That there's more to gain than there is to lose is what can be objectively measured. — Michael
I have no reason to believe that the former is more likely — Michael
But it can't be unbounded and uniform. So it is inconsistent in all possible cases.What you are saying is correct in any case (most cases?) where the prior probability distribution of the envelope values isn't unbounded and uniform. In the case where it is, then there is no inconsistency. — Pierre-Normand
What you are saying, is that if you postulate a distribution where (yes, I did reverse it) Pr(picked higher)=Pr(picked higher|V=v) and Pr(picked lower)=Pr(picked lower|V=v), then the results of the two conceptually-different formulas are the same. What I am saying is that one is conceptually incorrect, and one is conceptually correct. And I keep repeating this, because it is the error people make in the TEP when they perceive a paradox.[Exp(other) = (v/2)*Pr(picked higher) + (2v)*Pr(picked lower)] is correct in the special case where the prior distribution is uniform and unbounded, — Pierre-Normand
It cannot be equally likely without postulating a benefactor with (A) an infinite supply of money, (B) the capability to give you an arbitrarily-small amount of money, and (C) a way to select a random number uniformly from the set of all integers from -inf to inf.I know it's not necessarily equally likely. But we're assuming it, hence why Srap said we should just flip a coin. — Michael
sample(1:5, 1) #Pulls one random sample from 1 to 5 [1] 3
x <- rnorm(1, mean = 2.5, sd = 1) x <- abs(as.numeric(format(round(x, 0)))) x #Pulls a number at random then rounds it. #If I get 0 or greater than 5 then I just re-roll. #I could code the re-roll in but I am being lazy. [1] 2
It cannot be equally likely without postulating a benefactor with (A) an infinite supply of money, (B) the capability to give you an arbitrarily-small amount of money, and (C) a way to select a random number uniformly from the set of all integers from -inf to inf.
All three of which are impossible.
But the reason you should reject the solution you use is because it is not a correctly-formed expectation. You are using the probability of picking the smaller value, where you should use the probability that the pair of values is (v,2v) *AND* you picked the smaller, given that you picked v. — JeffJo
It cannot be equally likely without postulating a benefactor with (A) an infinite supply of money, (B) the capability to give you an arbitrarily-small amount of money, and (C) a way to select a random number uniformly from the set of all integers from -inf to inf. — JeffJo
Imagine you're given £100 and are offered the choice to pay £100 to play a game with a 5/6 chance of winning (say a dice roll). If you win then you win £1,000,000 and if you lose then you lose all the money you've won up to that point.
The average return for repeated games is 0, as you're almost certain to lose at some point. But playing it just once? That's worth it.
This is why I think talking about average returns over repeated games is a red herring. — Michael
So to make this a better analogy, let's say that some third party asks us both to play the game. He will roll two dice, and if I win then you give me £10 and if I lose then I give you £5. He doesn't tell us what result counts as a win for me and what counts as a win for you. It could be that 1-11 is a win for you, or it could be that 1-11 is a win for me, or it could be that 1-6 is a win for me.
I would be willing to play, as I have more to gain than I have to lose. You, presumably, wouldn't be willing to play, as you have more to lose than you have to gain. — Michael
This is why I think talking about average returns over repeated games is a red herring. — Michael
No, I agree it has to be constrained to operate in the real world. That's why there has to be a real-world maximum value, you can't have an arbitrarily-small real-world value, and you can't choose a uniformly-distributed integer in the range (-inf,inf) (which is not the same thing as choosing a uniformly-distributed real number in [0,1) and back-calculating a gaussian random value from it).I just want to note that we seem to be in agreement on everything. The only reason why we seemingly disagreed in our recent exchange is because you objected to my stated requirement that the game be conceived as a "real world" problem, — Pierre-Normand
You seem to think that it is only the highest-possible v where you have an expected loss. Maybe you are confused by the fact that it was the easiest example that shows it.I'm not saying that for every value in the chosen envelope it is equally likely that the other envelope contains twice as much as half as much. I'm saying that, given the £10 in my envelope, I will assume that the other envelope is equally likely to contain £20 as £5. I accept that there is an upper (and lower) bound. I'm just assuming that my £10 isn't it. — Michael
Then please, show us one that applies to the problem. And explain how it is a "well justified" anything, and not just a hypothetical.There can be a statistical analysis even without data. There can also be a well defined Bayesian prior. In fact I could exactly define a sigma, mu and range for Y that could be used as a well justified prior. — Jeremiah
Also just a note but probability theory IS a stats course. — Jeremiah
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