I never understood this constant obsession with the sex lives of others. Is it envy?

The sample space is [[10,20],[5,10]]. Notice how there are two 10s. Now show me the math which allows you to eliminate both of them.

Which 10 are you dropping 10=X or 10=2X? Also like I said I want to see the math.

But you're conflating different values of X: — Michael

No, that is what you are doing. You can't just drop the 10, it is still a possible outcome, as you don't know which 10 you are in. The sample space is [[10,20],[5,10]]. How are you eliminating both 10s and combining the two sets? You have to mathematically justify all of this.

I don't know why I waste my education and time.

Did you read the rest of my post?

So we have two cases here and have no clue which one we are in. Earlier I defined these cases as amount X case R, and and amount 2X case S. We have a lot of variables flying around so let's try to be consistent here.

Now that we have listed all possible outcomes we can define our sample space as [R,S], well call this sample space 2.

Now remember by our definitions an event is a subset of our sample space.

In event R X=10 and since in event R B must be 2X then B = 20.

So the sample space of event R is [10,20].

In event S 10=2X and since in event S B must be X then B= 5.

So the sample space of event S is [5,10]. (order does not matter)

So our sample space, which we named as sample space 2, is [R,S] where R is the set [10,20] and S is the set [5,10] or we can express it as [[10,20],[5,10]] — Jeremiah

The math definitively proves you wrong. This is not a matter of opinion, you are just flat out wrong.

Can B still be defined as [X, 2X]? — Michael

Your possible outcomes for A is 2X or X. In the outcome that A=X then B=2X, in the outcome that A=2X then B=X. So yes, it seems, B has two possible outcome of either X or 2X.

What you are saying is the equivalence of suggesting 2+2 =5. You can't setp outside the definitions simply because they are an inconvenience you, as then you are no longer doing math.

You need to follow the same steps, list all the possible outcomes and stick with the definitions.

Now let's work through this:

Remember the definition of our sample space is [A,B] where A is the set [X,2X] and B is the set [X,2X]. We'll call this sample space 1.

You open A and it has 10 bucks, but you don't know if that is 10=X or 10=2X so A is still defined as [X,2X]. If A is X then B is 2X and if A is 2X then B has to be X. So B is still defined as [X,2X]. That part remains the same.

So we have two cases here and have no clue which one we are in. Earlier I defined these cases as amount X case R, and and amount 2X case S. We have a lot of variables flying around so let's try to be consistent here.

Now that we have listed all possible outcomes we can define our sample space as [R,S], well call this sample space 2.

Now remember by our definitions an event is a subset of our sample space.

In event R X=10 and since in event R B must be 2X then B = 20.

So the sample space of event R is [10,20].

In event S 10=2X and since in event S B must be X then B= 5.

So the sample space of event S is [5,10]. (order does not matter)

So our sample space, which we named as sample space 2, is [R,S] where R is the set [10,20] and S is the set [5,10] or we can express it as [[10,20],[5,10]]


This is no different than any other math, we need to break it down and follow it step by step.

This right here does not go away simply because I miss defined a sample space in a different post. This is still true and proves you wrong. It is still something you have yet to actually address.

Let's review some definitions here:

If A and B are sets, then A is called a subset of B if, and only if, every element of A is also an element of B.

A sample space is the set of all possible outcomes of a random process or experiment. An event is a subset of a sample space.

Discrete Mathematics, An Introduction to Mathematical Reasoning, By Susanna S. Epp

Do you understand that these are not my definitions? They are established definitions that may be used in formal proofs.

Let's call the two envelopes A and B. Now envelope A could have X or A could have 2X and likewise B could have X or B could have 2X. Those are all the possible outcomes so by the definition of a sample space our sample space is [A,B] where A is the set [X,2X] and B is the set [X,2X], which means our sample space could also be written as [[X,2X],[X,2X]].

