I love this week's podcast. Skipping from a nice calm discussion at the beginning to a loud drunken slur-fight at the end is an excellent contrast. Note to self, stay quiet during conversations about math.
Yeah....as an Engineer that's taken a lot of Probability and Statistics, the math section of the podcast was very painful. Very. It's very hard to wrap your head around it, I agree. It has to do with exclusion, if you remove the picked item and reduce the pool of possibilities.
Like when I worked at pizza hut doing the cash at the end of the day, counting the register. I know that the probability of getting .00 cents is 1/100, but it was so cool when it happened! It seemed special cause I decided that 00 was a special case and I treated it different than if I had 23 cents after all the money was counted, cause who gives a shit about 23 cents? 00 cents is the only special case in the world to me, so when it happened I was like so rare, cause I only paid attention to that pairing.
Not just bring back video podcast but can you guys stop kidding yourselves and just go back to the name drunk tank? I remember someone saying that you did it for marketing...but c'mon I think at this point your ok with being the drunk tank again.
I guess as long as no one is discussing pancake-related holidays,foreign nicknames for male genitalia, or obscure saints that have killed off mythological creatures, your just bound to say something stupid Gavin.
*Instead of saying the probability of something all the time, i will denote the probability of something as P(). [EX: probability of red = P(red)]
Actually during the math part, everybody was right, INCLUDING GAVIN. Gavin was taking into account Dependence while everybody else were taking into account Independence. Dependencecares what happens in the past, but independencedoesn't cares what happens in the past. Or in other words, Events A and B are dependent if the fact that A occurs does affect the probability of B occurring. Events A and B are independent if the fact that A occurs does not affect the probability of B occurring.
Definition: 1) Events A and B are Dependent if P(A and B)= P(A) * P(B|A). - P(B|A) means the probability of B given A has already happened. 2) Events A and B are Independent if and only if P(A and B)= P(A) * P(B). - you can think about it as P(B|A) = P(B) because the probability of B is not affected if A occurred, P(B|A), or not.
Example: There are 5 colored balls in a cup: a red one, a blue one, a yellow one, a green one, and a black one. You want to find the probability of getting the red ball first, then the blue ball second, P(red and blue). Parts: A) Find P(red and blue) with Dependence B) Find P(red and blue) with Independence.
First: Regardless of dependence or Independence because nothing happened yet, P(red) = 1/5. This is always the case.
Second: the probability of getting the blue ball will depend if there is Independence or dependence (putting back the red ball in the cup or not).
A) Find P(red and blue) with Dependence If you want to find the P(blue) with dependence, you have to leave out the red ball because you do care that the red ball was taken out: you care what happen in the past. So because you care what happen to the red ball, you leave the red ball out and now there are 4 balls left in the cup. So the P(blue|red) = 1/4. So if we take the definition of dependence, P(red and blue) = P(red) * P(blue|red) = (1/5) * (1/4) = 1/20. So P(red and blue) = 1/20.
B) Find P(red and blue) with Independence. If you want to find the P( red and blue) with Independence, you have to put back the red ball back in the cup because you don't care you got the red ball before: you don't care what happens in the past. Because you put the red ball back in the cup, there are 5 balls in the cup. So P(blue) = 1/5. So if we take definition of independence, P(red and blue) = P(red) * P(blue) = (1/5) * (1/5) = 1/25. So P(red and blue) = 1/25. Summary: P(red and blue) with dependence = 1/20 and P(red and blue) with independence = 1/25
So yeah, I hope this clarifies things lol I love Rooster Teeth. <3 :D lol -J03 Ryd3r
CORRECTION And Added Notes: -Gavin Was Right about talking about dependence. Everybody else is getting independence and dependence mixed up. - P(A) * P(B|A) = P(AB) [ P(A and B) And P(AB) are the same thing], this a variation of conditional probability: Definition: Conditional probability is P(B|A) = P...
CORRECTION And Added Notes: -Gavin Was Right about talking about dependence. Everybody else is getting independence and dependence mixed up. - P(A) * P(B|A) = P(AB) [ P(A and B) And P(AB) are the same thing], this a variation of conditional probability: Definition: Conditional probability is P(B|A) = P(BA) / P(A)
PART 1: I can't fit all of the text on 1 comment.
Time And Comments: 1:02:21-1:02:26: That statement is correct because this weeks win doesn't care what happens in the past P(This weeks win | Last weeks win) = (This weeks win) : Independence
1:02:31 1:02:45: Assume total amount of numbers is 49. P(1)= 1/49. P(2|1)= 1/48, taking Dependence into account. P(2'|1) = 47/48. - ' means the compliment.
