E = a/2 - b/4
E = a/2 - a/2 = b/4 - b/4 = 0.
This still looks like you're considering what would happen if we always stick or always switch over a number of repeated games. I'm just talking about playing one game. There's £10 in my envelope. If it's the lower bound then I'm guaranteed to gain £10 by switching. If it's the upper bound then I'm guaranteed to lose £5 by switching. If it's in the middle then there's an expected gain of £2.50 for switching. I don't know the distribution and so I treat each case as equally likely, as per the principle of indifference. There's an expected gain of £2.50 for switching, and so it is rational to switch. — Michael
If we then play repeated games then I can use the information from each subsequent game and switch conditionally, as per this strategy (or in R), to realize the .25 gain.
This still looks like you're considering what would happen if we always stick or always switch over a number of repeated games. I'm just talking about playing one game. — Michael
If it's in the middle then there's an expected gain of £2.50 for switching — Michael
If we then play repeated games then I can use the information from each game to switch conditionally, as per this strategy (or in R), to realize the .25 gain over the never-switcher. — Michael
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
Since my sample space was a perfectly valid sample space, and I never mentioned events at all, it demonstrates your "very bad understanding" of those terms. It was a very bad choice of a sample space for this problem, for the reasons I was trying to point out and stated quite clearly. But you apparently didn't read that.That is a very bad understanding of what a sample space and an event is. — Jeremiah
Actually, I was applying it. Improperly with the intent to demonstrate why its restriction is important:You are not applying your Principle of Indifference there,
The Principle of Indifference places a restriction on the possibilities that it applies to: they have to be indistinguishable except for their names. You can't just enumerate a set of cases, and claim each is equally likely. — JeffJo
I didn't say we should (A) use a probability (B) density (C) curve. I stated correctly that there (A) must be a probability (B) distribution for the (C) set of possible values, and that any expectation formula must take this distribution into account. Even if you don't know it.Furthermore, it makes no sense to use a probability density curve on this problem, — Jeremiah
Mathematical Statistics with Applications, Wackerly, Mendenhall, ScheafferThe probabilist reasons from a known population to the outcome of a single experiment, the sample. In contrast, the statistician utilizes the theory of probability to calculate the probability of an observed sample and to infer from this the characteristics of an unknown population.
And it is even more obvious you want to use statistics anywhere you can, no matter how inappropriate. The lexicon of both probability and statistics is the same, since statistics uses probability. It applies it to the experimental data you keep talking about, and of which we have none.I already figured out that your field was not statistics, — Jeremiah
two.envelopes <- function(){ x <- (runif(1)) x <- (abs(as.numeric(format(round(x, 3)))))*10 #randomly selects a number #limits the number of decimal places x can have and muiltples x by 10 to simluate realistic dollar values. p <- c(x,2*x) A <- sample(p, 1, replace=F) #creates a vector with x and 2x then randomly selects one for A. if (A == x) { B <- 2*x } else { (B <- x) } return(c(A,B)) }
The puzzling part is about our understanding the mathematics, not how we use it to solve the problem. But that still makes it a probability problem. People who know only a little have difficulty understanding why the simple 5v/4 answer isn't right, and people who know more tend to over-think it, trying to derive more information from it that is there.My current, and I think "final", position is that this isn't really a probability puzzle at all. Here are my arguments for my view and against yours. — Srap Tasmaner
That's because the higher/lower question is the only one we can assign a probability to. There is on;ly one kind of probability that you can place on values. That's "valid," meaning there is a set of possibilities, and their probabilities sum to 1. Any other kind - frequentiest, bayesian, subjective, objective, informative, non-informative, or any other adjective you can think of - is outside teh scope of the problem.1. The only probability anyone has ever managed to assign any value to is the probability of choosing the larger (or the smaller) envelope -- and even that is only the simplest noninformative prior.
Correct.2. All other probabilities used in solutions such as yours are introduced only to immediately say that we do not and cannot know what their values are.
4. Much less the PDF on that space.[/quote
Careful. "PDF" usually refers to a "Probability Density Function," which means the sample space is continuous. We have a distribution for a discrete sample space.
Th only thing we can say about it (or the sample space ) is, is that it still must be valid. A valid sample space has a maximum value. A valid distribution implies there are values in the sample space where there is an expected loss.
This is a red herring. It only has meaning if we know the distribution, and we don't. So it has no meaning.5. By the time the player chooses, a value for X has been determined.
I assume you mean the amounts the benefactor puts in the envelopes (this isn't presented as a game show). That's why I usually speak generically about values. That can apply to the minimum value, which is usually what x refers to in this thread, the difference d which turns out to be the same thing as x but can be more intuitive, the value v in your envelope which can be x or 2x, and the total t (so x=t/3).6. We might also describe that as the host's choice of a value for X.
Then I'm not sure what you mean - it appears some of mine. If you are given v, and so have two x's, you have to consider the relative probabilities of those two x's.7. That choice is the very first step of the game and yet it appears nowhere in the probabilistic solutions, which in effect treat X as a function of the player's choices and what the player observes.
Please, get "updating" out of your mind here.10. The probabilistic model can safely be abandoned once it's determined that there will never be any evidence upon which to base a prior much less update one.
The point is that I'm saying both. You need to understand the various kinds of "variables."what is the advantage of saying that the variable X takes a value from an unknown and unknowable sample space, with an unknown and unknowable PDF, rather than saying X is not a variable but simply an unknown?
In the now canonical example of Michael's £10, he could say either:
(a) the other envelope must contain £20 or £5, but he doesn't know which; or
(b) there's a "50:50 chance" the other envelope contains £20 or £5, and thus the other envelope is worth £12.50.
I say (a) is true and (b) is false.
The fact that you use an expectation formula.What compels us to say that it is probabilistic ...
Here's my decision tree again (...) — Srap Tasmaner
If you are given v, and so have two x's, you have to consider the relative probabilities of those two x's. — JeffJo
Where JeffJo approach seems to me to be superior to yours is that it doesn't yield an incorrect verdict for the specific cases where the prior distribution is such as to yield envelope pairs where, conditionally on being dealt either the smaller or the larger amount from this pair, the expected gain from switching isn't zero. Your own approach seems to yield an incorrect result, in that case, it seems to me. — Pierre-Normand
Yours isn't really a decision tree that the player must make use of since there is no decision for the player to make at the first node. — Pierre-Normand
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