Do you know what the Law of Parsimony is? How about you do an empirical study and then apply the Law of Parsimony to your conclusions.

You obviously are not well informed or educated in the general philosophies of science and scientific investigation. Which means that you do not have a very solid position on which to assess the relation of science and God and instead it is you that does not understand the "scope of science".

If you are demanding evidence of an unfalsifiable claim you kind of just showed everyone it is you that needs to pick up a science text book.

Sicence also cannot disprove there are elfs living on the Sun. Guess that means it's real.

Clearly a function of the brain. Does the light come from the candle or is the candle channeling it from the 7th layer of Heaven?

Clearly you have never been delusional, you know where you have delusions that seem objectively real but they are not. Remember when you were talking about delusions. Your senses can and very often mislead you, even in a healthy brain. Reason is the bootstraps of the mind, not your senses.

Assuming that your brain is feeding you the truth. How do you know that it is?

If your defense of God is nothing but pure subjective belife that means you have no evidence of God.

There is no God people, rip the bandaid off and just accept the truth. The world will be better off when we can finally leave all this silly nonsense behind.

I was mocking your "millions" comment. I don't actually care about your personal belifes.

How do you differentiate anything at all from delusion?

Billions of people think you are not God, so I guess that means you are not God. Since billions is greater than millions.

Who understands God better? Those who study the natural world, which is said to be shaped by God himself or people who study scriptures which were clearly written by man.

If God really is the father of creation then it is the natural philosophers and scientists that truely study his work. Theologians study man, believing man to be God.

Theoretical or not you still shouldn't approach it without consideration of objective processes.

If people just treated x as undefined and ignored Y these conflicts would not exist and the model would be objectively congruent. People keep working on the assumption of one limit but objectively there are actually three sets of bounds.

It is the coin flip. There is a probability associated with the objective process and a probability associated with your chance of correctly guessing the outcome. If you did it right these two should match, but you can't just sweep aside the objective because it does not match your subjective expectations.

*On my phone sorry for any typos.

It is not the job of the objective to conform to your subjective expectations.

My equality is an objective truth and my math considering that equality is correct. So perhaps the problem is elsewhere.

Don't confuse subjective expectations with objective reality.

The OP says x and 2x. If the players wants to think that means 5 or 20, then that is not the fault of the game master.

So no one can meet my challenge?

You are of course entitled to your opinions.

Now,

Let $a$ represent the value you see when you open your envelope and let $b$ represent the value in the other envelope. They where both decided at the same time, by the same method. They are one outcome.

This means,

$P(a) = P(b)$

Always!

You consider the possible values for $b$,

$b=x_{i,1}$
$b=2x_{i,1}$
$b=x_{i.2}$
$b=2x_{i,2}$

Then the probability of each is,

$P(a)=P(b=x_{i,1})$
$P(a)=P(b=2x_{i,1})$
$P(a)=P(b=x_{i,2})$
$P(a)=P(b=2x_{i,2})$

Since they are all equal to $P(a)$ then this must be true:

$P(a)=P(b=x_{i,1})=P(b=2x_{i,1})=P(b=x_{i,2})=P(b=2x_{i,2})$

If your model does not meet this requirement then you are wrong.

Trying to completely stats-block the conversation is just petty. The truth is that you can't follow me, and instead of trying you want to wave it away as "statistics".

Also just a note but probability theory IS a stats course. If you are making probabilistic claims about an unknown population from a sample you are engaging in stats. $Y$, is our sample.

Let $a$ represent the value you see when you open your envelope and let $b$ represent the value in the other envelope. They where both decided at the same time, by the same method. They are one outcome.

This means,

$P(a) = P(b)$

Always!

You consider the possible values for $b$,

$b=x_{i,1}$
$b=2x_{i,1}$
$b=x_{i.2}$
$b=2x_{i,2}$

Then the probability of each is,

$P(a)=P(b=x_{i,1})$
$P(a)=P(b=2x_{i,1})$
$P(a)=P(b=x_{i,2})$
$P(a)=P(b=2x_{i,2})$

Since they are all equal to $P(a)$ then this must be true:

$P(a)=P(b=x_{i,1})=P(b=2x_{i,1})=P(b=x_{i,2})=P(b=2x_{i,2})$

If your model does not meet this requirement then you are wrong.

There can be a theoretical statistical assessment even without data. There can also be a well defined Bayesian prior. In fact I could exactly define a sigma, mu and domain for $Y$ that could be used as a well justified prior. The truth is when people are considering all the possible values of $Y$, they are very much setting up a Bayesian prior.

Let $a$ represent the value you see when you open your envelope and let $b$ represent the value in the other envelope. They where both decided at the same time, by the same method. They are one outcome.

This means,

$P(a) = P(b)$

Always!

You consider the possible values for $b$,

$b=x_{i,1}$
$b=2x_{i,1}$
$b=x_{i.2}$
$b=2x_{i,2}$

Then the probability of each is,

$P(a)=P(b=x_{i,1})$
$P(a)=P(b=2x_{i,1})$
$P(a)=P(b=x_{i,2})$
$P(a)=P(b=2x_{i,2})$

Since they are all equal to $P(a)$ then this must be true:

$P(a)=P(b=x_{i,1})=P(b=2x_{i,1})=P(b=x_{i,2})=P(b=2x_{i,2})$

If your model does not meet this requirement then you are wrong.

