/sci/ - Science & Math

The probability is 50%

>>5995256
1/2 = 50%

>>5995249
No, you get money either way.

>>5995266
but you get more money if you switch

switching makes no difference. unlike monty hall, you are given no new information prior to switch

>>5995279
What if you open the first envelope and then decide if you want to switch?

>>5995276
no, you don't

$1 and $2

below we use 1 and 2 to show first choice, and K and S to mean keep or switch

1K gives $1
2K gives $2

1S gives $2
2S gives $1

>>5995287
same. the only useful info is whether you have the larger or smaller amount

maybe if you have info of typical prize sizes you could judge from the amount

>>5995288
but we only know that the first envelope has $1 so the second could be $0.5 or $2
on average that's better than $1

>>5995293
but you are NOT getting the average. so that argument is invalid

>>5995298
there is a 50% chance to get $2 and 50% to get $0.5
that's better than a guaranteed $1

ITT

Statistics; not even once.

>>5995302
>>5995302
>there is a 50% chance to get $2 and 50% to get $0.5
on what planet do you think there is an envelope with four times the other, when the question says two times the other?

(nice troll material though)

>>5995325
if one is $1, both $0.5 and $2 are equally probable

Answer: run away with both envelopes.

>>5995328
no, one of them is impossible. you can't change reality by choosing.

>>5995361
which one is impossible?

>>5995365
we do not know.

you are conflating reality with information about reality. a schoolboy error.

>>5995365
we just don't know

>>5995371

>>5995368
so if i flip a coin i have a 100% chance to get one of the outcomes and 0% to get the other one but there is no way to know which one?

>>5995374
>Implying the envelopes are filled in the future.

apart from determinism, quantum probability etc, and keeping the discussion to conventional probability modelling, i give this troll 2/10

>>5995374
No.
This is not a coin flipping.
The money is already in the enveloppes.

>>5995374
nah, you misunderstand
supposing you get the $1 and have an option to switch
so if you do, you either get 50 cents, or $2
and theres no way to know which it is

it's already determined; so it aint a matter of probability. you basically just guess.

>>5995387
33.34%

>>5995395
twat
i know you're that millionaire troll guy

>>5995387
>it's already determined; so it aint a matter of probability.
Wrong. Probability is way of measuring YOUR uncertainty; it describes uncertainty in someone's mind, not uncertainty in reality (as reality is never uncertain). Thinking that probability is about reality is "conflating reality with information about reality", as >>5995368 said.

So yes, in this situation there is a 50% chance of the other envelope containing $0.50 and a 50% chance of it containing $2.

>>5995387
what if the coin has already been flipped and you have to guess?

does probability theory not apply in this case?

>>5995400
oh right. in which case, might as well switch

>>5995402
only if you're a retard who sees a coin is tails, and guesses heads anyway.

>>5995407
what if someone else does the coin flip, writes down the result and puts it in an envelope?

>>5995402

>does probability theory not apply
no one claimed it didn't apply before. but your model was wrong.

>>5995411
then, i still dont know, so as far as im concerned it's 50/50

>>5995416
but
>it's already determined; so it aint a matter of probability. you basically just guess.

How can so many of you be this fucking retarded? Please, try to explain it to me?

Let's start with TrollP:
You're (deliberately?) fucking up the math. There is one envelop with X and another with 2X. There is no X/2.

You select an envelop. The probability that you have selected X is 1/2 and 2X is 1/2, so the expected value (sum of (probabilities times values)) is 3X/2. The exact same probabilities apply for the contents of the other envelop, so both have the exact same expectation. Switching does not improve your chances in any way of getting more money.

You're either confused by the Monty Hall problem, or intentionally trying to confuse others with it. The difference is that in the Monty Hall problem, opening a door provides information that eliminates some of the possible outcomes. Probability is calculated from the number of possible outcomes so when that number changes, so do your chances.

Choosing an envelop here provides no additional information.

How do you cunts manage to sit around on a science and math board all fucking day without ever reading up on shit like basic fucking probability?

You're damn fucking right I'm mad. You mental niggers are giving my species a bad name.

>>5995443
<<<
faggot

>>5995443
You choose the envelope. It contains Y. Y = X or Y = 2X
If Y = X then after switching you get 2X = 2Y
if Y = 2X then after switching you get X = Y/2
on average switching gives you (2Y + Y/2) /2 = 5/4 Y which is greater than Y

Is there a problem?

