Philosophy Now Forum

vegetariantaxidermy wrote:You both appear to be determining the probability after the fact, which is a nonsense.

We are agreeing to negotiate how to proceed here. Thus dionisos just proposed a means to the language we may begin this process. I counter propose that (1) we have to avoid any method that places faith in repeated experiments in practice, and (2) to recognize that even while ONE of our views may be the absolute truth, to be fair, we should also include the possibility that both of us may be correct or even both of us are wrong (because their could be a third perspective too that is 'true' over both of ours).

Scott Mayers wrote:

vegetariantaxidermy wrote:You both appear to be determining the probability after the fact, which is a nonsense.

We are agreeing to negotiate how to proceed here. Thus dionisos just proposed a means to the language we may begin this process. I counter propose that (1) we have to avoid any method that places faith in repeated experiments in practice, and (2) to recognize that even while ONE of our views may be the absolute truth, to be fair, we should also include the possibility that both of us may be correct or even both of us are wrong (because their could be a third perspective too that is 'true' over both of ours).

I think you are just arguing for the sake of argument.

Scott. Clarify something for me. We agree that for every individual coin toss the odds of the coin landing heads or tails is exactly 50/50. Even if we have tossed the coin a hundred times already and it's come down heads every single time then on the 101st occasion the odds remain exactly 50/50 that it will come down heads again. This is what I understand probability to mean.

Firstly A. Do you share this understanding? and B. Do you therefore conclude that the odds of tossing heads 100 times in a row must be the same as it is for doing it once?

dionisos wrote:Scott Mayers,
If you don’t agree with my definition of what a probability is, please give your own.

I'm not disagreeing with you necessarily. I only question your need to include extending probabilities of multiple experiments because I fear that you might want to later use this to demonstrate a proof by an empirical method. I disagree with using any of the program codes we tried before because they do this in principle. It would default to favoring what is popularly accepted.

In the coin toss example above I was concerned that by proceeding to allow for multiple separate events, you could prove that 3/4 of the times one does the experiment of two tosses collectively, one increases their odds by playing the game multiple times. So if you follow this, do you at least accept that we cannot use the extended use of probabilities we normally may use in real life for trying to determine something using a statistical approach.

I WILL look over my texts to see how the notations are used for probabilities. I need some time though to do this. In the meantime, I want to demonstrate how what we perceive as a contradiction in reality (= paradox), this can be resolved by reinterpreting it in a different perspective.

To begin this point, let's see if we agree on certain factors of logic. I will only use the Propositional Calculus here for this sake. A propositional logic ignores the quantifiers "All" or "Some" that is later used for the Predicate (or Symbolic) Calculus. I'm guessing since you are familiar with computer programming that you at least know of Boolean logic and so we can use this too.

On Negations
Notice that while many think of negation as simply one 'kind' of reversal, there are actually many kinds we use in our languages. For instance, the term, "non-" differs from the term, "not-" when we use them to describe things. If I negate a noun as a proposition, A, this could mean "non-A" or simply, "not-A". If I prove that at least some "not-A" exists, then we are certain that the class, "non-A" exists. Since we are only using propositional logic here, we normally ignore these distinctions.

DeMorgan's Law states that to negate a conjunction or disjunction, we negate each term including the logical connective. So,

-(A & B) = (-A [-&] -B) = (-A v -B)

The "v" was traditionally used to represent the Latin word, "vel" which is the inclusive version of "or". We could use other symbols for negation where they conflict with understanding.

A.. B.. A v B.. -(A v B)
1.. 1.... 1......... 0
1.. 0.... 1......... 0
0.. 1.... 1......... 0
0.. 0.... 0......... 1

-A.. -B.. -A & -B.. -(A v B)
.0... 0...... 0......... 0
.0... 1...... 0......... 0
.1... 0...... 0......... 0
.1... 1...... 1......... 1

With this clarified, now consider the case where B = -A.

Then,
-(A & -A) = (-A [-&] --A) = (-A [-&] A) = (-A v A)

What this shows as that the solution to a contradiction is to apply the DeMorgan rule to find what IS true without being contradictory (or paradoxical). That is, if a world cannot have a contradiction (or paradox), we fix it by using this rule. If we think each truth as being unable to operate in the same reality, then we place the other one in an alternative 'world' or "dimension". Since a paradox cannot exist to nature in our contingent world, the fix is to add a "dimension".
As you should see, if we replace the values above for B to be equal to -A, this suggests how the contradiction in one place is resolved by looking at it in another way as either another dimension in one place or another 'world' if this isn't even allowed.