Did you follow that? They are sets in a set. In this case the elements of our sample space are the set [X, 2X]. So that means since your proposed solution does not follow the format of [X,2X] then, by the definition of a subset, it cannot be a subset of either A or B. — Jeremiah

I already admitted to the mistake, then corrected my position. You just get into rapid fire mode and you miss half of what is being said.

Ya, jumped the gun there and made a mistake myself (unlike you I am able to admit that). However, if you going to use my argument to justify your position, my argument, that you referenced there, still completely counters your statement of [5,20] and proves you 100% wrong. By using my argument as proof you have just admitted that you are wrong.

The sample space still would be [[X,2X],[X,2X]].

Which would be consistent with what I said here:

Intuitively this seems like new information which would call for you to update your prior; however, it still does not tell you if are you in case R or case S. You don't know if that is 10=X or 10=2X. So in truth you can't really update your prior, as your prior was based on the uncertainty of being in case R or case S. — Jeremiah

Your sample space becomes:

[[10=X,10=2X],[10=X,10=2X]].

Edit* This post contains an error see the next page for clarification.

Feigning ignorance does not resolve the conflict. As I have said many times over, the sample space of the other envelope is [X,2X]. That is not that hard to grasp, and since I don't believe you are an idiot, I am forced to conclude you feigning ignorance as a way to avoid this conflict.


Let's review some definitions here:

If A and B are sets, then A is called a subset of B if, and only if, every element of A is also an element of B.

A sample space is the set of all possible outcomes of a random process or experiment. An event is a subset of a sample space.

Discrete Mathematics, An Introduction to Mathematical Reasoning, By Susanna S. Epp

Do you understand that these are not my definitions? They are established definitions that may be used in formal proofs.

Let's call the two envelopes A and B. Now envelope A could have X or A could have 2X and likewise B could have X or B could have 2X. Those are all the possible outcomes so by the definition of a sample space our sample space is [A,B] where A is the set [X,2X] and B is the set [X,2X], which means our sample space could also be written as [[X,2X],[X,2X]].

Did you follow that? They are sets in a set. In this case the elements of our sample space are the set [X, 2X]. So that means since your proposed solution does not follow the format of [X,2X] then, by the definition of a subset, it cannot be a subset of either A or B.

You read it too fast and understood too little. I very clearly and fully addressed your concern.

You just said it yourself the sample space for both is X and 2X.

It went over this right here:

https://thephilosophyforum.com/discussion/comment/192854

Also, it should be made clear that an uninformative prior is a Classical apporach. A Baysian approach with an uninformative prior is philosophically the same apporach as a Classical apporach. Baysian divergence is when it introduces an informative prior.

I have no idea what a Frequentist would do. — andrewk

Flip a coin.

A simple random sample, which is what Classical statistics is based on, is a subset of a population which is obtained by a process which gives all sets of n distinct items in a population an equal chance of being selected. There are two n distinct items, so they would flip a coin.

Both Classical and Bayesian approach uses a random process to make selections from a sample space. Bayesian methods combine their prior with random samples from a sample distribution to create posterior distributions. Then they make their assessments off the posterior distributions.

That knowledge - based on seeing 10 in my envelope - allows me to rule out every other event (an event being a particular sum being in the other envelope) except for 5 and 20. So that's my sample space - just two events. — andrewk

Based on your knowledge you know that the sample space has to follow the algebraic form of [X,2X],and you know by the definition of a sample space, that your sample space can only contain possible outcomes (or at least I hope you do).

Therefore your sample space could never be [5,20], as it does not follow the form of [X, 2X]. Your sample space could be [5,10] or [10,20] but there is no way you can mathematically justify a sample space of [5,20] as it is not consistent with [X,2X].

Fraud.

Deleting my post doesn't make you right.

Classical and Bayesian analyses are both methods of statistics. Statistics is a data science, as in the analysis of data. They are not used or fit for simple probability problems like this one. Furthermore, as a science, they are empirically based, both of them.

I doubt you could even explain the differences.

If you have to limit your number of trials to leverage your prior then you are essentially doing the equivalence of p-hacking.

