The scams of Statistics...

[Scott Mayers]

Scott Mayers, thanks for trying something more clear.
But i don’t agree, and it is still not the kind for formalized answer i wanted. (I will let Obvious Leo try to convince you on it.)

I still wait for the definition of "probability". I said you i will not go outside of a step by step method.

I'm presenting to you to which you are ignoring where I prove correct and opt not to point out your perceived error (in my proof).
Then you simply keep demanding another further proof. The above clearly indicates the math I'm implicitly using as sufficient for a proof involving the probabilities involved. Point out precisely yourself where you believe I erred rather than simply stating you disagree without qualification. I don't see you demanding Leo here to prove he understands you with the same precision or qualification of needing math. You're proving your bias only to favor anyone who already agrees with you.

[Scott Mayers]

dionisos,

I need to test something. Can you describe what you see in the following image?

Test.png (8.12 KiB) Viewed 3164 times

[Obvious Leo]

Scott. Are you suggesting that your so-called proof remains valid despite the fact that it relates to no physically real scenario? You have yet to offer an answer to this question.

Why is it ALWAYS the case that switching guesses in the Monty Hall game DOUBLES the contestant's chance of winning?

This is an incontestable FACT, Scott, supported by millions of empirical experiments. How do you propose to deal with this problem, bearing in mind that the most magnificent of beautiful hypotheses can be destroyed by a single inconvenient fact?

[Scott Mayers]

Scott. Are you suggesting that your so-called proof remains valid despite the fact that it relates to no physically real scenario? You have yet to offer an answer to this question.

Why is it ALWAYS the case that switching guesses in the Monty Hall game DOUBLES the contestant's chance of winning?

This is an incontestable FACT, Scott, supported by millions of empirical experiments. How do you propose to deal with this problem, bearing in mind that the most magnificent of beautiful hypotheses can be destroyed by a single inconvenient fact?

Please, Leo, just link me to one such source...just even one. I've looked online and only find the very puzzle being argued using math. I also found 'experiments' which only offers us to test it in faith of some program I cannot see to determine whether it is fair or not. I also mentioned that you can't determine the odds in practice with the same precision you can in pure mathematics.

You didn't answer what the odds to win is in the example of the coin toss game being played out by the rules I mentioned previously. If you're not confident with the math I can spell it out but believe you may be able to intuit it. I wanted to use that to show you something.

Edit: I mistakenly wrote 'dice' when I meant 'coin'. Fixed.

[Obvious Leo]

I don't know how to post links and I want you to do the experiment for yourself. There was a time when most high school and many university undergraduate curricula had it as routine course-work but I'm not sure if this is still the case because I would have thought that there was nobody left in the world who doesn't understand the Monty Hall puzzle. It seems I'm wrong.

I didn't understand your question about the coin toss so run it by me again. If you toss a coin and I have to guess the outcome then I have a 50/50 chance of guessing correctly. I sincerely hope you're not going to try and prove me wrong with equations.

[Scott Mayers]

...

Why is it ALWAYS the case that switching guesses in the Monty Hall game DOUBLES the contestant's chance of winning?

This only appears to work once you are able to play the second part of the game beginning with the host opening a door. But once this is done, you can't actually go back to the 'stay' position accepting it as 1/3 because it too 'doubles'.

When you go forward and simply 'pretend' to switch, from that perspective looking back at your original choice, that selection turns out to appear 2/3 as well with the same logic. This means that you have to repair the odds to reflect this change which corrects it to 1/2.

Pretend that you just turned on the television after the Guest of one of these games had been revealed the goat by the host leaving only the two doors.

You would see two doors to which the guest is asked if they want to stay on their selection or switch. Knowing that a goat is behind one door and the winning car is behind the other without any further wisdom, this is 1/2 chance to win regardless from you as a viewer. Even if you heard that the host had previously revealed a goat and does this 100% of the time, it is irrelevant. In the second half of the game, you NEVER get two doors with goats! So these are canceled out, period!

[Obvious Leo]

OK. Let's try this.

I am the host and you are my subject. I hide my coin under a cup and then invite you to guess which cup. Do you agree that you have a 1/3 chance of being right and a 2/3 chance of being wrong?

[Scott Mayers]

I don't know how to post links and I want you to do the experiment for yourself. There was a time when most high school and many university undergraduate curricula had it as routine course-work but I'm not sure if this is still the case because I would have thought that there was nobody left in the world who doesn't understand the Monty Hall puzzle. It seems I'm wrong.