So my point here with regards to the Monty Hall Problem is that it is presented as both a REAL and paradoxical. Thus the solution is to have to accept that a dimension, not another world, exists that fixes this and is why we have to recognize that another perspective works better. Before trying to find this, dionisos, I want to see if you at least recognize this logic as sound before trying to use it in our analysis of the problem. Do you concur?

Obvious Leo wrote:Scott. Clarify something for me. We agree that for every individual coin toss the odds of the coin landing heads or tails is exactly 50/50. Even if we have tossed the coin a hundred times already and it's come down heads every single time then on the 101st occasion the odds remain exactly 50/50 that it will come down heads again. This is what I understand probability to mean.

Firstly A. Do you share this understanding? and B. Do you therefore conclude that the odds of tossing heads 100 times in a row must be the same as it is for doing it once?

No, the odds are dependent only upon a distinct logical argument without even needing to toss any number of times in practice.

My point is that each event is unique and so your preference to bother pointing to any experiments being done in reality does not count. That is, because a real case can occur such that all 100 times one tosses a Head, we can't use any statistical method to infer empirically that a probability is sufficiently confirmed as true or not. The truth is that it is 50/50 and no amount of empirical experiments will ever justify this. So we have to defer to math that focuses on only each event distinctively.

Earlier you appealed to some online ongoing experiment that you believed was good to prove your faith in the 2/3 factor. But this only means that you found just such an experiment out there to demonstrate this as 'fitting'. But it has to be true that there are other experiments out there that prove 1/3 or 1/2 as 'fitting' too. So we cannot trust those kinds of evidence to support any probability problem appropriately.

You CAN use a calculus approach to hopefully narrow down the possibility of errors, but in nature, even as unusual such experiments can be done, it is just possible that even where rare, such an outcome like ALL Heads could occur in 14 million times. (I picked this number because it is the approximate odds to win a 6/49 type lottery draw which assures us that although rare, somebody usually wins.)

Scott Mayers wrote:. But it has to be true that there are other experiments out there that prove 1/3 or 1/2 as 'fitting' too.

What does this mean? There are no such experiments so what do you mean when you say "it has to be true that there are".

Obvious Leo wrote:

Scott Mayers wrote:. But it has to be true that there are other experiments out there that prove 1/3 or 1/2 as 'fitting' too.

What does this mean? There are no such experiments so what do you mean when you say "it has to be true that there are".

So you were lying HERE?

Obvious Leo wrote:

Obvious Leo wrote:

Obvious Leo wrote:As I suspected there might be there is a Monty Hall game running on the internet.

As of now 278,324 contestants decided to stay with their first choice and won a total of 93,286 pretend cars, a success rate of just below 34%

168,244 contestants decided to change their choice and won a total of 111,535 pretend cars, a total of 66%.

Scott. Would you care to name the parties responsible for staging such a massive global conspiracy or would you prefer to apologise to the members of this forum for treating them like morons.

This experiment shows that switching guesses in the Monty Hall game yields a result where the contestant will win the car 66% of the time whereas by not switching guesses the contestant will win the car only 34% of the time.

Are you claiming that this is a statistical anomaly and that if the game is allowed to run longer then the true probability will level out to 50% for both alternatives, in accordance with your hypothesis. In other words are you claiming that a sample size of almost half a million guesses is insufficient to be regarded as statistically significant. If indeed you are claiming this then why is 100 guesses regarded as being of statistical significance by people who invite others to test the Monty Hall puzzle out for themselves? Once again I remind you that EVERY SINGLE TIME the Monty Hall game has EVER been played it has yielded the same outcome. Switching guesses doubles the contestant's chances of winning. Once again I remind you that you have yet to offer a satisfactory explanation for this.

Obvious Leo wrote:This experiment shows that switching guesses in the Monty Hall game yields a result where the contestant will win the car 66% of the time whereas by not switching guesses the contestant will win the car only 34% of the time.

