Regression toward the mean

[Ramius]

Too bad you didn't understand the links which you posted. A binomial distribution must have only two categories--"success" and "failure" if you will. Bungee Jumper was using eleven categories in his dice rolling example. Even you should have the mental capacity to see the difference between two categories and eleven categories.

867882[/snapback]

 

Pull holcombs rooster out of your ear and listen. I already stated the 2 possible outcomes.

 

1. rolling whatever you determine as a success (rolling a 2, or maybe a 5, or maybe rolling a double digit number)

2. not rolling those outcomes, hence a failure

 

By my count, that is 2 categories, success and failure, which works for a binomial distribution. Seems to me like i understood the links.

 

If you read my above post an continue to spew your sh--, that proves you cannot understand what someone types and clearly have no understanding of mathematics.

[Orton's Arm]

O RLY?

I see two outcomes:

 

1.) Less than eleven

2.) Greater than or equal to 11

 

Wheres the rest?

867888[/snapback]

Where's the rest? here, at the bottom of the page:

And it's really not that hard: You roll a pair of dice, it's probabilistic. When they stop moving, it's deterministic. The system has an expectation value (a "mean") of 7. Rolls subsequent to very low or very high rolls (2 or 3, or 11 or 12) will tend to regress toward the mean not because dice are error-prone or inaccurate - they're not, they're very accurate and not the least bit subject to error. It's because there's only 3 ways to roll a 2 or 3, and 33 other possibilities the next time you roll. Regression toward the mean happens because your current measure is deterministic, but your future measure is probabilistic.

 

That's probability, not error. It's really !@#$ing simple. Like I said, I can teach that to a three year old. Why the !@#$ do you have so much trouble with it? rolleyes.gif

 

And you can go here, first post of the page, to see this quote:

That's also why I used the two dice example before...it is a definitively discrete phenomenon that displays no inaccuracy, no error, and the binomial distribution of possibile values is a reasonable approximation of a normal distribution for the purpose of explaining regression toward the mean.

 

Again, Bungee Jumper didn't know that one of the requirements for a binomial distribution is that you group all your observations into only two categories. I had to tell him otherwise. Which would be fine, if he didn't incessantly (and falsely) accuse me of not understanding basic statistical terminology.

[Chilly]

Where's the rest? here, at the bottom of the page:

867917[/snapback]

 

He's saying the SAME EXACT THING there too.

 

It's because there's only 3 ways to roll a 2 or 3, and 33 other possibilities the next time you roll.

 

There are *still* only 2 outcomes:

 

1.) You roll a 2 or 3

2.) You roll anything else

 

Again - where's the rest?

 

And you can go here, first post of the page, to see this quote:

 

What he's saying there is correct as well. The distribution has a number of possible rolls - the 36 possibilities of the value of the roll. This is called the binomial coefficient, or, number of combinations that exist. However, even though there are a wide variety of different rolls, there are still only two outcomes:

 

1.) Rolling a 2 or 3

2.) Rolling anything else.

 

Again - where's the rest?

 

Again, Bungee Jumper didn't know that one of the requirements for a binomial distribution is that you group all your observations into only two categories. I had to tell him otherwise. Which would be fine, if he didn't incessantly (and falsely) accuse me of not understanding basic statistical terminology.

 

Yes, by defining the two categories. In this case, you have 36 possible combinations. However, we are only concerned about two outcomes, rolling a certain set of numbers, or NOT rolling a certain set of numbers.

[Ramius]

He's saying the SAME EXACT THING there too.

There are *still* only 2 outcomes:

 

1.) You roll a 2 or 3

2.) You roll anything else

 

Again - where's the rest?

What he's saying there is correct as well.  The distribution has a number of possible rolls - the 36 possibilities of the value of the roll. This is called the binomial coefficient, or, number of combinations that exist. However, even though there are a wide variety of different rolls, there are still only two outcomes:

 

1.) Rolling a 2 or 3

2.) Rolling anything else.

 

Again - where's the rest?

Yes, by defining the two categories.  In this case, you have 36 possible combinations.  However, we are only concerned about two outcomes, rolling a certain set of numbers, or NOT rolling a certain set of numbers.

867919[/snapback]

 

pwned

[Bungee Jumper]

 

Hey, HA, what's 3*4?

[Nervous Guy]

I think we really need a Statistics or Math forum so you clowns can really go at it...nerds.

[Orton's Arm]

He's saying the SAME EXACT THING there too.

There are *still* only 2 outcomes:

 

1.) You roll a 2 or 3

2.) You roll anything else

 

Again - where's the rest?

