise you cant start talking about probabilities.
For example fix the lower value and let the higher be dependant on it. Then you need a distribution of lower values in the experiment. There is no uniform dostribution over all numbers.
The choice of the distribution highly affects the outcome and makes any choice of envelope unsymmetric because depending on the distribution you will influence the choises switch and stay.

>>9954006
>Then he threw a dice. If the result was 1 or 2 he put $5 in the other envelope else he put $20 in it

That is NOT what the problem says. If the 'mechanism' for the problem were like that, it would be if he rolled a 1, 2, or a 3, not just a 1 or 2.

It is irrelevant how the envelopes were filled anyway, unless you're trying to turn this into a problem of trying to read the envelope filler's mind, instead of a raw math problem.

>>9954008
If you were doing this problem in a real life situation where you were attempting to "read" the person giving you the challenge, I don't think I'd disagree--I assume you're getting at the idea that if you picked an envelope and it had say $1,000 you would be hesitant to switch because you perceive it to be less likely that the other envelope has $2,000 as opposed to $500.

But if we're talking about raw numbers devoid of the human element (which I figured it was safe to assume in this context) it is irrelevant whether the envelope has $5 or $5,000,000,000

>>9954009
>it would be if he rolled a 1, 2, or a 3
In that case switching would still be better.
>It is irrelevant how the envelopes were filled anyway
It's not. Different fill algorithm lead to different expectation values.
>turn this into a problem of trying to read the envelope filler's mind, instead of a raw math problem.
There is not enough information given to be a raw math problem.

>>9954016
Yes, the amount is irrelevant. The $10 is just there to bait brainlets into calculating an incorrect EV

>>9954001
No, that's just you failing to see the real meat of the problem. Imagine being too stupid to even get the real 'paradox' here, and then boasting about it on a Taiwanese woodcarving BBS.

>>9954017
>In that case switching would still be better.
No

Can you explain why you still think switching is better?

>>9954019
Its not that the amount is irrelevant. Its that we know what is in the first envelope. The paradox arises in the possibility of constantly switching envelopes which is unnecessary. You know the amount, you switch and check the new amount. If the host allows you to switch back, then you pick whichever is higher. No contradiction.
What if he offers a new envelope every time you switch?
Clearly, the downside for every successive switch is -(1/2)^n • x and the upside is 2^n • x.
If either were randomly decided based on the current envelope, it would most definitely be profitable to keep switching until the fucker is bankrupt.

>>9954017
>Different fill algorithm lead to different expectation values.
In probability when something is unknown, it is called a random process. Probability is all about making the best use of all available information. If different filling algorithms lead to different EVs, then you should calculate the EV taking into account that the filing process is unknown

>>9954017
>It's not. Different fill algorithm lead to different expectation values.

Ok I see what you're getting at. Yes, if the fill algorithm for example were "okay, you chose the one with $10, now i'm going to fill the other envelope with a 50% probability of doubling it and a 50% probability of halving it" in which case I switching would be the correct strategy (in fact, you could keep switching, and if he kept "refiling" the envelopes, you could get infinite money). However I see nothing in the problem presented to suggest that that's how it works.

I was always assuming (correctly I think) that the values are predetermined and THEN you choose one, in which case the simulation above is correct and switching/not switching have equal EV's.

>>9954023
This explains it well:
http://consequently.org/papers/envelopes.pdf

It they are filled with the dice throw you have mechanism 2 from the pdf.

>>9954026
Knowing what's in the first envelope doesn't let you know that the second envelope is better

>>9954033
The envelopes are filled ahead of time. The filler can use whatever mechanism he wants, as long as he fills them ahead of time, like the picture suggests

>>9954033
I agree that the strategy is determined by the way the envelopes are filled. My assumption from reading the OP pic was that mechanism 1 was the correct way to look at it.

>>9954017
>Different fill algorithm lead to different expectation values
>>9954031
>if the fill algorithm for example were "okay, you chose the one with $10, now i'm going to fill the other envelope with a 50% probability of doubling it and a 50% probability of halving it"
That's not a valid filling algorithm in regard to the prompt. The only decision the filler can make is how much money he uses, which is irrelevant. If you think there's anything the filler can do to affect the outcome, let me know. Remember that the filler must fill the envelopes ahead of time

>>9954051
>Different fill algorithm lead to different expectation values
This is correct, but mechanism 1 as stated in the article is the correct mechanism based off of the OP picture.

