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Don't worry, this isn't a homework help thread. I have already solved the problem, being a statistics major and all  . This was just an assigned homework problem in a first undergraduate probability course and I thought it'd be interesting to see if anyone else could solve it. (there are spoilered hints for you too!)
The problem is as follows:
Granny chooses an amount X according to probability distribution p(x) and puts $X into one envelope and $2X into another. You choose one envelope at random (with probability 50% each) and open it to find $Y. You have the opportunity to keep the amount in the envelope, or to swap and keep whatever is in the other envelope (if you trade, you will keep either $Y/2 or $2Y, depending on whether Y=2X or Y=X).
- Under what conditions should you trade, after knowing how much money your envelope contains?
- For the naive estimate of 1/2 to be correct for that probability, what would Granny's distribution p(x) have to be? Any problems with that?
- If Granny used an exponential distribution with mean $100, what strategy optimizes the expected value of the amount of money you keep?
Some hints:
+ Show Spoiler [General hint] +The naive thinking that "it doesn't matter whether or not I should switch: I either have the envelope with X or 2X" is only valid if you don't open the envelope to find your $Y. Once you do, that argument no longer holds.
+ Show Spoiler [For first part] +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?
+ Show Spoiler [For first part] +You don't have to find the probability for any specific distribution, you just have to find it for a general distribution.
+ Show Spoiler [For second part] +You're golden once you arrive at the divergent definite integral 1/y(dy) from 0 to infinity. Hell, I think I just gave you guys the answer.
+ Show Spoiler [For third part] +Once you have the correct inequality for the first part, the third part should just be plugging in values.
To be honest, I was surprised this question was assigned in a first probability course. Granted, the prereqs for the course are multivariate calculus and linear algebra (neither of which you have to use for this problem, though you do need basic calculus) so there aren't that many freshmen in the class, but it seemed a bit difficult for an introductory undergraduate probability course.
That being said, I was also wondering your thoughts as to the value of a basic education in statistics and probability, no matter what you decide to do in life. I find that all too often, the general public has absolutely no clue what to do when presented with probabilities or statistics, and either doesn't care at all, or arrives at completely incorrect conclusions. Do you think everyone should learn at least the fundamentals of probability and statistics?
Edited for grammar.
 
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This problem seems really fun! I'd love to try my hand at it.
So what's a probability distribution?
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Wait, are you kidding?
T_T
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Before we know what the distribution function p(x) is, does it matter how much money is in the first envelope? We still don't know anything about what could be in the other bag, unless we find 1 cent in a bag, which must mean the other one has 2 cents.
I guess I'll spoiler the rest of my argument, it's based on spoiler #2:
+ Show Spoiler +The conditional probability takes into account the probability that you see $Y in the first envelope you open. Then you're looking at
P(X = x | X = Y) = (P(X = x, X = Y)/P(X=Y)
Since we don't know anything about the distribution (more importantly, the mean), we have no way to decide.
Also, it depends on what you call basic statistics. I'm sure everyone's learned the meaning of the mean of a set. Probably everyone's at least heard of the term "standard deviation" and has some idea as to what it means.
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On December 02 2009 02:15 Phrujbaz wrote:
This problem seems really fun! I'd love to try my hand at it.
So what's a probability distribution?
Uhh, this isn't the type of problem that you can just figure out without knowledge of the concepts... (or at least that's my understanding).
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Sounds just like a problem I had in my "first" stats class over here at UNC too. Granted I was being a douche and skipped other stats prereqs for it, cause my roommate took the class lol.
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yea I've never heard of the term "probability distribution" either but I haven't taken a probability course either.
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On December 02 2009 02:24 BottleAbuser wrote:Before we know what the distribution function p(x) is, does it matter how much money is in the first envelope? We still don't know anything about what could be in the other bag, unless we find 1 cent in a bag, which must mean the other one has 2 cents. I guess I'll spoiler the rest of my argument, it's based on spoiler #2: + Show Spoiler +The conditional probability takes into account the probability that you see $Y in the first envelope you open. Then you're looking at
P(X = x | X = Y) = (P(X = x, X = Y)/P(X=Y)
Since we don't know anything about the distribution (more importantly, the mean), we have no way to decide.
