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Media Does Match Thread Posting have a Correlation with Qooty Results?

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Loonerty said:
Sorry, I'll translate.

Coach, there is literally nothing more that I can say because you refuse to listen to facts about mathematics from someone with a ******* degree in the field.

Go back to asking people how they feel and leave statistics to people who understand it.
Click to expand...
Nice melt...
I was enjoying our little disputation. I thought you presented your argument really well. But like so many others here, when you’re challenged, questioned, and your word is not taken as gospel, you resort to getting personal. Disappointing end to a good chat. Thought you were better than that.

Thanks for your resume, by the way. What are you doing with that mathematics degree?
 
TheCoach16 said:
Nice melt...
I was enjoying our little disputation. I thought you presented your argument really well. But like so many others here, when you’re challenged, questioned, and your word is not taken as gospel, you resort to getting personal. Disappointing end to a good chat. Thought you were better than that.

Thanks for your resume, by the way. What are you doing with that mathematics degree?
Click to expand...

Wonder what he charges per hour??
 

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TheCoach16 said:
Nice melt...
I was enjoying our little disputation. I thought you presented your argument really well. But like so many others here, when you’re challenged, questioned, and your word is not taken as gospel, you resort to getting personal. Disappointing end to a good chat. Thought you were better than that.

Thanks for your resume, by the way. What are you doing with that mathematics degree?
Click to expand...
What advice do you give your poor clients?

Fingers in their ears, "na na na I can't hear you"? Seems to be your life motto.

I have explained to you, as someone who has studied mathematics and statistics at a university level, why your reasoning is incorrect.
But you refuse to believe it.

I don't try and tell you you're incorrect about any psychology theories or practices. You know why? Because I don't have the faintest idea about it. You'd be well served to do the same.
 
It's all pretty simple really

Probability of a Single Event

If you roll a six-sided die, there are six possible outcomes, and each of these outcomes is equally likely. A six is as likely to come up as a three, and likewise for the other four sides of the die. What, then, is the probability that a one will come up? Since there are six possible outcomes, the probability is 1/6. What is the probability that either a one or a six will come up? The two outcomes about which we are concerned (a one or a six coming up) are called section on conditional probabilities on this page to see how to compute P(A and B) when A and B are not independent.

Probability of A or B

If Events A and B are independent, the probability that either Event A or Event B occurs is:

P(A or B) = P(A) + P(B) - P(A and B)

In this discussion, when we say "A or B occurs" we include three possibilities:

  1. A occurs and B does not occur
  2. B occurs and A does not occur
  3. Both A and B occur
This use of the word "or" is technically called inclusive or because it includes the case in which both A and B occur. If we included only the first two cases, then we would be using an exclusive or.

(Optional) We can derive the law for P(A-or-B) from our law about P(A-and-B). The event "A-or-B" can happen in any of the following ways:

  1. A-and-B happens
  2. A-and-not-B happens
  3. not-A-and-B happens.
The simple event A can happen if either A-and-B happens or A-and-not-B happens. Similarly, the simple event B happens if either A-and-B happens or not-A-and-B happens. P(A) + P(B) is therefore P(A-and-B) + P(A-and-not-B) + P(A-and-B) + P(not-A-and-B), whereas P(A-or-B) is P(A-and-B) + P(A-and-not-B) + P(not-A-and-B). We can make these two sums equal by subtracting one occurrence of P(A-and-B) from the first. Hence, P(A-or-B) = P(A) + P(B) - P(A-and-B).



Now for some examples. If you flip a coin two times, what is the probability that you will get a head on the first flip or a head on the second flip (or both)? Letting Event A be a head on the first flip and Event B be a head on the second flip, then P(A) = 1/2, P(B) = 1/2, and P(A and B) = 1/4. Therefore,

P(A or B) = 1/2 + 1/2 - 1/4 = 3/4.

If you throw a six-sided die and then flip a coin, what is the probability that you will get either a 6 on the die or a head on the coin flip (or both)? Using the formula,

P(6 or head) = P(6) + P(head) - P(6 and head)
= (1/6) + (1/2) - (1/6)(1/2)
= 7/12

An alternate approach to computing this value is to start by computing the probability of not getting either a 6 or a head. Then subtract this value from 1 to compute the probability of getting a 6 or a head. Although this is a complicated method, it has the advantage of being applicable to problems with more than two events. Here is the calculation in the present case. The probability of not getting either a 6 or a head can be recast as the probability of

(not getting a 6) AND (not getting a head).