What he's saying there is correct as well.  The distribution has a number of possible rolls - the 36 possibilities of the value of the roll. This is called the binomial coefficient, or, number of combinations that exist. However, even though there are a wide variety of different rolls, there are still only two outcomes:

 

1.) Rolling a 2 or 3

2.) Rolling anything else.

 

Again - where's the rest?

Yes, by defining the two categories.  In this case, you have 36 possible combinations.  However, we are only concerned about two outcomes, rolling a certain set of numbers, or NOT rolling a certain set of numbers.

867919[/snapback]

You've surely heard the saying, "My country. Right or wrong, but my country." Sometimes I get the feeling that there are people on this thead saying, "Bungee Jumper. Right or wrong, but Bungee Jumper."

 

In the quotes I posted above, Bungee Jumper announced he was using "the binomial distribution of possible values [for rolling a pair of dice] as a reasonable approximation of a normal distribution." You can't do that unless you're dividing up those dice rolling results into eleven categories. Which means his conception of a binomial distribution was mistaken.

[Ramius]

 

Hey, HA, what's 3*4?

867972[/snapback]

 

Well that all depends. Lets say that you take a group of people. This group of people multiplies 3*4. When the results come back, you find that on average, the group determined that 3*4 = 12. So you decide to test the group again. One would think that the group would again determine that 3*4 = 12. But this isnt the case.

 

After the first test, you have some people who determined that 3*4 = 12. But you also have a % of people that got lucky or unlucky when doing their math. There are going to be a bunch of "lucky" folks who thought that 3*4 = 11, but due to error, got lucky and determined the answer as 12. There is also going to be an "unlucky" group who thought 3*4 = 13, but got unlucky and determined the answer is 12. Given a normal distribution of answers, there are going to be a larger number of lucky 11's than there unlucky 13's. When taking the test again, the true values of the participants answers will appear. This is called regression to the mean.

 

So while you may THINK that 3*4 = 12, regression to the mean due to error shows us that in reality, when you multiply 3*4, the answer will be less than 12.

[Chilly]

In the quotes I posted above, Bungee Jumper announced he was using "the binomial distribution of possible values [for rolling a pair of dice] as a reasonable approximation of a normal distribution." You can't do that unless you're dividing up those dice rolling results into eleven categories. Which means his conception of a binomial distribution was mistaken.

867998[/snapback]

 

Here are the possible values when you roll a pair of dice. This is the distribution that makes up the BINOMIAL COEFFICIENT and is what he was referring to:

 

2: 1&1

3: 2&1, 1&2

4: 1&3, 2&2, 3&1

5: 2&3, 3&2, 4&1, 1&4

6: 5&1, 1&5, 4&2, 2&4, 3&3

7: 6&1, 1&6, 5&2, 2&5, 4&3, 3&4

8: 2&6, 6&2, 5&3, 3&5, 4&4

9: 6&3, 3&6, 4&5, 5&4

10: 6&4, 4&6, 5&5

11: 6&5, 5&6

12: 6&6

 

THIS DOES NOT PREVENT YOU FROM HAVING A BINOMIAL DISTRIBUTION. If we convert that chart above into:

 

0 - Rolled a 2 or 3

1 - Rolled a 4-12

 

2: 0

3: 0, 0

4: 1, 1, 1

5: 1, 1, 1, 1

6: 1, 1, 1, 1, 1

7: 1, 1, 1, 1, 1, 1

8: 1, 1, 1, 1, 1

9: 1, 1, 1, 1

10: 1, 1, 1

11: 1, 1,

12: 1

 

You see how many freaking 1s there are? If you get a 0, you have a MUCH greater chance of getting a 1 on your next roll than a 0. Thats what he was saying.

[Wraith]

He's saying the SAME EXACT THING there too.

There are *still* only 2 outcomes:

 

1.) You roll a 2 or 3

2.) You roll anything else

 

Again - where's the rest?

What he's saying there is correct as well.   The distribution has a number of possible rolls - the 36 possibilities of the value of the roll. This is called the binomial coefficient, or, number of combinations that exist. However, even though there are a wide variety of different rolls, there are still only two outcomes:

 

1.) Rolling a 2 or 3

2.) Rolling anything else.

 

Again - where's the rest?

Yes, by defining the two categories.  In this case, you have 36 possible combinations.  However, we are only concerned about two outcomes, rolling a certain set of numbers, or NOT rolling a certain set of numbers.