Are you claiming that this is a statistical anomaly and that if the game is allowed to run longer then the true probability will level out to 50% for both alternatives, in accordance with your hypothesis. In other words are you claiming that a sample size of almost half a million guesses is insufficient to be regarded as statistically significant. If indeed you are claiming this then why is 100 guesses regarded as being of statistical significance by people who invite others to test the Monty Hall puzzle out for themselves? Once again I remind you that EVERY SINGLE TIME the Monty Hall game has EVER been played it has yielded the same outcome. Switching guesses doubles the contestant's chances of winning. Once again I remind you that you have yet to offer a satisfactory explanation for this.

Sources please?

Scott Mayers wrote:Sources please?

I'm not doing your research for you. You can look them up for yourself. They're all over the internet and thousands of academic papers have been published on the subject. it is a completely and utterly uncontroversial proposition that switching guesses doubles the contestant's chances of winning in the Monty Hall game and nobody in the whole world other than you is claiming otherwise. The burden of proof lies with you.

Obvious Leo wrote:

Scott Mayers wrote:Sources please?

I'm not doing your research for you. You can look them up for yourself. They're all over the internet and thousands of academic papers have been published on the subject. it is a completely and utterly uncontroversial proposition that switching guesses doubles the contestant's chances of winning in the Monty Hall game and nobody in the whole world other than you is claiming otherwise. The burden of proof lies with you.

I might perhaps call you a pedophile too. Would this require that you'd have the burden to prove me wrong?

Scott. If you google "Monty Hall Problem" it will yield 375,000 search results. If you can find just ONE site amongst all of these which supports your argument I'll walk backwards to Sydney stark naked whistling Yankee Doodle Dandy.

You can call me whatever you like and I would regard myself under no obligation in the matter.

What you're forgetting, Scott, is that this is not an abstract mathematical problem which can be resolved by the correct use of equations. This is a real physical prediction about a real experimental scenario which anybody in the whole world can duplicate. You can actually test this for yourself and have an answer inside ten minutes.

Scott Mayers wrote: For two tosses, we have the following possibilities. Each of these represent real possible outcomes should one do any such experiment:
....Game 1...... Game 2
(1)....H..............H
(2)....H..............T
(3)....T..............H
(4)....T..............T


If we did an empirical experiment, in practice each of the above rows represent possible REAL results. As such, you could see that in (1) such an outcome would be probable as 1/4. This particular possible outcome also represents the truth "At least one H" is probable. As such, the probability to have "At least one H" is also probable 3/4 times given multiple worlds where these two coins could be tossed. In other words, if we had different groups do two coin tosses independently, it would appear that 3/4 of them would support that at least 1/2 of the time, the proper 1/2 odds of any given event could be confirmed.

P(at least one H) = 3/4

First, yes it is true that : P(at least one H) = 3/4 when you toss two coins.
This is not a matter of representation.
If someone want to toss two coin, toss one and get a tail, then he get new information, and the probability to get at least one H change.
P(at least one H | the first toss is a tail) = 1/2 when you toss two coins.

Second, i don’t understand your text, i will show you why:

This particular possible outcome also represents the truth "At least one H" is probable

probable 3/4 times

it would appear that 3/4 of them would support that at least 1/2 of the time

This is exactly why, i don’t want we continue with this method, i want English sentences only when they are necessary, otherwise i want math.
Because i believe you confuse yourself with natural language, in a way you can’t with math.

In your other message, i accept anything you said up to this(included):

-(A & -A) = (-A [-&] --A) = (-A [-&] A) = (-A v A)

But it is even more clear to show the actual value of "-(A & -A)"
-(A & -A)=-(False)=True
(-A v A)=True

I don’t agree with everything you said after this, that i consider to be some sort of new age thinking, and something you certainly can’t deduce from these simple equations.
I consider it is just you, confusing yourself with natural language.

This is to explain to you, why this is useless to continue like this, to go outside the general strategy i gave.
Don’t try to respond to what i didn’t understand, or do it only with math, because i know from experience, that this with lead to nonstop messages and misunderstanding if you do. (and this is why i will stop to respond to these kind of messages).

Now, or you accept my definition of what a probability is, and any conclusion this definition could lead to, or you don’t accept it, and give your own.
I don’t see any reason to continue any discussion about probability with you, before this first step: we should before all, agree on what a probability is.