Expected gains are an average over the long run. Probability is the frequency of occurrences of repeated random events and that includes only possible outcomes.

That is far from a proof and the only thing it displayed is your lack of understanding of what expected gains are.

If you simulated a data set for the two envelopes, then did a Classical analysis and a Baysian analysis they would both come out to the same solution.

Even, and this is important, if your prior was a loss of 5 and a gain of 20. The MCMC chain would drag the misleading prior back on track to reflect the data. A Baysian method is not an excuse to skip empirical verification.

That is not a Classical statistics interpretation. Classical statistics is based on using estimated variance.

However, while we are on this topic, there are five steps in the Baysian method. The last step in a Baysian method is to check your model against the data. It requires empirical verification.

The very definition of probability is the frequency of occurrences over repeated random events.

The first time is no different than the 10,000th time. We could set this up 10,000 different times, bring in 10,000 different people so that each event is a "first" time and it still would not reflect your model.

when there were Nazis permitted to march the streets, was that we should respect another person's opinion, even if we hate it. — FreeEmotion

I value freedom and the right of free expression and only in that interest do I tolerate their presence but in no way do I respect their views. There is such a thing as being too open minded and I draw that line at respecting the views of Nazis.

If that was your take away from the USA then you failed to understand what was really happening. That has nothing to do with respecting the views of Nazis, it was about respecting the Constitution.

1) Insult the person when you do not agree with his policies
2) Assume that (1) will provide validity the the argument against his policies. — FreeEmotion

You missed.

You have to deal with the fact than many many people voted for this 'idiot'. What is the explanation for that, well maybe I have to accept the fact that they too, were 'clueless idiots'. — FreeEmotion

Never said that anywhere. I think you are actually driving to make me worst than I am so that I fit your caricature better.

Do you realize that expected gain is an average between gains and losses? It is stuff like this that makes me doubt your vague claim about your employment.

Not buying it, as you have displayed a large lack of comprehension here.

That distribution is (1) unknown, — andrewk

It is not unknown at all, as I just laid it out for you, and as I already pointed out 1/2X+X reflected the expected gains from the real world example.

I have come to the conclusion that you just spit out jargon but you really have no clue what you are talking about. I actually came to that conclusion several threads ago. I don't think you understand the concept of a probability distribution and, in relation to statistics, I don't think you know what those density curves are for.

Well technically this is a probability distribution: 1/2(X) + 1/2(2X). 50% is distributed to X and 50% is distributed to 2X.

But what you are talking about is a probability density curve, which is also a probability distribution, but one we would use in estimating parameters of a population from sample data. Typically we shorten the reference by just calling them distributions, but yes you are right, they are all technically distributions. However, your uses of probability density curves is still misplaced.

No one said a probability density curve was used to select X. And it won't matter anyway, as X is a real positive value, 2X will always be greater than X, the midpoint will always be 1/2X+X, and your possible outcomes will always be either 2X or X. This has nothing at all to do with distributions. Also the expected value being calculated is not of the density curve, it is the expected value of your take from the contents of the envelopes.

Statistics is simply not designed for a problem like this; it is better to just use basic mathematics. Statistics is for analyzing data.

It is an interesting problem nonetheless.

Let me see if I can sum it up more simply.

You have two envelopes, A and B.

There are two possible amounts for the contents of the two envelopes either X or 2X. You don't know which is which.

We'll call amount X case R, and and amount 2X case S.

You are handed envelope A.

Before you open it, you know that A could be case R or it could be case S. You have no clue which it is, so as an uninformative prior you give each case a 50% likelihood.

Now for the tricky part, you open envelope A and see it has 10 bucks in it.

Intuitively this seems like new information which would call for you to update your prior; however, it still does not tell you if are you in case R or case S. You don't know if that is 10=X or 10=2X. So in truth you can't really update your prior, as your prior was based on the uncertainty of being in case R or case S.

So your prior stands.