867919[/snapback]

 

Bungee Jumper is correct that the binomial distribution for discrete variables is somewhat analogous to the normal distribution for continuous variables. That being said, some of the explanations that have resulted trying to rationalize what he said are reeeaaaally stretching logical thought.

 

The binomial distribution, as HA initially asserted, is used when there are only two discrete outcomes from a process. Where some of the rationalizations have departed from logic is by lumping "anything else" together as a "failure." For the binomial distribution to be useful, what is defined as a success or failure has to be meaningful. Randomly selecting a possible outcome such as two or three to be "success" and making everything else "failure" is not meaningful. If I were to do that, I would not get a useful picture of the underlying probability distribution. In fact, I would need ELEVEN seperate binomial distributions to get a useful discription of the pair-of dice-rolling process (2 is successful, all else failure, then 3 is successful, all else failure, then 4....and so forth).

 

Using the definition supplied by some of you, ANYTHING can be represented using the binomial distribution (which is true) and that representation would be useful (which is not true). The binomial distribution is VERY useful for determining the probability of getting repeated values (I rolled a 12 the first time, what are the odds I get a 12 the second time...?). In reality, the pair of dice scenario is MULTINOMIAL.

 

The normal distribution is much more useful for describing the pair-of-dice scenario despite having to make the concession to "not-quite-continuous variables." After all, not much, if anything, in the real world can actually be measured on a continuous basis, yet the normal distribution works just fine.

 

So, in conclusion, BJ's original comment was not wrong. But I really have to laugh at other peoples attempts to explain what he meant (as if he needs their help...). OWNED? Please! Hahahahaha.

[Chilly]

The binomial distribution, as HA initially asserted, is used when there are only two discrete outcomes from a process. Where some of the rationalizations have departed from logic is by lumping "anything else" together as a "failure." For the binomial distribution to be useful, what is defined as a success or failure has to be meaningful. Randomly selecting a possible outcome such as two or three to be "success" and making everything else "failure" is not meaningful. If I were to do that, I would not get a useful picture of the underlying probability distribution. In fact, I would need ELEVEN seperate binomial distributions to get a useful discription of the pair-of dice-rolling process (2 is successful, all else failure, then 3 is successful, all else failure, then 4....and so forth).

 

Using the definition supplied by some of you, ANYTHING can be represented using the binomial distribution (which is true) and that representation would be useful (which is not true). The binomial distribution is VERY useful for determining the probability of getting repeated values (I rolled a 12 the first time, what are the odds I get a 12 the second time...?).

 

Correct me if I'm wrong, that was the whole idea there big guy - the odds that you'd get a 2 or 3 vs a different number.

 

So, in conclusion, BJ's original comment was not wrong. But I really have to laugh at other peoples attempts to explain what he meant (as if he needs their help...). OWNED? Please! Hahahahaha.

868024[/snapback]

 

Right, thats it. I'm doing it because I feel BJ needs help explaining himself to the world.

[Orton's Arm]

Here are the possible values when you roll a pair of dice.  This is the distribution that makes up the BINOMIAL COEFFICIENT and is what he was referring to:

 

2: 1&1

3: 2&1, 1&2

4: 1&3, 2&2, 3&1

5: 2&3, 3&2, 4&1, 1&4

6: 5&1, 1&5, 4&2, 2&4, 3&3

7: 6&1, 1&6, 5&2, 2&5, 4&3, 3&4

8: 2&6, 6&2, 5&3, 3&5, 4&4

9: 6&3, 3&6, 4&5, 5&4

10: 6&4, 4&6, 5&5

11: 6&5, 5&6

12: 6&6

 

THIS DOES NOT PREVENT YOU FROM HAVING A BINOMIAL DISTRIBUTION.  If we convert that chart above into:

 

0 - Rolled a 2 or 3

1 - Rolled a 4-12

 

2: 0

3: 0, 0

4: 1, 1, 1

5: 1, 1, 1, 1

6: 1, 1, 1, 1, 1

7: 1, 1, 1, 1, 1, 1

8: 1, 1, 1, 1, 1

9: 1, 1, 1, 1

10: 1, 1, 1

11: 1, 1,

12: 1

 

You see how many freaking 1s there are?  If you get a 0, you have a MUCH greater chance of getting a 1 on your next roll than a 0.  Thats what he was saying.

868017[/snapback]

Your second distribution would look more like this

2,3: 0, 0, 0

4 - 12: 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1

 

Is the above a reasonable approximation for the normal distribution? I think not. Bungee Jumper clearly had your first chart in mind when he talked about his so-called "binomial distribution," and did not intend it to be converted into the single real binomial distribution I've described above, nor the series of binomial distributions Wraith described.