Anyone remember the discussion of the coin flip?

You flip a coin, before it lands it has a 50/50 chance for H or T. After it lands what is its chance for H or T?

Remember that?

You flip a coin, it has a 50/50 chance of being H or T, it lands on H that event is over and done with; physically the coin cannot be T. However, you don't look at it, so you guess that it could be H or T, and you give each a 50/50 chance. That is your guess and the probability applies to your guess. You guess can never change the actual value on the coin.

You randomly selected $x_i$, and the chance of selecting that value is $P(x_i)$, which is unknown, but after it is selected what is its chance of being selected? 0, as the event is already decided. The contents of the envelopes cannot change no matter what you guess at, that event is over and done with and both envelopes have been filled.

By the time you are looking at your 10 bucks the contents of both envelopes has already been decided and is set in stone. Since the contents of both envelopes is determined at the same time by the same method then the probability of both must always be equal. If your model does not conform to this truth then you are wrong.

The contents of A has the same exact chance of occurring as the contents of B, this relationship should hold for every single consideration of possible outcomes for $Y_i$, if it doesn't then you did something wrong.

If you are mixing case one and case two you already confounded your unknown limits. Whatever you consider for the possible contents of the unopened envelope the probability of its occurrence must equal the probability of the occurrence in your envelope, which means every consideration of possible outcomes should be consider as equally likely because that relations holds in every single case. If your model cannot fit to that reality then your model is wrong, as it does not reflect the problem in the OP.

Cash limit is 15 bucks only one dollar bills.

I could have consulted the Ouija board, examined the start charts, went on a deep spiritual journey and only after downing an insane amount of LSD shoved some numbers into a hat and drew the number 3. Then jammed 3 bucks in A and 6 bucks in B.

Anyone want to guess at the probability of getting 3 in A in such a case ? I can tell you exactly what it is: It is the same probability of getting 6 bucks in B. As they are both selected in the same manner, at the same time from the same sample space.

Cash limit is 15 bucks only one dollar bills.

Heck, I could have reached into my wallet and just pulled out 10 ones realized that I don't have enough for 20 bucks and just then arbitrarily decided to stuff 4 bucks in one and 8 in the other.

No random selection at all, but the contents of both envelopes still came from one event from the same sample space at the same time. Whatever the probability of such a method was they both share it.

I think people are confusing the actual cash with the selection method.

Anyone want to see me make a random selection from a continuous normal distribution under the same conditions?

Cash limit is 15 bucks only one dollar bills.


2 bucks.

So I put 2 in A and 4 in B. Once again they are the same event pulled from the same sample space at the same time.

I don't know why people keep making all these assumptions about the unknown domain, distribution, and selection methods when there are an uncountable number of ways to accomplish this task.

The problem with making assumptions about what you think is and what is not, is that you can't think of everything and there could be something you overlooked. This is why the Law Parsimony is important. Unconfirmed assumptions always carry unknown uncertainty even if they seem perfectly reasonable, that is why we should try to limit them.

Let's do this, an example with limits and a defined scale

Say I have a 15 bucks, that is my cash limit and it is all in one dollar bills.

There now we know what the game master has.

I decide to use my computer to randomly select a value,


It selects 3.

I stuff 3 bucks in one envelop and 6 bucks in the other other. Both envelopes are filled from the same chance mechanism and sample sample space; it is one event.

I don't think I need to to follow through on the rest. No one ever said the domain of $X_i$ has to be the same as my cash limit. That was an assumption people made.

It absolutely necessarily has to be equally likely as the same chance mechanism is used to decide the contents of both envelopes. The contents of both envelopes is one event from the same sample space.

If a certain someone here had a greater understanding of statistics they would realize why this is must be true. You have to consider the whole process from start to finish and model the whole thing.

Consider this,

You open envelope $A$ and you see it has 10 bucks ($x_i$).

Now you consider the other envelope, envelope $B$.

You decide $B$ could have 5 bucks ($1/2x_i$) or it could have 20 bucks ($2x_i$).

You don't know what the chance of either 5 or 20 bucks is but you do that the probability of 20 is equal to the probability of 10 and the probability of 5 is equal to the probability of 10, which means the probability of 5 is equal to the probability of 20.

In other terms:

$P(x_i) = P(1/2x_i)$

$P(x_i) = P(2x_i)$

Which means:

$P(2x_i) = P(x_i) = P(1/2x_i)$

$P(2x_i) = P(1/2x_i)$

They necessarily have the same chance of occurring.

That there's more to gain than there is to lose is what can be objectively measured. — Michael

I have already made enough comments on this aspect. No point going over it again.

I have no reason to believe that the former is more likely — Michael

It is precisely as likely. They are equal in probability.

My point is that your personal cretria is something that we cannot objectivly measure.

It's not necessarily equally likely — Pierre-Normand

I know it's not necessarily equally likely. — Michael

Actually it is. Whatever is in the other envelope had the same chance to occur as what is in your envelope. If you have 10 and that had 75% chance to occur then any possible value in the other envelope must also have a 75% chance to occur. If the other envelop has 5 bucks then it has 75% chance to occur. If it is 20 then it has a 75% chance to occur. The envelopes will always share this relationship, always. So whatever you consider as the possible outcome, all considerations are necessarily equally likely