>That's not a valid filling algorithm in regard to the prompt. The only decision the filler can make is how much money he uses, which is irrelevant. If you think there's anything the filler can do to affect the outcome, let me know. Remember that the filler must fill the envelopes ahead of time
I agree on this as well, I was simply stating what the answer would be with the other filling algorithm.

>>9954049
Why do you assume? The picture quite clearly states that the envelopes are prefilled

>>9953207
Wow OP. You troll. Here is the actual problem from wiki.

>You are given two indistinguishable envelopes, each containing money, one contains twice as much as the other. You may pick one envelope and keep the money it contains. Having chosen an envelope at will, but before inspecting it, you are given the chance to switch envelopes. Should you switch?

By knowing I get $10 on the first pick it's to my advantage to switch.

>>9954056
Mechanism 1 from the article posted above is equivalent to the prefilling mechanism you're talking about. No disagreement here.

>>9954049
>>9954051
>>9954055
>>9954056
Mechanism 1, 2 and 3 are all prefilled and nothing in OP's pic suggest that mechanism 1 is more likely than mechanism 2.

>>9953207
I am feeling like this whole thing is sophistry. The paradox, at least the one stated in the wiki, arises out of something like Buridan's ass. The problem is almost not even related to the math, but we turn to math to try to resolve some intuitively bad decision like constantly switching envelopes. I am reading probability textbook right now. I will come back and look at this once I get a better grasp and understand the scenarios if there is indeed meaningfully different interpretations.

>>9954060
>Mechanism 1, 2 and 3 are all prefilled
I just looked, and you're wrong. There is no prefilling algorithm that can affect the outcome of this senario. 2 and 3 and not prefilling mechanisms

>>9954060
Tell me, how are you going to prefill my envelope with x and the other envelope with x/2 or 2x? Will you magically predict with of the 2 envelope I will choose first?

>>9954064
They are prefilling mechanisms.
Another one would be to deterministically fill one envelope with $10 and the other with $20. It's still consistent with OP's pic.

>>9954060
Mechanism 1 is the only one which is prefilled before you make a decision on which envelope you choose. Mechanism 2 and 3 specifically mention "your envelope" and "the other envelope" respectively which means that they are filled after your decision has already been made. This is what causes the switching/not switching to yield higher expected values in mechanism 2 and 3 respectively.

>>9954066
>>9954068
No, you're fucking retards that can't read.

>>9954057
see part 5 in >>9954033
Why is /sci/ so retarded? Did /g/ invade us or something?

>>9954069
"A number, x, is chosen in any way one likes. That is put
in your envelope. Either 2x or x/2 is then put in the other envelope,
depending on the toss of a fair coin."

>>9954067
Prefilling one with x and the other with 2x is the way I assumed OP's pic worked, which is equivalent to mechanism 1.

>>9954069
Please enlighten us

>>9954071
Let x=$10. You get the one with $10 inside it.

>>9954075
In the case of mechanism 1 where the envelopes are prefilled, you can either get x or 2x. You cannot assume that x=$10.

>>9954075
>You get the one with $10 inside it
How will the filler know which envelope to place the $10 in ahead of time?

>>9954076
It does work with mechanism 2 however, in which case switching is the correct strategy.

>>9954080
If he placed the $10 in the other one then you wouldn't have gotten the one with $10.

How about trying to apply the goat game logic here:
Suppose we are using mechanism one, but instead of two envelopes we have ten envelopes each one filled with money uniquely filled from the set {n, 2n, ... , 2^10n}
n is such that if known it does not reveal any information about the value or ranking of any envelope.
Opening the envelope reveals some value m, then the option to switch with any of the envelopes is presented. What is the best course of action?

>>9954082
So you're assuming that the filler will only ever use $10 and $20?

>>9954084
I'm not assuming shit.

>>9954083
>ten envelopes
>{n, 2n, ... , 2^10n}
What does the "..." stand for because to me it looks those are eleven envelopes.