This follows because if you did not get a 6 and you did not get a head, then you did not get a 6 or a head. The probability of not getting a six is 1 - 1/6 = 5/6. The probability of not getting a head is 1 - 1/2 = 1/2. The probability of not getting a six and not getting a head is 5/6 x 1/2 = 5/12. This is therefore the probability of not getting a 6 or a head. The probability of getting a six or a head is therefore (once again) 1 - 5/12 = 7/12.

If you throw a die three times, what is the probability that one or more of your throws will come up with a 1? That is, what is the probability of getting a 1 on the first throw OR a 1 on the second throw OR a 1 on the third throw? The easiest way to approach this problem is to compute the probability of

NOT getting a 1 on the first throw
AND not getting a 1 on the second throw
AND not getting a 1 on the third throw.

The answer will be 1 minus this probability. The probability of not getting a 1 on any of the three throws is 5/6 x 5/6 x 5/6 = 125/216. Therefore, the probability of getting a 1 on at least one of the throws is 1 - 125/216 = 91/216.

Easy peasy amiright?
 
Yakker said:
It's all pretty simple really

Probability of a Single Event

If you roll a six-sided die, there are six possible outcomes, and each of these outcomes is equally likely. A six is as likely to come up as a three, and likewise for the other four sides of the die. What, then, is the probability that a one will come up? Since there are six possible outcomes, the probability is 1/6. What is the probability that either a one or a six will come up? The two outcomes about which we are concerned (a one or a six coming up) are called section on conditional probabilities on this page to see how to compute P(A and B) when A and B are not independent.

Probability of A or B

If Events A and B are independent, the probability that either Event A or Event B occurs is:

P(A or B) = P(A) + P(B) - P(A and B)

In this discussion, when we say "A or B occurs" we include three possibilities:

  1. A occurs and B does not occur
  2. B occurs and A does not occur
  3. Both A and B occur
This use of the word "or" is technically called inclusive or because it includes the case in which both A and B occur. If we included only the first two cases, then we would be using an exclusive or.

(Optional) We can derive the law for P(A-or-B) from our law about P(A-and-B). The event "A-or-B" can happen in any of the following ways:

  1. A-and-B happens
  2. A-and-not-B happens
  3. not-A-and-B happens.
The simple event A can happen if either A-and-B happens or A-and-not-B happens. Similarly, the simple event B happens if either A-and-B happens or not-A-and-B happens. P(A) + P(B) is therefore P(A-and-B) + P(A-and-not-B) + P(A-and-B) + P(not-A-and-B), whereas P(A-or-B) is P(A-and-B) + P(A-and-not-B) + P(not-A-and-B). We can make these two sums equal by subtracting one occurrence of P(A-and-B) from the first. Hence, P(A-or-B) = P(A) + P(B) - P(A-and-B).



Now for some examples. If you flip a coin two times, what is the probability that you will get a head on the first flip or a head on the second flip (or both)? Letting Event A be a head on the first flip and Event B be a head on the second flip, then P(A) = 1/2, P(B) = 1/2, and P(A and B) = 1/4. Therefore,

P(A or B) = 1/2 + 1/2 - 1/4 = 3/4.

If you throw a six-sided die and then flip a coin, what is the probability that you will get either a 6 on the die or a head on the coin flip (or both)? Using the formula,

P(6 or head) = P(6) + P(head) - P(6 and head)
= (1/6) + (1/2) - (1/6)(1/2)
= 7/12

An alternate approach to computing this value is to start by computing the probability of not getting either a 6 or a head. Then subtract this value from 1 to compute the probability of getting a 6 or a head. Although this is a complicated method, it has the advantage of being applicable to problems with more than two events. Here is the calculation in the present case. The probability of not getting either a 6 or a head can be recast as the probability of

(not getting a 6) AND (not getting a head).

This follows because if you did not get a 6 and you did not get a head, then you did not get a 6 or a head. The probability of not getting a six is 1 - 1/6 = 5/6. The probability of not getting a head is 1 - 1/2 = 1/2. The probability of not getting a six and not getting a head is 5/6 x 1/2 = 5/12. This is therefore the probability of not getting a 6 or a head. The probability of getting a six or a head is therefore (once again) 1 - 5/12 = 7/12.

If you throw a die three times, what is the probability that one or more of your throws will come up with a 1? That is, what is the probability of getting a 1 on the first throw OR a 1 on the second throw OR a 1 on the third throw? The easiest way to approach this problem is to compute the probability of

NOT getting a 1 on the first throw
AND not getting a 1 on the second throw
AND not getting a 1 on the third throw.

