|
16936 Posts
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.
 
|
This problem seems really fun! I'd love to try my hand at it.
So what's a probability distribution?
|
16936 Posts
Wait, are you kidding?
T_T
|
|
|
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.
|
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).
|
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.
|
yea I've never heard of the term "probability distribution" either but I haven't taken a probability course either.
|
16936 Posts
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.
|
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).
|
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.
|
16936 Posts
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.
|
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)
|
16936 Posts
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 
|
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.
|
I don't understand anything you're saying.
|
oh my goodness, a stat problem!
Where's Gretorp? I'm sure he would enjoy this problem very much.
|
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?
|
|
|
16936 Posts
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.
|
|
|
|
|
|
EPT