I didn't understand your question about the coin toss so run it by me again. If you toss a coin and I have to guess the outcome then I have a 50/50 chance of guessing correctly. I sincerely hope you're not going to try and prove me wrong with equations.

In one game, you could toss the coin and get either a H or a T. I defined the game as one in which you 'win' only if you get a head.

But if I added that you could have a second chance if you lose (getting a T), then to win, you need the second game to get a head at least to win.

First game:
H or T

1) If H, you win [game ends first possible win] <--- 1/2 win

2) If T, you get to toss a second game. Go to Second Game

Second game:
H or T

3) If H, you win [game ends second win] <--- 1/2 x 1/2 = 1/4 win

4) If T, you lose [game ends for just two rounds]

You only lose 1/3 of the time on the case above but win 2/3 of the time.

This too suggests that you increase your odds for simply playing a second round, right? But how can this actually mean this unless the more rounds you play increases your odds if you keep getting another chance? So if you were to extend this to even more rounds where for every time you got a T, you get to try again, eventually in an infinite number of games, you'd approach a guaranteed 100% chance to win.

For a better picture of this, I'll use the 'code' thing here so that I can indent and allow you to see this easier:

Code: Select all

(1)
(a)H [WIN]
(b)T --> (2)
	(a) H [WIN]
	(b) T --> (3)
			(a) H [WIN]
			(b) T --> (4)
					 (a) H [WIN]
					 (b) T --> ... 
					      			 ...
					      				    ...
					      				    		(b)T --> (∞)
					      				    		(a) H [WIN]		  		   

In other words, technically, if you play such a game with more rounds, you always increase your quantity of WINs up to 100% the longer you play. Obviously, if this was done by experiment, this would and should prove 'true' too. Notice that the game stops when you get a 'H' once.

In analogy, if you get a Car (like the 'H') in the first round, it stops. So this is where they get the 1/3 to Stay from in the game and 2/3 if you Switch. I'm showing you here that I really DO understand the apparent logic. This is because when you play on, the game treats the first round as REAL.

So far, do you follow this logic and see that I really DO understand this from the traditional perspective that you and dionisos agree to?

[Obvious Leo]

I have no idea what any of this last offering has to do with the Monty Hall scenario under discussion.

Will you answer this question.

I am the host and you are my subject. I hide my coin under a cup and then invite you to guess which cup. Do you agree that you have a 1/3 chance of being right and a 2/3 chance of being wrong?

[Scott Mayers]

I have no idea what any of this last offering has to do with the Monty Hall scenario under discussion.

Will you answer this question.

I am the host and you are my subject. I hide my coin under a cup and then invite you to guess which cup. Do you agree that you have a 1/3 chance of being right and a 2/3 chance of being wrong?

Your demanding to take the lead here. If you want me to try to prove that I already understand your position, read the above. You're trying to prove something I already agree to. That is, I already agree that the 2/3 is a real answer. The problem is that I can also show that other answers follow too. Thus it is my argument that you have to pay attention to because it is you who is not getting what I'm saying here. You aren't supposed to prove what I already know. If anything, you need to prove that the other perceptions are NOT equivalently valid.

But this means first that I show you how I understand with clarity your position. I extend the argument to a game with coins alone to show you why we cannot trust using repeated games to prove that theory 'true'. In other words, from the above you should see that as you do 'experiments' it should by default only confirm the logic being used even if in error.

[Obvious Leo]

You're trying to prove something I already agree to. That is, I already agree that the 2/3 is a real answer. The problem is that I can also show that other answers follow too.

Are you taking the piss? I'll remind you that this is a philosophy forum. Either the probability of switching one's guess doubles or it doesn't but it can't both double and not double.

[Scott Mayers]

I have no idea what any of this last offering has to do with the Monty Hall scenario under discussion.

Will you answer this question.

I am the host and you are my subject. I hide my coin under a cup and then invite you to guess which cup. Do you agree that you have a 1/3 chance of being right and a 2/3 chance of being wrong?

The weight of the 1/3 odds stay 1/3 until you go on to another round. As soon as you begin the second round, the odds of 'staying' too go up to 2/3. This is because if the switch is only 1/3 for the remaining goat, to stay means you increase your odds of getting that goat 1/2 the time on the stay. This is

a 1/2 - 1/3 = 3/6 - 2/6 = 1/6 chance to get that goat if you stay. The other 1/6 chance for the other goat is lost because it was revealed. Thus you have 1/6 chance for remaining goat if you stay 0/6 chance for the other goat [since it is revealed], and the remaining 4/6 chance to get the car = 2/3 for staying too!