>>5995443
>quite ancient and difficult problem
>"hurr guize ur retarded this is obvious"
http://en.wikipedia.org/wiki/Two_envelopes_problem
You've said nothing that solves the paradox.
Imagine you're able to look at the enveloppe content before you take the decision. The content is now a certain value a. The other enveloppe content is either a or a/2.

Congratulation on being the biggest failure of this thread.

>>5995443
calm down

>>5995464
spergers always get pretty emotional about math.

>>5995457
>cites wiki article
>article contains academic solution to "paradox"
>solution is same as the one posted
>megaderp

>Imagine you're able to look at the enveloppe content before you take the decision. The content is now a certain value a. The other enveloppe content is either a or a/2.

No, the other one is either 2a or a/2, but knowing the contents of one envelop gives you no information that affects the probabilities, as you don't know if "a" is X or 2X. It's the same as knowing that one envelop is green and the other is blue.

>Congratulation on being the biggest failure of this thread.
Go look up the Dunning-Kruger effect. Your lack of understanding is so great that you cannot even recognize correct reasoning here.

>>5995528
Imagine the following situation
>I prove A and not A, post proof on /sci/ and ask to find the mistake
>you copy my proof of A and say that not A is obviously false so that makes no sense
well I can prove not A and say that A is false

do you not see the problem?

>>5995455
The error is that the value of Y is not constant in your calculation.

Case 1
Y = X, then you get 2Y

Case 2
Y = 2X, then you get Y/2

That's correct, but when you try to calculate the expectation value, you can't say
(2Y + Y/2) because the Y in "2Y" is equal to X, while the Y in "Y/2" is equal to 2X.

Replacing the values, (2X + X)/2 = 3X/2.
>>5995447
>>5995457
who's mad now, faggots?w

>>5995552
the value of Y is constant, it's the money in the first envelope.
I can even open it and see what it is

>>5995552
>who's mad now, faggots
erm, still you?

>>5995547
see >>5995552 then go get some ointment for your butthurt

>>5995552
why are you evaluating the value in terms of X?
Y is what we get if we don't switch so we should compare the value to Y

>>5995555
Ok, let's take baby steps here.

One envelope contains $1 (X), the other $2 (2X)

Case 1
Y = $1, then you switch and get $2 (2Y)

Case 2
Y = $2, then you switch and get $1 (Y/2)

See how Y is different? The amount in the envelopes is fixed, and the value is Y is fixed IN EACH CASE, but the value is depends on the case.

So, expected value for switching ($2 + $1)/2 = $1.50.
Expected value for not switching ($1 + $2)/2 = $1.50.

>>5995566

One envelope contains X, the other 2X
You open the first envelope and see Y = $1 in it

Case 1
X = $1, then you switch and get $2 (2Y)

Case 2
X = $0.5, then you switch and get $0.5 (Y/2)

See how X is different? The amount in the first envelope is fixed, and the value is X is fixed IN EACH CASE, but the value is depends on the case.

So, expected value for switching ($2 + $0.5)/2 = $1.25.
Expected value for not switching $1.

>>5995249
Consider two possible values for X, prior to opening the envelope: A or B. Both A and B are positive numbers. But what is the prior probability, P(A) or P(B)? Is P(A)=P(B) for all A,B? If so, then let the probability P(A) = c. The total probability must equal 1, but since probability is constant, 1=c*infinity, implying c=0.

From this we conclude that either P(X)=0 for all X, meaning that every positive number is equally likely, or there exist A,B such that P(A) !=P(B). Is it really plausible that it is equally likely that the envelope contains $1 or $10^(10^100) or $(34858+73/1024)? What justification do we have for making this assumption about the distribution of values of X?

But if we reject the assumption that P(X) is flat, then measuring some particular X by opening the envelope gives us information about the likelihood of the other envelope containing 2X and X/2. They need no longer be equally likely, so your reasoning breaks down.

>>5995563
Because that's how probability works. You need to calculate based on actual possible outcomes. Your use of Y to designate the value that you currently have leads to confusion when you try to evaluate the expected value, because you end up with an equation where you use the variable Y to represent two different real values simultaneously, and that is where the "paradox" comes from.

Please, reread the previous posts carefully and try to understand this distinction. It is the key to understanding the problem.

>>5995577
Let's assume that the numbers are powers of two and the probability of <span class="math">2^n[/spoiler] is <span class="math">\frac{2^n}{3^{n+1}}[/spoiler]. Do the math in this case

>>5995575
The values are fixed before you choose. There is either $1 and $0.5, or there is $2 and $1. Why can't you understand this.