>>9953207

question isn't very clear about what we're taking the expectation over.

if the two configurations are {5,10} and {10,20}, and we are shown a 10, then either is equally likely, but we always stand to gain more by switching.

if it were {5, 10} {10,15}, and we drew a 10, then there's no benefit to switching, but that's not the case.

>>9954083
No need, the correct solution has already been posted and supplied with further reading about the subject. It's just retards who were wrong going maximum damage control -mode now.

>>9954084
If we're using mechanism 2 or 3, one envelope has $x in it. In mechanism 2, your envelope has $x in it. In mechanism 3, the other has $x in it.

If you're using mechanism 2, you switch because the other envelope has an expected value 25% higher than your envelope (which has $x). This is because you multiply the other envelope's value by either 2x or 1/2x, each with a 50/50 chance, resulting in an expected value of 1.25x

If you're using mechanism 3, you stay because your envelope has an expected value 25% higher (the other envelope always has $x).

In mechanism 1, which is the "prefilling" situation, it is irrelevant whether you stay or switch.

The mechanism used determines the strategy. Mechanism 1 is the one I inferred the OP was using, based off of the language it used. Mechanisms 2 and 3 are similar to the Monty Hall paradox, since choice affects the game.

>>9954087
ten of the eleven values are picked

>>9954089

Why are you assuming the two outcomes are equally likely? All we are told is that one envelope contains twice the other.

>>9954093

well it's tough to answer the question otherwise.

>>9954093
>>9954094

and that's also why i said that it's unclear what we're taking the expectation over.

>>9954093

but i guess you're right. it's hard to compute an expectation if you don't have a distribution.

>>9954082
You don't know if x=10, it could've been x=5 or x=20, but you didn't pick the envelope he was hoping you'd choose

>>9954097
The expected value of the other envelope is also $10

>>9954099

how do you get that from the original wording?

>>9954101
My reasoning is that if I have 2 envelopes that appear the same, then I expect them to have equal value

>>9954086
You are assuming that you picked the envelope the filler intended for you to choose >>9954098

>>9954104

the only random variable in the OP is your first choice of envelope, which is uniform, so i guess you're right then. my mistake for assuming a distribution that wasn't there. is there a name for that fallacy?

>>9953677
What I'm saying is that if you always switch, it's no better than just picking one of the envelopes at random. By always switching, you're basically just "picking" the envelope you don't initially choose.

>>9954108
>is there a name for that fallacy?
Bertand paradox

>>9954104
>My reasoning is that if I have 2 envelopes that appear the same, then I expect them to have equal value
But the envelope you opened has 10$, and the only options for the other is 5 or 20. Why would you expect it to have 10?

>>9954112

so there's not enough info to come up with an expected value then?

>>9954114
I don't expect a $10 bill

>>9954119
Then why do you expect the first envelope, which is known to have 10$, to not have 10$?

>>9954121
I do expect the first envelope to have $10, after opening it

>>9953207
You don't know the distribution of ($5,$10) pairs and ($10,$20) pairs. Without this information, computing the expectation is impossible.

>>9954114
What do you expect the value of the second envelope to be?

>>9954123
But then you don't expect the two envelopes to have equal value.

>>9954129
I do, I expect both to be $10

There are two possible games the (5, 10) and the (10, 20) game.
Let p be the probability that we're in the (10, 20) game then the expectation value for switching is

p * 20 + (1-p) * 5 = 15 * p + 5

>>9954083
If we assume we are dealing with integer values of money, you could first try to guess what the range of the whole envelope money progression is, based off the amount of money in the envelope you pick.

In the case of your example, it must be a geometric progression with a common ratio of 2, which means at least 10 of the 11 envelopes have an even amount of money. If you pick an envelope with odd number of money, then it MUST be the lowest number in the envelope set. You can then extrapolate the entire progression. In which case you can be 100% sure switching is a good idea. However it's slightly more complicated when you pick an even number, or a number that might be in the middle or higher in the progression.

If your envelope for example is $512, you may infer that the progression went something like

>2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048

and would be hesitant to switch since the expected value of a random envelope choice (which would exclude the envelope you chose initially) would be lower than $512. On the other hand, there is always the possibility that the progression started at $512 and goes way up.