Actually, this isn't true. There is an optimal solution for any probability distribution with defined first and second central moments. We don't consider those distributions for which they're undefined because that would imply that Granny has an infinite amount of money that she could put into the envelope.
EDIT: Wait, it actually might not matter. There should be the same solution for all distributions.
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If P(x) is simply the probability that Granny picks $x for the amount, then you have to switch if P(y) > P(0.5y).
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Okay, let me get this straight. We have no idea how much money Granny has, except that she's putting a finite number of money into one bag and twice that into another. We don't know the average or variation of this distribution. We just know that one envelope has a certain amount in it. And we can determine how likely it is that the other one has twice the amount we see?
Are we supposed to answer this in terms of the mean and variance of the distribution?
All right, let's give this one a go, but it seems way too simple of a solution:
+ Show Spoiler + Let's say you see 100 dollars in the first bag. There's a 50% chance that there is only 50 dollars in the other bag, and 50% chance that there are 200 dollars in the other bag. Then the expected value of money in the other bag is 50*.5 + 200*.5 = 125. So you'll always want to switch.
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On December 02 2009 02:32 BottleAbuser wrote:
Okay, let me get this straight. We have no idea how much money Granny has, except that she's putting a finite number of money into one bag and twice that into another. We don't know the average or variation of this distribution. We just know that one envelope has a certain amount in it. And we can determine how likely it is that the other one has twice the amount we see?
Are we supposed to answer this in terms of the mean and variance of the distribution?
No, we're only to determine under what conditions we should switch, given any arbitrary probability distribution.
For the first question, the answer pretty much goes "Switch if _____ > _____". For the third, you just plug in values to the answer you got for the first (you KNOW the mean and variance for the distribution in this case). The answer to the second is pretty much given in the hint, you just need to realize what this case implies about Granny's wealth.
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Oh I think I get the idea. If you find an even number of money, then keep it. Otherwise, if you have an odd number of dollars, choose the other bag. (because doubling something makes it an even number)
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On December 02 2009 02:37 Assault_1 wrote:
Oh I think I get the idea. If you find an even number of money, then keep it. Otherwise, if you have an odd number of dollars, choose the other bag. (because doubling something makes it an even number)
An interesting thought, but very far from the answer 
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If switching exactly half the time is always optimal, then p(y) always equals p(0.5y). That implies that granny's wealth is infinite, since she must always put in twice the amount of money with equal probability.
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I don't understand anything you're saying.
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oh my goodness, a stat problem!
Where's Gretorp? I'm sure he would enjoy this problem very much.
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Ugh, probability. My least favorite area of mathematics. Oh well, I guess I'll think about this until EtherealDeath responds to my post in the other thread. I'll have to teach myself the basics of statistics, though, so bear with me. 
+ Show Spoiler +
Probability distribution is a function p(x) that shows the probability that she chose that amount of money, right?
Okay, so we open the first envelope, and it contains y dollars, which we know to be either x or 2 * x. So either y/2 = x, or y = x. So we can compare p(y/2) and p(y), and we switch envelopes when p(y) > p(y/2).
Assuming I'm right. Do we know what the probability distribution is, or is it unknown?
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The same solution holds for any probability distribution.
And you can't approach the problem correctly unless you use conditional distributions.
I'm going to lunch now...I'll check back on this thread in a bit.
I can post a partial solution if you want (spoilered of course) when I come back. Let me know if you want one.
EDIT: It's not the two envelopes problem.
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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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On December 02 2009 03:27 Mastermind wrote:Show nested quote +On December 02 2009 02:56 JeeJee wrote: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.
yeah yeah and you can trace it all back to addition
bdares bottleabuser got it right.
oops, namechange
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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
Try studying engineering. It's pure calculus applications... pretty much every machine or structure made for general public use in the past hundred years has been a product of the application of calculus.
