I have much to say about myself, but I don’t consider it worth most people’s time, so I’ll spare most of it.
I am currently going through the sequences and had no intention of commenting on any post or writing anything until I had finished all of them. I have to admit, though, it’s really quite difficult to “stay on” the sequences. I have a hundred-something tabs of lesswrong open right now, and it has come to a point where I am understanding them all pretty fully.
EDIT: Issue is under control. All is going well.
I would have preferred a better welcome for myself, but this is acceptable!
Nonetheless, I offer an entirely different way of explaining it. The explanation goes with the visual: http://www.freeimagehosting.net/4dd80(color coding: A is red and B is blue, so when A and B are the case this is represented in purple, when neither are it is black).
The four possibilities are: A&B, -A&B, A&-B, and -A&-B. They are represented in the grid as 1, 2, 3, and 4. Added together they will sum to 100%.
What’s the probability that out of 1, 2, 3, and 4, scenario 1 (A&B) is the case? (1) / (1+2+3+4)
Whatever it is, it’s the same as the probability that out of 1, 2, 3, and 4, either 1 or 2 is the case, and it is 1 that is the case out of 1 and 2. ((1) / (1+2)) * ((1+2) / (1+2+3+4))
Likewise, it’s the same as the probability that out of 1, 2, 3, and 4, either 1 or 3 is the case, and it is 1 that is the case out of 1 and 3. ((1) / (1+3)) * ((1+3) / (1+2+3+4))
The numerators and the denominators cancel in the two above cases to amount to (1) / (1+2+3+4), just like in the first of the three ways of phrasing the probability of 1 out of 1, 2, 3, and 4. We see that they are equal and set them against each other. We won’t actually use the first part of the equation, just the last two.
We replace ((1) / (1+2)) with P(A|B) (probability that A is the case, given that B is the case), ((1+2) / (1+2+3+4)) with P(B) (probability that B is the case), ((1) / (1+3)) with P(B|A) (probability that B is the case, given that A is the case), and ((1+3) / (1+2+3+4)) with P(A) (probability that A is the case). (1) / (1+2+3+4) is P(A&B) (probability that A and B are the case), but this won’t be used in the usual way of writing the rule, which isolates one term by dividing both sides of the equation by P(B).
We divide by P(B) so that when we know the three terms on the right, we can solve for the one on the left:
P(A|B)=P(B|A)*P(A)/P(B)
So, suppose we know that P(A)=40%, P(B|A)=30%, P(B|-A)=10%, and we want to find P(A|B).
Looking at the grid shows us at a glance that P(A)=(1+3), P(B|A)=(1)/(1+3), P(B|-A)=(2)/(2+4), and P(A|B)=(1)/(1+2). Remember Bayes’ Rule expresses P(A|B) as (1)/(1+3)*(1+3)/(1+2), the (1+3)’s cancel each other out, that shows why P(B|A) is multiplied by P(A). We always know that (1+2+3+4)=100%
ahoy.
I have much to say about myself, but I don’t consider it worth most people’s time, so I’ll spare most of it.
I am currently going through the sequences and had no intention of commenting on any post or writing anything until I had finished all of them. I have to admit, though, it’s really quite difficult to “stay on” the sequences. I have a hundred-something tabs of lesswrong open right now, and it has come to a point where I am understanding them all pretty fully.
EDIT: Issue is under control. All is going well.
I would have preferred a better welcome for myself, but this is acceptable!
This is unfortunately not me yet.
Nonetheless, I offer an entirely different way of explaining it. The explanation goes with the visual: http://www.freeimagehosting.net/4dd80 (color coding: A is red and B is blue, so when A and B are the case this is represented in purple, when neither are it is black).
The four possibilities are: A&B, -A&B, A&-B, and -A&-B. They are represented in the grid as 1, 2, 3, and 4. Added together they will sum to 100%.
What’s the probability that out of 1, 2, 3, and 4, scenario 1 (A&B) is the case? (1) / (1+2+3+4)
Whatever it is, it’s the same as the probability that out of 1, 2, 3, and 4, either 1 or 2 is the case, and it is 1 that is the case out of 1 and 2. ((1) / (1+2)) * ((1+2) / (1+2+3+4))
Likewise, it’s the same as the probability that out of 1, 2, 3, and 4, either 1 or 3 is the case, and it is 1 that is the case out of 1 and 3. ((1) / (1+3)) * ((1+3) / (1+2+3+4))
The numerators and the denominators cancel in the two above cases to amount to (1) / (1+2+3+4), just like in the first of the three ways of phrasing the probability of 1 out of 1, 2, 3, and 4. We see that they are equal and set them against each other. We won’t actually use the first part of the equation, just the last two.
We replace ((1) / (1+2)) with P(A|B) (probability that A is the case, given that B is the case), ((1+2) / (1+2+3+4)) with P(B) (probability that B is the case), ((1) / (1+3)) with P(B|A) (probability that B is the case, given that A is the case), and ((1+3) / (1+2+3+4)) with P(A) (probability that A is the case). (1) / (1+2+3+4) is P(A&B) (probability that A and B are the case), but this won’t be used in the usual way of writing the rule, which isolates one term by dividing both sides of the equation by P(B).
We divide by P(B) so that when we know the three terms on the right, we can solve for the one on the left:
P(A|B)=P(B|A)*P(A)/P(B)
So, suppose we know that P(A)=40%, P(B|A)=30%, P(B|-A)=10%, and we want to find P(A|B).
Looking at the grid shows us at a glance that P(A)=(1+3), P(B|A)=(1)/(1+3), P(B|-A)=(2)/(2+4), and P(A|B)=(1)/(1+2). Remember Bayes’ Rule expresses P(A|B) as (1)/(1+3)*(1+3)/(1+2), the (1+3)’s cancel each other out, that shows why P(B|A) is multiplied by P(A). We always know that (1+2+3+4)=100%
So:
Intuitively:
Or the long way:
(I can’t help you in any useful way, but I can say “Welcome!”)
Welcome!