[Bill from NYC]

I am going to enter and leave in one post.

 

This thread is absolutely the most boring one that I have ever had the displeasure of reading, and I have been here for the better part of 9 years.

 

Hey, it IS my fault for even looking at it, but I don't think that some of you really can grasp just how flat out dull this is.

[Wraith]

Correct me if I'm wrong, that was the whole idea there big guy - the odds that you'd get a 2 or 3 vs a different number.

868036[/snapback]

 

No, that wasn't the whole idea.

[Ramius]

I am going to enter and leave in one post.

 

This thread is absolutely the most boring one that I have ever had the displeasure of reading, and I have been here for the better part of 9 years.

 

Hey, it IS my fault for even looking at it, but I don't think that some of you really can grasp just how flat out dull this is.

868060[/snapback]

 

did you read the other 50+ pages in the 2 previous threads?

[Chilly]

Your second distribution would look more like this

2,3: 0, 0, 0

4 - 12: 1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1

 

Yeah, I was just trying to show that in relation to the first part of my post.

[Bungee Jumper]

You've surely heard the saying, "My country. Right or wrong, but my country." Sometimes I get the feeling that there are people on this thead saying, "Bungee Jumper. Right or wrong, but Bungee Jumper."

 

In the quotes I posted above, Bungee Jumper announced he was using "the binomial distribution of possible values [for rolling a pair of dice] as a reasonable approximation of a normal distribution." You can't do that unless you're dividing up those dice rolling results into eleven categories. Which means his conception of a binomial distribution was mistaken.

867998[/snapback]

 

Most of these people don't !@#$ing like me, you git.

[Bungee Jumper]

Bungee Jumper is correct that the binomial distribution for discrete variables is somewhat analogous to the normal distribution for continuous variables. That being said, some of the explanations that have resulted trying to rationalize what he said are reeeaaaally stretching logical thought.

 

The binomial distribution, as HA initially asserted, is used when there are only two discrete outcomes from a process. Where some of the rationalizations have departed from logic is by lumping "anything else" together as a "failure." For the binomial distribution to be useful, what is defined as a success or failure has to be meaningful. Randomly selecting a possible outcome such as two or three to be "success" and making everything else "failure" is not meaningful. If I were to do that, I would not get a useful picture of the underlying probability distribution. In fact, I would need ELEVEN seperate binomial distributions to get a useful discription of the pair-of dice-rolling process (2 is successful, all else failure, then 3 is successful, all else failure, then 4....and so forth).

 

Using the definition supplied by some of you, ANYTHING can be represented using the binomial distribution (which is true) and that representation would be useful (which is not true). The binomial distribution is VERY useful for determining the probability of getting repeated values (I rolled a 12 the first time, what are the odds I get a 12 the second time...?). In reality, the pair of dice scenario is MULTINOMIAL.

 

The normal distribution is much more useful for describing the pair-of-dice scenario despite having to make the concession to "not-quite-continuous variables." After all, not much, if anything, in the real world can actually be measured on a continuous basis, yet the normal distribution works just fine.

 

So, in conclusion, BJ's original comment was not wrong. But I really have to laugh at other peoples attempts to explain what he meant (as if he needs their help...). OWNED? Please! Hahahahaha.

868024[/snapback]

 

That's why I've been staying out of it.

[Bungee Jumper]

Well that all depends. Lets say that you take a group of people. This group of people multiplies 3*4. When the results come back, you find that on average, the group determined that 3*4 = 12. So you decide to test the group again. One would think that the group would again determine that 3*4 = 12. But this isnt the case.

 

After the first test, you have some people who determined that 3*4 = 12. But you also have a % of people that got lucky or unlucky when doing their math. There are going to be a bunch of "lucky" folks who thought that 3*4 = 11, but due to error, got lucky and determined the answer as 12. There is also going to be an "unlucky" group who thought 3*4 = 13, but got unlucky and determined the answer is 12. Given a normal distribution of answers, there are going to be a larger number of lucky 11's than there unlucky 13's.  When taking the test again, the true values of the participants answers will appear. This is called regression to the mean.

 

So while you may THINK that 3*4 = 12, regression to the mean due to error shows us that in reality, when you multiply 3*4, the answer will be less than 12.

868001[/snapback]

 

 

But wouldn't the true value of 3*4 be the average of 3 and 4...i.e. three and a half?

[Wraith]

That's why I've been staying out of it. 

868077[/snapback]

 

I figured as much.