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actually meta.... communication engineering or anything as a matter of fact, you need to have some random variables involved in it
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I think the problem in layman terms is this:
There are two envelopes of money with one envelope having exactly twice as much money as the other. You choose one envelope and then decide if you want to switch envelopes and keep what is in the other envelope.
The "trick" is that most people will always want to switch because they think they stand to gain more than they stand to lose.
For example, you open the first envelope and it has $100 in it. You know the second envelope either has $200 or $50, since one envelope is double the other. So if you switch you risk to gain $100 or lose $50 which seems like a nice risk.
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It's very clear that both probability and calculus are absolutely fundamental areas of math that are absolutely indispensable. However, for the average man I would say that probability is a somewhat more useful thing to know than calculus
Here's what I've done with the problem. I really don't know if I'm correct or not though.
+ Show Spoiler +
Given that grandma put X in one envelope and 2X in another envelope:
v = quantity that we observed in envelope 1
p1 = probability that v=X given envelope 1 has v, i.e. P(v=X | e1=v)
EV[keep] = x
EV[trade] = p1*0.5*x + (1 - p1)*2*x
trade if :
x < p1*2*x + (1 - p1)*0.5*x
x < 1.5*p1*x + 0.5*x
0.5*x < 1.5*p1*x
p1 > 1/3
Now to determine p1:
q(x) = the probability distribution from which Grandma decides X
Using Bayes' Theorem:
P(v=X | e1=v) = P(v=X, e1=v) / P(e1=v)
Solving for P(e1=v):
P(e1=v) = 1/2 * q(v) + 1/2 * q(v/2)
Solving for P(v=X, e1=v)
P(v=X, e1=v) = 1/2 * P(v=X)
P(v=X) = q(v)
P(v=X, e1=v) = 1/2 * q(v)
Putting it all together:
P(v=X | e1=v) = P(v=X, e1=v) / P(e1=v)
P(v=X | e1=v) = 1/2 * q(v) / (1/2 * q(v) + 1/2 * q(v/2))
p1 = 1/2 * q(v) / (1/2 * q(v) + 1/2 * q(v/2))
Going back to the original inequality:
p1 > 1/3
1/2 * q(v) / (1/2 * q(v) + 1/2 * q(v/2)) > 1/3
3/2 * q(v) > 1/2 * q(v) + 1/2 * q(v/2)
2 * q(v) > q(v/2)
This means that when we observe the amount v in envelope 1, we should switch if 2*q(v) > q(v/2), where q(x) is the probability distribution that grandma uses.
I don't know what's being asked with the second bullet point.
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On December 02 2009 06:53 BlackJack wrote:
I think the problem in layman terms is this:
There are two envelopes of money with one envelope having exactly twice as much money as the other. You choose one envelope and then decide if you want to switch envelopes and keep what is in the other envelope.
The "trick" is that most people will always want to switch because they think they stand to gain more than they stand to lose.
For example, you open the first envelope and it has $100 in it. You know the second envelope either has $200 or $50, since one envelope is double the other. So if you switch you risk to gain $100 or lose $50 which seems like a nice risk.
Actually that's just the two envelopes problem.
The major way in which this problem differs is that the amount X is generated by a probability distribution. It changes the problem completely.
And as for applications, there's so much calculus in statistics you wouldn't even believe it. Calculus is such a useful field of mathematics and finds itself in every day life. For example, you could use calculus to argue that if it rains at a constant rate, you'll get the same amount of wet no matter what speed you move through it (though you'll probably end up colder if you walk leisurely). The argument can basically be summed up as the fact that you're displacing the same amount of water, but it takes some calculus to show exactly why.
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Realistic approach: Granny being a cheapskate who has a strange kick out of such silly games, X probably has a normal distribution with a mean of 2 bucks and a high deviation, like 40 or 50 cents.
It should be easy from this on.