The answer will be 1 minus this probability. The probability of not getting a 1 on any of the three throws is 5/6 x 5/6 x 5/6 = 125/216. Therefore, the probability of getting a 1 on at least one of the throws is 1 - 125/216 = 91/216.

Easy peasy amiright?
Click to expand...
Nah mate it's all random.
 
Yakker said:
It's all pretty simple really

Probability of a Single Event

If you roll a six-sided die, there are six possible outcomes, and each of these outcomes is equally likely. A six is as likely to come up as a three, and likewise for the other four sides of the die. What, then, is the probability that a one will come up? Since there are six possible outcomes, the probability is 1/6. What is the probability that either a one or a six will come up? The two outcomes about which we are concerned (a one or a six coming up) are called section on conditional probabilities on this page to see how to compute P(A and B) when A and B are not independent.

Probability of A or B

If Events A and B are independent, the probability that either Event A or Event B occurs is:

P(A or B) = P(A) + P(B) - P(A and B)

In this discussion, when we say "A or B occurs" we include three possibilities:

  1. A occurs and B does not occur
  2. B occurs and A does not occur
  3. Both A and B occur
This use of the word "or" is technically called inclusive or because it includes the case in which both A and B occur. If we included only the first two cases, then we would be using an exclusive or.

(Optional) We can derive the law for P(A-or-B) from our law about P(A-and-B). The event "A-or-B" can happen in any of the following ways:

  1. A-and-B happens
  2. A-and-not-B happens
  3. not-A-and-B happens.
The simple event A can happen if either A-and-B happens or A-and-not-B happens. Similarly, the simple event B happens if either A-and-B happens or not-A-and-B happens. P(A) + P(B) is therefore P(A-and-B) + P(A-and-not-B) + P(A-and-B) + P(not-A-and-B), whereas P(A-or-B) is P(A-and-B) + P(A-and-not-B) + P(not-A-and-B). We can make these two sums equal by subtracting one occurrence of P(A-and-B) from the first. Hence, P(A-or-B) = P(A) + P(B) - P(A-and-B).



Now for some examples. If you flip a coin two times, what is the probability that you will get a head on the first flip or a head on the second flip (or both)? Letting Event A be a head on the first flip and Event B be a head on the second flip, then P(A) = 1/2, P(B) = 1/2, and P(A and B) = 1/4. Therefore,

P(A or B) = 1/2 + 1/2 - 1/4 = 3/4.

If you throw a six-sided die and then flip a coin, what is the probability that you will get either a 6 on the die or a head on the coin flip (or both)? Using the formula,

P(6 or head) = P(6) + P(head) - P(6 and head)
= (1/6) + (1/2) - (1/6)(1/2)
= 7/12

An alternate approach to computing this value is to start by computing the probability of not getting either a 6 or a head. Then subtract this value from 1 to compute the probability of getting a 6 or a head. Although this is a complicated method, it has the advantage of being applicable to problems with more than two events. Here is the calculation in the present case. The probability of not getting either a 6 or a head can be recast as the probability of

(not getting a 6) AND (not getting a head).

This follows because if you did not get a 6 and you did not get a head, then you did not get a 6 or a head. The probability of not getting a six is 1 - 1/6 = 5/6. The probability of not getting a head is 1 - 1/2 = 1/2. The probability of not getting a six and not getting a head is 5/6 x 1/2 = 5/12. This is therefore the probability of not getting a 6 or a head. The probability of getting a six or a head is therefore (once again) 1 - 5/12 = 7/12.

If you throw a die three times, what is the probability that one or more of your throws will come up with a 1? That is, what is the probability of getting a 1 on the first throw OR a 1 on the second throw OR a 1 on the third throw? The easiest way to approach this problem is to compute the probability of

NOT getting a 1 on the first throw
AND not getting a 1 on the second throw
AND not getting a 1 on the third throw.

The answer will be 1 minus this probability. The probability of not getting a 1 on any of the three throws is 5/6 x 5/6 x 5/6 = 125/216. Therefore, the probability of getting a 1 on at least one of the throws is 1 - 125/216 = 91/216.

Easy peasy amiright?
Click to expand...
rfctiger74 can you read this and let me know what it all means.
 
TheCoach16 said:
Surprising or just random?

If the 100th head doesn’t surprise you, the first 99 shouldn’t either. Because the first toss had no bearing on the second toss, the previous two had no bearing on the third, the previous 73 had no bearing on the 74th and the previous 99 had no influence in the 100th. You agreed with me on this point.