I take two envelopes. I put $X in one and $2X in the other, and seal them. Then you choice one, and I offer you the choice to switch. That's what this problem is.

What you describe is different:

I put $X in an envelope and I give it to you. I then put either $2X OR $X/2 in another envelope and seal it, then offer you the chance to switch. In that case, if I randomly chose the amount in the second envelope, then you could indeed expect $1.25 for switching, but again, it's a different question.

>>5995586
to make it more clear: that is the probability of 2^n being in the smaller envelope and 2^(n+1) in the larger one

>>5995588
From my point of view there is no difference between two formulations

That's like saying that you flip two coins and I have to guess the second flip after you tell me the result of the first one

>>5995586
So you're saying that the smallest possible value is 2^0 = $1? (Otherwise the series doesn't converge to 1).

If that's the case, then if I open the envelope to find $1, then I should certainly switch, since I know the other envelope has $2. So you see, I can gain information by opening the envelope.

>>5995599
but if you find $2 then the other one is $1 with probability (1/3) / (1/3 + 2/9) and $4 with probability (2/9) / (1/3 + 2/9) so you should also switch

OP forgot an important part, and people are adding in information you don't have...

You get to see what's in the envelope before you have to decide whether to switch or not.

You also don't know how much the amounts are. This is why people doing calculations using example values of $1 & $2 are approaching this wrong.

You don't know that. Yes, when you work out the expected value blind, they have the same expected value. But due to the setup, the expected value is always higher than the revealed value, always prompting a switch based on this.

Say you do pick an envelope, and you open it, and it's $2. You now know the other envelope has either $1 or $4. There's no reason to believe you picked the higher or lower value envelope, so both are equally likely. And so the expected value of the other envelope is 50%*$1 + 50%*$4 = $2.50. Which is higher than the $2 you're currently holding.

>>5995606
True, but ask yourself, what is the expected value of money in a given envelope? <n> = sum(2^n*2^n/3^(n+1),{n=0 to infinity}) = sum(4^n)/3^(n+1),{n=0 to infinity}) = infinity.

So you find upon opening an envelope that the money inside is infinitely less than you had expected to find. So the amount in your envelope is *really small* compared to what you expected. Is it any surprise that in that situation you'd be better off switching?

>>5995618
but if it always makes sense to switch regardless of the value it doesn't matter if we open the envelope or not

>>5995622
so on average the second envelope is better than the first one
and the first one is better than the second one

Let's say you did it twice:
Staying both times: $2, average $2
Switching both times:
$4x1/4
$2.50x1/2
$1x1/4
Average: $1
Deal w/ it, nerds

>>5995629
No, switching back doesn't make sense. When you open an envelope, you get information. If you switch, you expect (p>.5) to lose money, but that the amount you'll lose will be sufficiently small that it will be balanced by the smaller chance of gaining money. Once you've switched once, you have an envelope containing an unknown amount, where that amount is more likely than not to be small, but its expected value is higher because, in the unlikely event that it contains more money, it will have quite a lot. So you shouldn't switch back.

The discrepancy between the expected outcome [p(gain) vs p(lose)] and the expectation value [p(gain)*$_gain vs p(lose)*$_lose] is due to the unusual case of a prior distribution of money that has an infinite expectation value.

>>5995598
>your point of view
>reality

lol, that has got to be THE WORST statement I have seen on /sci/ in a loooong time.

>>5995759
>When you open an envelope, you get information
the information always tells us to switch so we don't have to open

>>5995785
>guy flips a coin but does not tell you the outcome
>from your point of view its 50/50
>in reality it's heads

>>5995789
you just switched your argument. your first statement was that two different situations, a and b, were "in your point of view" equivilant.

and im just telling you that your point of view can kindly go fuck itself for being so retarded.

>>5995794
>I take two envelopes. I put $X in one and $2X in the other, and seal them. Then you choice one, and I offer you the choice to switch. That's what this problem is.

>What you describe is different:

>I put $X in an envelope and I give it to you. I then put either $2X OR $X/2 in another envelope and seal it, then offer you the chance to switch

there is no difference

>>5995787
But they both have infinite expected value prior to opening. It's not exactly surprising that you can get cases where Infinity_1 > Infinity_2 and Infinity_2 > Infinity_1 when you do naive comparisons.

