To remove that option, I recast the problem with the envelopes containing IOUs rather than cash — andrewk
I was being terse. A longer version of what I said is "So '2X' is meaningless if you try to use it as a value." This thread has gone on too long, and I didn't want to have to explain the mathematics of probability theory any more than I already have.Isn't 2X just a transformation of X that doubles the possible values in X? — Andrew M
The amount you have is $x. The other envelope contains either $2x or $x/2. If it's $2x then you gain $x by switching. If it's $x/2 then you lose $x/2 by switching. — Michael
Jeremiah only got into the sims & distributions business because everyone was talking about these things and it was his intention to put an end to all the speculation and get the discussion back on track. It seemed to me he did this reluctantly with the intention of showing that even if you made some assumptions you shouldn't -- this has always been his view -- it might not help in the way you think it does. — Srap Tasmaner
Do we need to assume that X is not continuous? If it is, all these probabilities are just 0, aren't they? — Srap Tasmaner
I'm still confused. This makes it sound like the switching argument isn't fallacious -- it just makes an unwarranted assumption. — Srap Tasmaner
The result that Pr(X=$10) must be zero is not an issue then, because F(X=$10) might not be. — JeffJo
You can't simply treat a random variable as an unknown. You can consider a set of unknown values from its range, but only if you couple that with their probabilities. — JeffJo
In this expression, Michael is using two new random variables, but doesn't recognize them as such. R represents the palyer's choice, and can be {2,1/2} with probability 50% each. Y is the other envelope, defined by Y=R*X (remember, these are random variables). His error is only recognizing one and not treating it as carefully as he should. So he missed that the distribution for Y is found by the transformation methods in your link, and that he needs to know the probabilities of the pair being ($x/2,$x) and ($x,$2x) to do this. — JeffJo
I just flipped a coin on my desk. I can see whether it is Heads or Tails. So "there is never a case where there is a non-zero chance of Heads, and a non-zero chance of Tails."
But If I ask you to assess the two cases, you should say that each has a 50%. The fact that the outcome is determined does not change how probability works for those who do not know the determination.
If I'm told that one envelope contains twice as much as the other, and if I pick one at random, am I right in saying before I open it that there's a 50% chance that my envelope contains the smaller amount? If so, I must also be right in saying after I open it and see the amount that there's a 50% chance that my envelope contains the smaller amount. — Michael
The simple truth is that you have been misinterpreting me since you joined the conversation. I saw it from your first response to me. I looked at your post and realized you were making false assumptions based on viewing post out of context of the thread. I knew if I enegaged you on that level the conversation would consistent of me untangling all of your misconceptions. — Jeremiah
You can't just enumerate a set of cases, and claim each is equally likely. If you could, there would be a 50% chance of winning, or losing, the lottery. — JeffJo
That is a very bad understanding of what a sample space and an event is. You are not applying your Principle of Indifference there — Jeremiah
Ya, great math there. — Jeremiah
Yes, my point was that the lottery example is a very bad description of a sample space. — JeffJo
It makes no sense to use a probability density curve[1] on this problem, considering X would only be selected ONCE[2], which means X<2X ALWAYS[3] (given that X is positive and not 0). That means no matter what X is the expected value will always be 1/2X+X[4], in every single case.
If you try to fit X to a statistical[5] distribution you are just piling assumptions on top of assumptions[6]. You are making assumptions about the sampling[7] distribution and the variance[8]. Assumptions in which you do not have the data to justify. You are also making assumptions about how X was even selected.[9] — Jeremiah
Then, despite the fact that I tried to address only those posts that had a smidgen of relevancy, or ones you pointed out as significant (and later claimed were not), you continued to insist you wouldn't read what I wrote. And it's quite clear you didn't; or at least that you didn't understand any of it.I am not doing this, not until you actually read all of my posts in this thread. — Jeremiah
you became belligerent started saying you would not read anything I wrote until I had read all of the irrelevant posts. — JeffJo
How could anyone who has read this thread have possibly concluded that I ever made this conclusion? When all I said was that any use of statistics - which you did advocate repeatedly - was inappropriate for a probability problem or a thought problem?How could anyone who has read this thread possibly concluded I was ever advocating for a statistical solution. I have been very clear that a statistical approach is incorrect. — Jeremiah
So that "observational data set" is the "experimental data set," isn't it? With each sample being an instance of the experiment "how does a single member of population X behave in circumstances Y?""Statistics is used on an experimental data set from repeated trials." — JeffJo
Yes, and it is also used on observational data sets to make generalized inferences about a population. — Jeremiah
So read the statement in its context, where I said exactly that. You are removing it from its context to make it look bad:Then go ahead and switch envelopes in the OP. — JeffJo
There is not enough information to calculate expected gain. — Jeremiah
But if you don't care about chances, only the possibility of gain? ... Then go ahead and switch envelopes in the OP. Just don't expect a gain. That can't be determined from the information.
So read the statement in its context, where I said exactly that. You are removing it from its context to make it look bad:it gives you a strategy that works on your assumed prior — JeffJo
Assuming your prior is correct, that is. — Jeremiah
Even in that more general case, the Bayesian approach can give a switching strategy with a positive expected net gain. — andrewk
No, it gives you a strategy that works on your assumed prior, not necessarily on reality.
Yes, as I have said repeatedly. And if you read the entire thread, you will see that this has been my point all along. Even though you don't know what the distribution is, you still have to treat whatever value you are using as a random variable with a probability distribution, and not simply "as an unknown." Which is what you have advocated.The point is that there must be a prior distribution for how the envelopes were filled — JeffJo
True, but you will have no knowledge of what that may be. — Jeremiah
And what people did I criticize this way? I simply pointed out that this problem is controversial because of an error that is routinely made everywhere the controversy exists.And yet you didn't read the posts, did you? Not then, maybe you read a few more after I pushed you. I may be an ass, but at least I read a thread before criticizing people. — Jeremiah
The purpose is to show why the formula (v/2)/2 + (2v)/2 = 5v/4 is wrong. The approach behind the formulation is indeed correct; it just makes a mistake that doesn't show up in the formula. And can't, if you accept the assertion "it is pointless to consider the conditional probability."The limit does not need to be specified, as the envelopes will never step outside the limit. Mathematically you cannot determine if you have the the smaller amounts or larger amounts as you can never rule out which case you are in. You can speculate on such things, but you can't quantify them. It is pointless to consider the conditional probability since both cases are subjectively equal in probability, it would still boil down to a coin flip. You can do it for completeness, but it really makes no difference. — Jeremiah
The solution has always been what I posted on the first page of this thread in post number 6, which has also been my stance this entire thread. A statistical solution has never been a viable option, which has also been my stance this entire thread. The truth is this problem has always been really simple to solve, it is untangling all the speculations and assumptions that confounded it. — Jeremiah
You could have X or 2X. If you have X and you switch then you get 2X but lose X so you gain X; so you get a +1 X. However, if you have 2X and switch then you gain X and lose 2X; so you get a -1 X. — Jeremiah
When all I said was that any use of statistics - which you did advocate repeatedly - — JeffJo
Statistics is simply not designed for a problem like this; it is better to just use basic mathematics. Statistics is for analyzing data. — Jeremiah
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