Therefore, If I did toss 100 heads in a row, you may find it “surprising”, someone else may not. But either way, you’d accept that despite your expectation, it can happen, because each toss is random.
Click to expand...
You are kidding right?

I would bet all my money that I ever made and ever will make that you will NEVER flip heads 100 times in a row.
Your lack of mathematical understanding is appalling.
 

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Yakker said:
It's all pretty simple really

Probability of a Single Event

If you roll a six-sided die, there are six possible outcomes, and each of these outcomes is equally likely. A six is as likely to come up as a three, and likewise for the other four sides of the die. What, then, is the probability that a one will come up? Since there are six possible outcomes, the probability is 1/6. What is the probability that either a one or a six will come up? The two outcomes about which we are concerned (a one or a six coming up) are called section on conditional probabilities on this page to see how to compute P(A and B) when A and B are not independent.

Probability of A or B

If Events A and B are independent, the probability that either Event A or Event B occurs is:

P(A or B) = P(A) + P(B) - P(A and B)

In this discussion, when we say "A or B occurs" we include three possibilities:

  1. A occurs and B does not occur
  2. B occurs and A does not occur
  3. Both A and B occur
This use of the word "or" is technically called inclusive or because it includes the case in which both A and B occur. If we included only the first two cases, then we would be using an exclusive or.

(Optional) We can derive the law for P(A-or-B) from our law about P(A-and-B). The event "A-or-B" can happen in any of the following ways:

  1. A-and-B happens
  2. A-and-not-B happens
  3. not-A-and-B happens.
The simple event A can happen if either A-and-B happens or A-and-not-B happens. Similarly, the simple event B happens if either A-and-B happens or not-A-and-B happens. P(A) + P(B) is therefore P(A-and-B) + P(A-and-not-B) + P(A-and-B) + P(not-A-and-B), whereas P(A-or-B) is P(A-and-B) + P(A-and-not-B) + P(not-A-and-B). We can make these two sums equal by subtracting one occurrence of P(A-and-B) from the first. Hence, P(A-or-B) = P(A) + P(B) - P(A-and-B).



Now for some examples. If you flip a coin two times, what is the probability that you will get a head on the first flip or a head on the second flip (or both)? Letting Event A be a head on the first flip and Event B be a head on the second flip, then P(A) = 1/2, P(B) = 1/2, and P(A and B) = 1/4. Therefore,

P(A or B) = 1/2 + 1/2 - 1/4 = 3/4.

If you throw a six-sided die and then flip a coin, what is the probability that you will get either a 6 on the die or a head on the coin flip (or both)? Using the formula,

P(6 or head) = P(6) + P(head) - P(6 and head)
= (1/6) + (1/2) - (1/6)(1/2)
= 7/12

An alternate approach to computing this value is to start by computing the probability of not getting either a 6 or a head. Then subtract this value from 1 to compute the probability of getting a 6 or a head. Although this is a complicated method, it has the advantage of being applicable to problems with more than two events. Here is the calculation in the present case. The probability of not getting either a 6 or a head can be recast as the probability of

(not getting a 6) AND (not getting a head).

This follows because if you did not get a 6 and you did not get a head, then you did not get a 6 or a head. The probability of not getting a six is 1 - 1/6 = 5/6. The probability of not getting a head is 1 - 1/2 = 1/2. The probability of not getting a six and not getting a head is 5/6 x 1/2 = 5/12. This is therefore the probability of not getting a 6 or a head. The probability of getting a six or a head is therefore (once again) 1 - 5/12 = 7/12.

If you throw a die three times, what is the probability that one or more of your throws will come up with a 1? That is, what is the probability of getting a 1 on the first throw OR a 1 on the second throw OR a 1 on the third throw? The easiest way to approach this problem is to compute the probability of

NOT getting a 1 on the first throw
AND not getting a 1 on the second throw
AND not getting a 1 on the third throw.

The answer will be 1 minus this probability. The probability of not getting a 1 on any of the three throws is 5/6 x 5/6 x 5/6 = 125/216. Therefore, the probability of getting a 1 on at least one of the throws is 1 - 125/216 = 91/216.

Easy peasy amiright?
Click to expand...
TLDR
 

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TheCoach16 said:
Well that’s all I’m saying. Granted, in a very argumentative, frustrating manner...
Click to expand...
So why constantly refute everything I was saying, which was fact?
 

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