Anyone help me find the flaw in this argument?

Envelope I has X payoff, envelope II is (2x + x/2)/2 = 1.25x.

There is 50% probability to pick either (hence no switch), so the payoff is (1.25+1)/2 = 1.125

This is at odds with (2+1)/2 = 1.5

>>5995817
so let's consider the following case
>man A and man B get one envelope each
>both open them and decide they should switch
>they exchange envelopes and each walks away with a profit

>>5995833
But the men hold envelopes with infinite expected value. The difference in the envelopes values is Infinity - Infinity = undefined. So you can get the result that each is greater than the other, since Infinity does not obey the normal rules for ordering numbers.

>>5995833
Sorry, misread your post in my response (>>5995839). They don't walk away with a profit. One wins and one loses. The only way they could each really believe that they would profit from a switch regardless of what they saw, however, is for them to have had envelopes of infinite expected value.

>>5995528
>solution is same as the one posted
It isn't, at least try to read a fucking wiki article to the end.

Is 50% of the money I get enough to pay off debts? yes switch and hope for more money. No keep with current ammount.

>>5996180
oh before you open. I misread. nvm

>open envelope
>$0

>>5995249
Hmm, my 50% chance of grabbing the better envelope has been replaced with... the same 50% chance of grabbing the better envelope.

Firstly, research the motivations of the person giving you money, most importantly whether they would prefer you to gain the greater or lesser amount.

One could then experiment on a large number of people to discover whether more of them switched or stayed. Assuming the original donor had some similar idea (they likely did, or they wouldn't have bothered offering the switch), you would then base your decision on that data.

For example, if the donor wanted to minimise the amount given and 90% of people people switched, you should keep the original envelope.

For a more simple and practical method, one could assume that the donor would only offer a switch if it gave the potential for them to lose less money. Therefore, without more information it would make sense to keep the starting envelope (or switch if the donor would prefer to give away the greater amount)

Same guy as
>>5996314
Skimmed the numbers first time around, should have taken into account that switching always has a higher expectation. One should always switch unless there is some external knowledge stating otherwise (knowing the person giving away the money, knowing the amounts, needing a certain amount of money only etc.)

>>5996185
>implying zero is positive

>>5996887
> implying it's not

>>5996898
zero is only half positive

>>5996887
assume it is negative

as 0*0 = 0 this breaks the negative multiplied by negative = negative

therefore 0 is positive

>>5996909
>negative multiplied by negative = negative

I like how this problem is a disguised case of "it can either be true or false so it's 50/50" kind of fallacy.

You don't switch.
Mathematically it wouldn' matter if you switch if he always offered you the choice.
But psychologically it is more likely , that he offers you a switch when you have the envelope containing the high amount of money.

SO DONT SWITCH

>>5996925
>I missed the whole point of the problem, please rape my face

>>5996909
>negative multiplied by negative = negative

-i*-i=-1
>muh real numbers

>>5996964
>-i is negative
Because there is a minus sign in front of it?
So if a = -1, -a is negative too?

always switch, winning chance will increase from 50% to 66%

OP's problem is retarded. It doesn't matter unless there's three doors and one is revealed http://www.youtube.com/watch?v=9vRUxbzJZ9Y

>>5998907
That video sucks, it doesn't state the problem correctly.

quote from your video
"You make your choice and the host decides to reveal one of the doors"

If it's stated like this, you cannot calculate the probability of winning after you switch. If the guy can choose to open or not open a door, he could for example offer you that option only if you guess correctly, in which case probability to win after swicthing is 0%.

In order to the paradox to work, the rules of the game should explicitly say that one door always gets revealed and you are always given the choice to switch.

>>5998919
Of course we have to assume that and the video explains the concept fine enough. Stop being such a huge fucking asshole

>>5998929
The only assholes are people who state the paradox incorrectly. This makes people who are not good at maths misunderstand the problem and spread misinformation.

The difference between two ways of understanding how the game goes should be the main focus of the video because that's exactly what makes this a paradox.

If you do not have a priori information that the host will reveal one of the doors, the winning strategy is "flip a coin and switch with probability 1/2" which will give you a 50% chance to win.

The 2/3 chance after swiching exists only if you have that information which is definitely not the case if you imagine a real life situation when the host can open or not open a door (which the statement strongly suggests).

TL;DR STOP MISINFORMING PEOPLE AND STATE PROBLEMS CORRECTLY

>>6000000