Richard, obviously if F does not imply S due to other dangers, then one must use method 2:
P(W|F,S) = P(F|W,S)P(W|S)/P(F|S)
Let’s do the math.
A comet is going to annihilate us with a probability of (1-x) (outside view) if the LHC would not destroy the Earth, but if the LHC would destroy the Earth, the probability is (1-y) (I put this change in so that it would actually have an effect on the final probability)
The LHC has an outside-view probability of failure of z, whether or not W is true
The universe has a prior probabilty w of being such that the LHC if it does not fail will annihilate us.
I leave it as an exercise to the reader to show that there is no change in P(W|F,S) if the chance of the comet hitting depends on whether or not the LHC fails (only the relative probability of outcomes given failure matters).
Really though Richard, you should not have assumed in the first place that I was not capable of doing the math. In the future, don’t expect me to bother with a demonstration.
Allan: you’re right, I should have thought that through more carefully. It doesn’t make your interpretation correct though...
I have really already spent much more time here today than I should have...
Richard, obviously if F does not imply S due to other dangers, then one must use method 2:
P(W|F,S) = P(F|W,S)P(W|S)/P(F|S)
Let’s do the math.
A comet is going to annihilate us with a probability of (1-x) (outside view) if the LHC would not destroy the Earth, but if the LHC would destroy the Earth, the probability is (1-y) (I put this change in so that it would actually have an effect on the final probability)
The LHC has an outside-view probability of failure of z, whether or not W is true
The universe has a prior probabilty w of being such that the LHC if it does not fail will annihilate us.
Then:
P(F|W,S) = 1
P(F|S) = (ywz+x(1-w)z)/(ywz+x(1-w)z+x(1-w)(1-z))
P(W|S) = (ywz)/(ywz+x(1-w)+x(1-w)(1-z))
so, P(W|F,S) = ywz/(ywz+x(1-w)z) = yw(yw+x(1-w))
I leave it as an exercise to the reader to show that there is no change in P(W|F,S) if the chance of the comet hitting depends on whether or not the LHC fails (only the relative probability of outcomes given failure matters).
Really though Richard, you should not have assumed in the first place that I was not capable of doing the math. In the future, don’t expect me to bother with a demonstration.
Allan: you’re right, I should have thought that through more carefully. It doesn’t make your interpretation correct though...
I have really already spent much more time here today than I should have...