Definition: The complement of an event A is the set of all outcomes in the sample space(all possible possibilities) that are not included in the outcomes of event A. -A key word is NOT - P(A') = 1 - P(A) OR P(A) + P(A') =1 -1 is the whole sample space
Example 1: You have fair six sided die. Parts: Find 1) P(1) 2) P(1') 3) Prove P(1) + P(1') = 1
1) P(1) The probability of getting a 1 is 1/6. So P(1) = 1/6.
2) P(1') The probability of NOT getting a 1 is 5/6. So P(1')= 5/6.
3) Prove P(1) + P(1') = 1 The probability of getting a 1 plus the probability of Not getting a 1 is 1, the sample space. P(1) + P(1') = (1/6) + (5/6) = 6/6 = 1. So P(1) + P(1') = 1
Example 2: There are 4 colored balls in a bag. A red one, a blue one, a purple one, and a green one. Parts: Find 1) P(red | blue) with Dependence 2) P(red' | blue) with Dependence *Going to do part 1 and 2 quickly since I already explained it before
1) P(red | blue) with Dependence You leave the blue ball out of the bag. The total number of balls in the bag is 3. So P(red | blue) = 1/3
***2) P(red' | blue) with Dependence [What Gavin Was Talking About, but simplified] 2 solutions: A) You leave the blue ball out of the bag. The total number of balls in the bag is 3. So the probability of NOT getting a the red ball given you took out the blue ball is 2/3, P(red' | blue) = 2/3. So P(red' | blue) = 2/3. B) P(red' | blue) = 1 - P(red | blue) = 1 - (1/3) = 2/3. So P(red' | blue) = 2/3.
1:02:46-1:02:48: Only true if events are independent
1:02:54-1:03:03: Gus is WRONG here. The first and every number after that has the same probability if the events were independent, you put the "picked" ball back into the "bag". The "bag" is the whole sample space. What Gus Is describing is dependence not independence. Example 3: There are 4 colored balls in a bag. A red one, a blue one, a purple one, and a green one. Parts: Find 1) P(blue) 2) P(red | blue) with Dependence 3) P( red | blue) with independence 4) Prove that dependent events have different probability 5) Prove that independent events have the same probability
1) P(blue) = 1/4
2) P(red | blue) with Dependence You leave the blue ball out of the bag. The total number of balls in the bag is 3. So P(red | blue) = 1/3
3) P( red | blue) with independence You put back the blue ball in the bag. The total number of balls in the bag is 4. P( red | blue) = P(red) = 1/4. So P(red) = 1/4
4) Prove that dependent events have different probability [Proof That Gus Is Wrong] P(blue) and P(red|blue) [from part 2] are dependent since part 2 said with dependence. P(blue)= 1/4 and P(red|blue) = 1/3. Both P(blue) and P(red) have different probability. So P(blue) and P(red) are dependent events and they have different probability.
5) Prove that independent events have the same probability [Proof That Gus Is Wrong] P(blue) and P(red|blue) [from part 3] are independent since part 3 said with independence. P(blue)= 1/4 and P(red|blue) = P(red) = 1/4. Both P(blue) and P(red) have the same probability. So P(blue) and P(red) are independent events and they both have the same probability.
1:03:04:-1:03:15 numbers ranging from 1 to 1000, total number of digits are 1000. P(123) = 1/1000 The probability is the same for every number. Or in other words every digits is equally likely to get picked. Meaning This is a uniform distribution in a discreet case.
Definition: Uniform Distribution (discreet) is a finite number of equal distant values that are equally likely to be picked -every one of n values,the total number, has equal probability 1/n. -X is Uniform on [1,n], P(X=x) =1/n Within the range: 1<x<n -P(X=x) is the probability of X when it is equal to some number x. A discrete variable is a variable which can only take a countable number of values.
Example: You have a fair 6 headed die. Let X be the number appeared. Parts: Find 1) P(X=1) 2) P(X=2) 3) P(X=x), x is some number
First: All events, numbers, are equally likely so X is uniform on [1,6] with the range from 1<x<6 and it is discrete variable because you can count the numbers from 1 to 6.
1) P(X=1) Probability of getting a 1 is 1/6. So P(X=1) = 1/6
2) P(X=2) Probability of getting a 2 is 1/6. So P(X=2) = 1/6
3) P(X=x), x is some number Probability of getting x is 1/6. So P(X=x) = 1/6
1:03:19-1:03:24: Gavin is right saying out of 50 different numbers because one of the numbers could be 123456. Gavin is saying its unique number because it is a sequential number, but other than that, the probability of all the 50 numbers are the same because the numbers are equally likely to appear in a uniform distribution.