Edit: (2/3 + 2/3)/2 = 1/3 by the way.

[Scott Mayers]

You're trying to prove something I already agree to. That is, I already agree that the 2/3 is a real answer. The problem is that I can also show that other answers follow too.

Are you taking the piss? I'll remind you that this is a philosophy forum. Either the probability of switching one's guess doubles or it doesn't but it can't both double and not double.

"Real answer" means that it is valid by the math being used by perception. Probability is NOT an exact part of mathematics. It is like comparing the soft science of Sociology to the hard science of chemistry. On the Wikipedia page on "probability", it says this:

Interpretations[edit]

When dealing with experiments that are random and well-defined in a purely theoretical setting (like tossing a fair coin), probabilities can be numerically described by the number of desired outcomes divided by the total number of all outcomes (tossing a fair coin twice will yield head-head with probability 1/4, because the four outcomes head-head, head-tails, tails-head and tails-tails are equally likely to occur). When it comes to practical application however, there are two major competing categories of probability interpretations, whose adherents possess different views about the fundamental nature of probability:

Objectivists assign numbers to describe some objective or physical state of affairs. The most popular version of objective probability is frequentist probability, which claims that the probability of a random event denotes the relative frequency of occurrence of an experiment's outcome, when repeating the experiment. This interpretation considers probability to be the relative frequency "in the long run" of outcomes.[5] A modification of this is propensity probability, which interprets probability as the tendency of some experiment to yield a certain outcome, even if it is performed only once.

Subjectivists assign numbers per subjective probability, i.e., as a degree of belief.[6] The degree of belief has been interpreted as, "the price at which you would buy or sell a bet that pays 1 unit of utility if E, 0 if not E."[7] The most popular version of subjective probability is Bayesian probability, which includes expert knowledge as well as experimental data to produce probabilities. The expert knowledge is represented by some (subjective) prior probability distribution. The data is incorporated in a likelihood function. The product of the prior and the likelihood, normalized, results in a posterior probability distribution that incorporates all the information known to date.[8] Starting from arbitrary, subjective probabilities for a group of agents, some Bayesians[who?] claim that all agents will eventually have sufficiently similar assessments of probabilities, given enough evidence (see Cromwell's rule).

I take the "Subjectivist" view above here on this problem.
Actually, I take both partially here except in the "objective" basis, I hold that only ONE experiment is needed...the logical test to prove the case. Subjectively, this holds in that it points to the way we can subjectively interpret alternative solutions depending upon one's perspective.

[Scott Mayers]

This reminds me of a time in my youth when a roommate of mine was trying to tell me about a 'system' to assure he could win lottery games more frequently. I was troubled by this and had to figure out how he determined this. Apparently he was investing in one source material from a magazine which claimed that the person selling the system used it to win the lottery with unusual frequency.

I thought that it had to be luck. But my friend just laughed at me as he continued to prove me wrong as he actually kept winning these games. So I refreshed my Statistics and Probability from high school Algebra and then solved the problem. I then used this to demonstrate to my roommate the actual logic behind it and why this person selling this system reasonably won.

You can guarantee certain wins depending upon how much money you invest in a single draw to cover all extra odds in one single draw. I also remembered that even in your own country (or was it New Zealand?) in which a similar draw was going on (a 6/49 draw) to which a company set up a company to sell shares to the public for this one draw. The odds for the game is 1/(14 Million) and so if you had 14 million dollars AND the draw was greater than this number, you could guarantee a win by simply buying up each of the 14 Million different tickets. Note that you also 'win' multiple times for all different combinations! The only major risk is if it gets 'split'. The odds of one person is alone great. But for two, this requires a lot more people and comparing a 100% chance to win over the other potential winners who have an almost zero chance to win, their odds was great. And, in fact, they won. The problem was that for all the shares that people invested in them, it split the prize between them all. So a $1 investment was assured to make you money but it could be only $0.30, for instance.

P.S. My friend was still not convinced. I don't know where he is today. But if he held to that view, I'm almost certain he's now suffering some gambling addiction somewhere without a penny to his name at most times.

[Obvious Leo]

Scott. Why do you keep referring to scenarios which are not the Monty Hall scenario under discussion? If you would confine yourself to simple yes/no answers to my questions we might be able to get somewhere but once you start qualifying those answers we are no longer talking about the same thing.
I want to do this step by step in the simplest way possible.

Do you agree that in MY experiment if I place a single coin under a single cup you have a 1/3 chance of guessing which cup I placed it under. Forget about what comes next and just answer yes or no.