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i could definitely solve it, but i agree that its a bit challenging for an introducing question of a first course on probability. (im a statistics major currently writing my m.sc. thesis^^)
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On December 02 2009 08:01 Black Gun wrote:
i could definitely solve it, but i agree that its a bit challenging for an introducing question of a first course on probability. (im a statistics major currently writing my m.sc. thesis^^)
What's your thesis about? If you don't mind my asking :D
What was your experience as an undergraduate statistics major like? I find it really interesting.
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On December 02 2009 08:02 Empyrean wrote:Show nested quote +On December 02 2009 08:01 Black Gun wrote:
i could definitely solve it, but i agree that its a bit challenging for an introducing question of a first course on probability. (im a statistics major currently writing my m.sc. thesis^^) What's your thesis about? If you don't mind my asking :D What was your experience as an undergraduate statistics major like? I find it really interesting.
well, its a really cool subject, but the part of statistics which is taught to non-majors is rather boring.
here in germany, we start with these "ordinary statistics topics" in the first 2 semesters, which include only the basic results of probability, and only in the 3rd and 4th we are taught the rigorous foundations of probability, measure theory, and so on.
the really problematic courses are analysis I and II and linear algebra which are hold during the first 2 semesters. there, about 50% of all freshmen fail already, and about half of them (25% of the freshmens) fail so hard that they give it up already. (the math basics are taught on a really high lvl here...)
basically, they intentionally want to purge the weak by these math courses. therefore the introduction into statistics and probability is rather easy and practically-oriented during the first 2 semesters.
edit: whats ur major btw?^^
i havent started yet with my master´s thesis, but im currently getting into the literature and finding a topic. i plan to write about the estimation of response densities in regression settings via boosting techniques. if that tells u anything
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I'm a statistics major here.
For my school, you aren't even allowed to become a statistics major until you pass multivariate calculus and linear algebra. My path through the major looks like
1st semester:
Probability, Regression Analysis
2nd semester:
Statistics, Bayesian/Modern Statistics, Survey Methodology
3rd semester:
Statistical Decision Analysis, Design and Analysis of Causal Studies
4th semester:
Independent Study, Computational Data Analysis
5th semester:
Independent Study (continued).
After your independent study (a year long project), if you still have time, you can take some higher level electives. Some examples are Probability/Measure Theory, Applied Stochastic Processes, Generalized Linear Models, Data Mining, Spatial Statistics, etc., it really depends on what they feel like teaching.
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* Under what conditions should you trade, after knowing how much money your envelope contains?
* For the naive estimate of 1/2 to be correct for that probability, what would Granny's distribution p(x) have to be? Any problems with that?
* If Granny used an exponential distribution with mean $100, what strategy optimizes the expected value of the amount of money you keep?
sorry, i don't understand most of this since i've had no formal edumication.
However to me it seems like it always makes sense to always swap the envelope. Since you have 50% chance of gaining double the amount of your current envelope, and 50% chance of gaining only half. I'm not sure if this is the only thing you ask for though (:
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On December 02 2009 06:56 Empyrean wrote:
And as for applications, there's so much calculus in statistics you wouldn't even believe it. Calculus is such a useful field of mathematics and finds itself in every day life. For example, you could use calculus to argue that if it rains at a constant rate, you'll get the same amount of wet no matter what speed you move through it (though you'll probably end up colder if you walk leisurely). The argument can basically be summed up as the fact that you're displacing the same amount of water, but it takes some calculus to show exactly why.