1:03:28-1:03:30: Quiz: Answer Gavin's question: X is the number picked, n = 50, & X is uniform on [1,50]. Parts Find 1) P(X=2| X=1) with dependence 2) P(X=2|X=1) with independence
i just became a member so im not sure were to go to ask questions, but in your podcasts i hear you guys talk about your school lives. you guys have a lot of underage listeners and viewers who might not be the most popular in their communities, i know im not. i was wandering if you guys used to be bullied at school, if those of you that have kids worry about that, or have anything to say about bullying in general? i know its not the most comical subject but i would love to hear you guys talk about that
The thing is that you were right Gavin in the lottery no number can be repeated so the odds of getting a certain outcome is less than the odds of getting any other outcome besides that one.
An excellent podcast, I wish everything was sponsored by Tequila. You were nearly almost half right at the maths, Stats and probabilities was definitely the worst part of my a-level. Also, you've spelled maths wrong.
P.S. NEVER Get Comcast! It's bullshit! I'm in Mexico but I worked at Comcast, and fuck it, I don't really know how AT&T and Sprint work, but the promotions in Comcast most of the time are not respected, installation costs basically the same as the month rental, nonono FUCK IT.
Gavin is completely right. He is thinking of Permutations, or the chance at getting a series of items in specific order. Everyone else is thinking of combinations, or the chance of getting any random series of items.
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Like when I worked at pizza hut doing the cash at the end of the day, counting the register. I know that the probability of getting .00 cents is 1/100, but it was so cool when it happened! It seemed special cause I decided that 00 was a special case and I treated it different than if I had 23 cents after all the money was counted, cause who gives a shit about 23 cents? 00 cents is the only special case in the world to me, so when it happened I was like so rare, cause I only paid attention to that pairing.
Comment
Actually during the math part, everybody was right, INCLUDING GAVIN. Gavin was taking into account Dependence while everybody else were taking into account Independence. Dependence cares what happens in the past, but independence doesn't cares what happens in the past. Or in other words, Events A and B are dependent if the fact that A occurs does affect the probability of B occurring. Events A and B are independent if the fact that A occurs does not affect the probability of B occurring.
Definition:
1) Events A and B are Dependent if P(A and B)= P(A) * P(B|A).
- P(B|A) means the probability of B given A has already happened.
2) Events A and B are Independent if and only if P(A and B)= P(A) * P(B).
- you can think about it as P(B|A) = P(B) because the probability of B is not affected if A occurred, P(B|A), or not.
Example:
There are 5 colored balls in a cup: a red one, a blue one, a yellow one, a green one, and a black one. You want to find the probability of getting the red ball first, then the blue ball second, P(red and blue).
Parts: A) Find P(red and blue) with Dependence B) Find P(red and blue) with Independence.
First:
Regardless of dependence or Independence because nothing happened yet, P(red) = 1/5. This is always the case.
Second: the probability of getting the blue ball will depend if there is Independence or dependence (putting back the red ball in the cup or not).
A) Find P(red and blue) with Dependence
If you want to find the P(blue) with dependence, you have to leave out the red ball because you do care that the red ball was taken out: you care what happen in the past. So because you care what happen to the red ball, you leave the red ball out and now there are 4 balls left in the cup. So the P(blue|red) = 1/4. So if we take the definition of dependence, P(red and blue) = P(red) * P(blue|red) = (1/5) * (1/4) = 1/20. So P(red and blue) = 1/20.
B) Find P(red and blue) with Independence.
If you want to find the P( red and blue) with Independence, you have to put back the red ball back in the cup because you don't care you got the red ball before: you don't care what happens in the past. Because you put the red ball back in the cup, there are 5 balls in the cup. So P(blue) = 1/5. So if we take definition of independence, P(red and blue) = P(red) * P(blue) = (1/5) * (1/5) = 1/25. So P(red and blue) = 1/25.
Summary:
P(red and blue) with dependence = 1/20 and P(red and blue) with independence = 1/25
So yeah, I hope this clarifies things lol
I love Rooster Teeth. <3 :D lol
-J03 Ryd3r
-Gavin Was Right about talking about dependence. Everybody else is getting independence and dependence mixed up.
- P(A) * P(B|A) = P(AB) [ P(A and B) And P(AB) are the same thing], this a variation of conditional probability:
Definition: Conditional probability is P(B|A) = P...
You were nearly almost half right at the maths, Stats and probabilities was definitely the worst part of my a-level. Also, you've spelled maths wrong.