Basically
>try to figure out the progression based off of the common ratio you are given and your first envelope
>estimate what "percentile" of the progression the envelope you chose is in, based off of what numbers you would expect to find given the common ratio and your initial choice
>if your envelope money is an odd number, you should switch, assuming your common ratio is even.

>>9954126
In fact, you don't even know which probability axioms are being used, which number system the 10$ belongs to, or even the logical system.

>>9954115

if there's only one set {10,5}, then you always lose 5 by switching. if it's {10,20}, then you always gain 10 by switching. the fact that you know the value in one of the envelopes removes the only source of randomness in the question statement.

>>9954130
>I do, I expect both to be $10
Why would you expect an envelope known not to contain $10 to be $10?

>>9954135
It's either what you assume or it isn't.
You have a 50/50 chance, right?

>>9954140
>You have a 50/50 chance, right?
Depends on your probability axioms. It could even be a -50/-50 chance.

>>9954137
If I tell you that a bag contains either 1 ball or 3 balls and that the distribution is 50% 1 ball, 50% 3 balls. You expect the bag to contain 2 balls even though it cannot contain 2 balls

>>9954142
So what makes you expect the second envelope has 5$ with a probability of 2/3 and 20$ with a probability of 1/3?

>>9954146
I don't know. It's either $10 or undefined

>>9954132
If I see the dollars in the envelope and I am given the choice to switch, then in the worst case I have the highest value (1/x chance) and I am switching. In other cases, I have 1/(x-r) chance of improving where r is the rank of my current envelope which I do not know. We know that if we assume the highest value envelope is the new car and the rest are goats, then we should switch envelopes. Distributing the value of the car unevenly among the goats by some algorithm seems like a twist on it. Is there already some generalization of this problem?

>>9954150
>It's either $10 or undefined
What makes you expect 10$ to be an option?

>>9954141
depending on your axioms it could be penisy/fleventy chances

>>9954112
>>9954115
What about using the "maximum ignorance" principle?

>>9954154
Because anything other than $10 would make switching meaningful

>>9954165
But anything other than $12.5 would mean that
[eqn]P(\text{Value of picked envelope} > \text{Value of other envelope}) \neq P(\text{Value of other envelope} > \text{Value of picked envelope}) [/eqn]

>>9954168
Why?

>>9954168
Then it's undefined

>>9954136
>the fact that you know the value in one of the envelopes removes the only source of randomness in the question statement
Wrong. See >>9954126 for your blind spot

>>9954152
First off, I'm assuming you could look at the first envelope and then switch only once, so switching initially is the only thing that makes a difference. If you switched without looking at the second envelope you haven't done anything in terms of probability.

You have to consider that switching has both potential benefits and losses, except in the case where you know that you have the lowest envelope (where the common ratio is even and you picked an envelope with an odd numbered amount of money). This is where expected value comes in.

Expected value in this context is simply the sum each envelope multiplied by 1/x, where x is the number of envelopes. This isn't totally reliable since you only know one envelope value and you have to extrapolate the other values. But if you know the progression, then if you randomly selected an envelope then put it back, selected another, etc. and then took the average amount of money on each pick, you'd get approximately the expected value.

Therefore you need to estimate what the range of numbers in the progression is, then find the expected value based off of those numbers, then decide whether the envelope you have is higher or lower than the expected value. It is impossible to know for sure what the exact progression is, so there's some logic involved.

>I have 1/(x-r) chance of improving where r is the rank of my current envelope
I don't think that's right. You have an (r-1)/10 chance of improving, where r is the rank of your current envelope (i.e. the number 1 envelope is the best envelope, 11 is the worst)

>which I do not know
This can't be allowed. There is no way to do anything useful to get more money if you aren't allowed to look at the envelope you pick at first.

>>9954178

i already know this. i'm saying that your original choice of envelope is the only source of randomness in the question. i'm pretty much agreeing with that other anon you quoted.

this paradox can only be solved with utility theory. do i gain more by winning 10$ than i suffer when losing 5$

>>9953217
The situation is not symmetrical.

>>9953429
Its 12.5 just like you said.

>>9954126
What if you assume it's 50/50 so your NEET ass will actually provide a useful, non-masturbatory answer?