is that actually true? doesn't make intuitive sense to me.. lets say its raining out, and you have to walk from your car door to apartment door (say 5 meters). you can sprint that, and you'll get some raindrops but you won't be soaked
otoh, walk very slowly (say 3 hours for those 5m..) and you'll def. be soaked (and probably suffering from extreme hypothermia)
or am i misunderstanding what you mean? (the "same amount of wet" seems .. ambiguous)
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On December 02 2009 11:38 JeeJee wrote:Show nested quote +On December 02 2009 06:56 Empyrean wrote:
And as for applications, there's so much calculus in statistics you wouldn't even believe it. Calculus is such a useful field of mathematics and finds itself in every day life. For example, you could use calculus to argue that if it rains at a constant rate, you'll get the same amount of wet no matter what speed you move through it (though you'll probably end up colder if you walk leisurely). The argument can basically be summed up as the fact that you're displacing the same amount of water, but it takes some calculus to show exactly why. is that actually true? doesn't make intuitive sense to me.. lets say its raining out, and you have to walk from your car door to apartment door (say 5 meters). you can sprint that, and you'll get some raindrops but you won't be soaked
Yeah this doesn't really make any sense to me either. The way empyrean explained it only makes sense if the raindrops were stationary, and there were a constant amount of raindrops per volume of air.
In the real world situation, taking 5 hours to walk 5 meters would end up with a lot more rain landing on your head than taking 5 seconds to walk 5 meters. The amount of rain landing on your head would be constant no matter what speed you are moving at though, it just matters how much time you are in the rain.
Not really important though, most likely just a mistake from empyrean
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ur own velocity changes the effective angle in which the raindrops hit u, the faster u are the more skewed the angle is. therefore a greater percentage of ur total body surface gets directly hit by raindrops when running through the rain instead of walking.
imagine 2 extreme cases: u stand still, ie zero velocity. only ur head and ur shoulders get wet. (and ur belly, depending on how fat u are ).
the other extreme would be moving so fast that the velocity of the raindrops is negligible compared to it, in this case ur whole front would get wet while the back half of ur head and shoulders wouldnt.
empyrean is right though that the amount of raindrops per exposed area is unaffected by ur velocity.
edit: omfg, whats going on? my english is so horrible and full of mistakes today. wtf!
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Haha, don't worry about it, you're very understandable. Your English is much better than my German
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On December 02 2009 14:05 Black Gun wrote:
ur own velocity changes the effective angle in which the raindrops hit u, the faster u are the more skewed the angle is. therefore a greater percentage of ur total body surface gets directly hit by raindrops when running through the rain instead of walking.
imagine 2 extreme cases: u stand still, ie zero velocity. only ur head and ur shoulders get wet.
that's not true, consider the rain at a slight angle. it still falls under empyrean's assumptions of constant rate yet renders this argument moot
and even if it's purely vertical, it'll still trickle down and you get completely soaked if you just stand there ^_^
edit:
empyrean is right though that the amount of raindrops per exposed area is unaffected by ur velocity.
ok, this i could see as being true
not the original "same wetness" claim though
and yes your english is very good :O i don't know what you're complaining about :p
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On December 02 2009 11:10 nttea wrote:
* Under what conditions should you trade, after knowing how much money your envelope contains?
* For the naive estimate of 1/2 to be correct for that probability, what would Granny's distribution p(x) have to be? Any problems with that?
* If Granny used an exponential distribution with mean $100, what strategy optimizes the expected value of the amount of money you keep?
sorry, i don't understand most of this since i've had no formal edumication.
However to me it seems like it always makes sense to always swap the envelope. Since you have 50% chance of gaining double the amount of your current envelope, and 50% chance of gaining only half. I'm not sure if this is the only thing you ask for though (:
No. Granny has either 3Y or 1.5Y between the two envelopes. The thing is one value is more likely than the other. If you work out how more likely one is than the other you can make a decision.
Has been solved in a previous post I think. 1.5Y has to be less than 2 times as likely as 3Y for the switch to be profitable.
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For example, you could use calculus to argue that if it rains at a constant rate, you'll get the same amount of wet no matter what speed you move through it (though you'll probably end up colder if you walk leisurely). The argument can basically be summed up as the fact that you're displacing the same amount of water, but it takes some calculus to show exactly why.
The amount of water that hits your front (assuming vertical rain and stomach) is always the same.
But the amount that hits your top depends on how long you're in the rain.
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EPT