You consider swapping. You look at envelope B and realize it could be in case R or case S. You don't know which, so just as before you give it an uninformative prior and you give each case a 50% likelihood.

Now you could swap, but really probabilistically it changes nothing as your initial uncertainty would still stand. You would just trade the uncertainty of A for the uncertainty of B which is the same uncertainty. And the math? Well the math is indifferent to all options. So the decision to swap really comes down to if you feel lucky or not.

But if we try to model this from the participant's point of view — Srap Tasmaner

Up until I introduced the real world example, we both were looking at as if we were participates. Michael and I were looking at the same problem in the same context based on the same information.

we use Y. Then the expected gain loss sample spaces are [-Y/2, 0] and [2Y, 0], aren't they? — Srap Tasmaner

He was looking at it as 2X or X/2. Using Y does not change anything at all as there are still two possible values for Y and that is Y=X or Y=2X. You don't change anything by using Y, the algebra would still come out the same way. By the definition of a function it has to and if it doesn't then you did something wrong.

I'm tempted to think andrewk's point is relevant here: — Srap Tasmaner

It is binary, either it is X or 2X. So if one really needs to use a distribution as a sampling model then use a Bernoulli distribution, with alpha equal to one and beta equal to one as your priors. That is if you want to use a statistical nuke as a fly swatter. It is over the top and completely uncalled for.

Is the objection to Michael's approach not the assignment of the prior but simply that he is using sample spaces he cannot justify using? — Srap Tasmaner

Michael's error is not treating X as an unknown variable in accordance to the definition of a function. You have to trust the algebra and established proofs/definitions more than your intuition.

When he saw the 10 bucks he forgot to consider the uncertainty of his starting position and rewrote the values of Y to Y=X and Y=X/2. Then he creates two possible cases on known values of X. However the 10 bucks does not give us the value of X, as we still don't know if our starting point is 2X or X.

The tick to this problem is trusting the algebra over your eyes, which makes it less than intuitive. In this case the new information you receive can actually mislead you, but if you follow the math it leads you down the right path.

could show why even the uninformative prior does not lead directly to your conclusion. — Srap Tasmaner

I already did. Recall that my initial algebraic model was based on all the same information as Michael and I used an uninformative prior, and my approach leads to an expected value of 1/2(X) + 1/2(2X) = 1/2X+X. My initial approach modeled the real world example accurately, all based on the same information and situation as Michael.

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

Recall that event L is when you start with 2X and event K is when you start with X

Since we don't know which it is upon seeing the money we will consider this an uninformative prior and give each a fair 50% likelihood. Then our sample space is [K,L]

In the event of L our expected gain loss sample space is [-X, 0]

In the event of K our expected gain loss sample space is [X,0]

That is the same even if you go the 1/2 route.

Let's try running a simulation on that structure.

K <- c("X",0)
L <- c("-X",0)
q <- c(K,L)
w <- sample(q, 10000, replace = TRUE)
sum(w == "X")
sum(w == "-X")
sum(w == 0)

The Result are:

x: 2528
-x: 2510
0: 4962
``` — Jeremiah

If you follow the algebra it leads you to a solution which reflects a real world situation and when it comes to modeling probability this should be the goal.

In the absence of knowing the distribution of X, any calculations based on expected values prior to opening the envelope are meaningless and wrong. Since the claimed paradox relies on such calculations, it dissolves. — andrewk

Not true.

Even after seeing the first envelope you still don't know which case you are in, you could have 2X or X, you also don't know what case the other envelope is in, it could be 2X or X. This is also ture before you take your first peak. The fact is opening the envelope does not give us the vaule of X. So if you switch you switch to the same unknown. The errors is in treating X as a known after you open the the first envelope.

In both cases our expected value then is:

1/2(X) + 1/2(2X) = 1/2X+X.

Which is the mid point between X and 2X.

I picked this version to generate more discussion. I think this version of the problem forces more conflict of ideas, which in turn generates more discussion.