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On December 02 2009 02:52 roflcopter420 wrote:Show nested quote +On December 02 2009 02:06 Empyrean wrote:I have already solved the problem, being a statistics major and all  O RLY? Good SIR, I do believe you have made a great discovery then! I shall inform my professor at once, considering he has failed to solve the problem even after putting in many many hours. http://en.wikipedia.org/wiki/Two_envelopes_problemShow nested quote +This is still an open problem among the subjectivists as no consensus has been reached yet
This problem is very different. In that problem, we do not know the amount of money in the envelope we choose, and we can switch infinite times.
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On December 02 2009 02:50 vAltyR wrote:
Ugh, probability. My least favorite area of mathematics.
=( =(
i loooove stats
its by far the most fun and the most USEFUL area of math that i've learned in like 3+ years of uni. unlike, say, calculus where applications of that are limited to pretty much.. more calculus :S
hell, once i get to pick some courses, i'm taking more stats, no questions asked
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Wait, I've been doing assumptions. I should know better than to trust my intuition. Let's calculate the EV of switching:
ev(switch)= p(y)*2y + p(0.5y)*0.5y.
you should switch if p(y)*2y + p(0.5y)*0.5y > y
What does this imply? Hum, I don't know.
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Considering the third part:
+ Show Spoiler + Well, as before, we want to switch if E[second bag] > Y. Given the amount Y, the second bag has 2Y with probability p(Y), and has Y/2 with probability p(Y/2). Switch if 2Y*p(Y) + Y/2*p(Y/2) > Y.
Eh, Phrujbaz beat me to it. I think there's something a bit wrong with it though... something about having p(Y) + p(Y/2) having to equal 1, given Y...
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Whether you want to switch or not will very from person to person because it depends on how risk averse a certain person is, once you get past EV, it's an economics problem :p.
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I just realized the following has to be true, since we have EITHER x or 2x:
1 = p(y) + p(0.5y)
2y * p(y) + 0.5y * (1- p(y)) > y
2y * p(y) + 0.5y - 0.5p(y) > y
1.5p(y) + 0.5y > y
1.5p(y) > 0.5y
3p(y) > y
This is when we have to switch. That seems such a stupid result. I must have made a mistake somewhere.
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That looks right to me.. maybe the setup is wrong:
1 = p(x2 = 2y | x1 = y) + p(x2 = y/2 | x1 = y)
Where x2 denotes the other envelope, and x1 denotes the first opened envelope.
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In school this is always the introduction to probabilities.
Exciting stuff. Going back to sleep...
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haha good job BottleAbuser. p(y) + p(0.5y) doesn't have to equal one. I don't remember why I thought that it did anymore. What does have to equal one is p(x2=2y) + p(x2=0.5y), something entirely different 
Edit: The probability that grandma chose y for the first envelope changes once you know y, so you cannot use p(y) anymore. That's the source of my original confusion.
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Grrr... I've done probability questions like this, but I've always found probability in general frustrating compared to number theory and calculus, where I feel much more at home.
I agree that fundamentals of probability and statistics should always be taught, there are far too many playing the lottery these days.
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So we have the probability that we see y when we open the envelope:
0.5(p(y) + p(0.5y))
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Probability and Statistics are the only math classes I have ever taken that I hated(I have a minor in math). Having said that, I think it is important for everyone to know basic probability and statistics, but the depth they go into those subjects in introductory courses in university is more than what the layman needs. I would say the first 1/3 of the stats and probability courses I took are what every person should know by the time they are an adult(in an ideal world).
I can remember my probability prof going over that envelope question(or a similar one) in my class, but dont expect me to even attempt it because I will not do that kind of math ever again unless I am forced to.
On December 02 2009 02:56 JeeJee wrote:Show nested quote +On December 02 2009 02:50 vAltyR wrote:
Ugh, probability. My least favorite area of mathematics. =( =( i loooove stats its by far the most fun and the most USEFUL area of math that i've learned in like 3+ years of uni. unlike, say, calculus where applications of that are limited to pretty much.. more calculus :S hell, once i get to pick some courses, i'm taking more stats, no questions asked
I guess you have never heard of science or economics.
edit: calculus is also used in probability lol.
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What is the probability that grandma chose y once we know that one of the envelopes contains y?
She either chose 0.5y or y for x, if we see y in one of the envelopes . Which is it? It depends on the relative probabilities of choosing 0.5y or y. If she chooses x to be y twice as often as she chooses x to be 0.5y, then the probability she chose y is 2/3 !
In general, if she chooses x to be y a times as often as she chooses x to be 0.5y, the probability she chose y is a / (a+1).
p(grandma chose y) = (p(y) / p(0.5y)) / ((p(y) / p(0.5y) + 1)
Don't know how to simplify xD
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I dunno, you could argue that in using calculus you are also using results from geometry... or arithmetic.... I think he's talking about the direct application of a field.
I think the inequality we came up with first is right. We can easily calculate p(y) and p(y/2) given that p is exponential with mean 100.
Which reminds me: the exponential function is memoryless, so given Y... oh, fuck..
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The inequality we came up with is the answer for question number 1.
We distracted ourselves by calculating expected value, which doesn't depend on the probability p(y) but on the ratio p(y)/p(y/2). My original answer was right: for switching to make no difference, p(y) must equal p(y/2), which means grandma's wealth is infinite, as she must choose twice the amount of money with equal probability as any other amount of money.
As for the answer to question 3, maybe I should first read what a probability distribution is. And what an exponential probability distribution is.
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Wikipedia page ftw. In short: where the mean of an exponential distribution is 1/L, p(x) = Le^(-L*x) (ofc this is for x>0)
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Ah hehe classic problem, made me furious in the same way the Monty Hall problem did ^^
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Sounds a bit like the monty hall problem, yes?
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Here's what I got from the hint.
What is the conditional probability that your envelope holds $X (and hence it would be advantageous to trade), conditional on the observed value of $Y?
Using Bayes' Theorem here:
Notation-- let x be the observed value, Y be the event of observing amount x in the open envelope, and X be the event that 2x dollars is in the other, unopened, envelope
P(X|Y) = P(Y|X)P(X) / P(Y)
P(X|Y) = (0.5)*f(x) / .5*f(x)+.5*f(x/2)
P(X|Y) = f(x) / ( f(x) + f(x/2) )
We are asked to maximize our EV.
EV(keep) = x
EV(switch) = 2x*P(X|Y) + 0.5x(1 - P(X|Y) )
EV(switch) = ( 2xf(x) + 0.5xf(x/2) ) / ( f(x) + f(x/2) )
Third part: normalizing from 100 to 1.
Assuming you know where parentheses are-
EV(switch) = 2xe^-x + .5xe^(-x/2) / e^-x + e^(-x/2)
Should switch if
2e^-x + .5e^(-x/2) / e^-x + e^(-x/2) > 1
2e^-x + .5e^(-x/2) > e^-x + e^(-x/2)
e^-x > .5e^(-x/2)
e^(-x/2) > .5
-x/2 > ln(.5)
x < -2ln(.5)
x < 1.39
We are switching if we see less than $1.39 in the envelop, for the $1 average case; for the mean of $100 dollars case we switch if we see less than $138.63.
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EPT
