Hm, thanks for making me really think about it, and not letting me slide by without doing calculation. It seems to me, given my preferences, about which I am not logically omniscient, and given my structural uncertainty around these issues, of which there is much, I think that my 50 percent confidence interval is between .00001%, 1 in 10 million, to .01%, 1 in ten thousand.
Oh, should they? I’m the first to admit that I sorely lack in knowledge of probability theory. I thought it was better to give a distribution here to indicate my level of uncertainty as well as my best guess (precision as well as accuracy).
Contra Roko, it’s OK for a Bayesian to talk in terms of a probability distribution on the probability of an event. (However, Roko is right that in decision problems, the mean value of that probability distribution is quite an important thing.)
This would be true if you were estimating the value of a real-world parameter like the length of a rod. However, for a probability, you just give a single number, which is representative of the odds you would bet at. If you have several conflicting intuitions about what that number should be, form a weighted average of them, weighted by how much you trust each intuition or method for getting the number.
For small probabilities, the weighted average calculation is dominated by the high-probability possibilities—if your 50% confidence interval was up to 1 in 10,000, then 25% of the probability probability mass is to the right of 1 in 10,000, so you can’t say anything less than (0.75)x0 + (0.25)x1 in 10000 = 1 in 40,000.
I wasn’t using a normal distribution in my original formulation, though: the mean of the picture in my head was around 1 in a million with a longer tail to the right (towards 100%) and a shorter tail to the left (towards 0%) (on a log scale?). It could be that I was doing something stupid by making one tail longer than the other?
It would only be suspicious if your resulting probability were a sum of very many independent, similarly probable alternatives (such sums do look normal even if the individual alternatives aren’t).
It seems to me, given my preferences, about which I am not logically omniscient, [...]
I’d say your preference can’t possibly influence the probability of this event. To clear up the air, can you explain how does taking into account your preference influence the estimate? Better, how does the estimate break up on the different defeaters (events making the positive outcome impossible)?
Sorry, I should have been more clear: my preferences influence the possible interpretations of the word ‘save’. I wouldn’t consider surviving indefinitely but without my preferences being systematically fulfilled ‘saved’, for instance; more like damned.
Hm, thanks for making me really think about it, and not letting me slide by without doing calculation. It seems to me, given my preferences, about which I am not logically omniscient, and given my structural uncertainty around these issues, of which there is much, I think that my 50 percent confidence interval is between .00001%, 1 in 10 million, to .01%, 1 in ten thousand.
shouldn’t probabilities just be numbers?
i.e. just integrate over the probability distribution of what you think the probability is.
Oh, should they? I’m the first to admit that I sorely lack in knowledge of probability theory. I thought it was better to give a distribution here to indicate my level of uncertainty as well as my best guess (precision as well as accuracy).
Contra Roko, it’s OK for a Bayesian to talk in terms of a probability distribution on the probability of an event. (However, Roko is right that in decision problems, the mean value of that probability distribution is quite an important thing.)
This would be true if you were estimating the value of a real-world parameter like the length of a rod. However, for a probability, you just give a single number, which is representative of the odds you would bet at. If you have several conflicting intuitions about what that number should be, form a weighted average of them, weighted by how much you trust each intuition or method for getting the number.
Ahhh, makes sense, thanks. In that case I’d put my best guess at around 1 in a million.
For small probabilities, the weighted average calculation is dominated by the high-probability possibilities—if your 50% confidence interval was up to 1 in 10,000, then 25% of the probability probability mass is to the right of 1 in 10,000, so you can’t say anything less than (0.75)x0 + (0.25)x1 in 10000 = 1 in 40,000.
I wasn’t using a normal distribution in my original formulation, though: the mean of the picture in my head was around 1 in a million with a longer tail to the right (towards 100%) and a shorter tail to the left (towards 0%) (on a log scale?). It could be that I was doing something stupid by making one tail longer than the other?
It would only be suspicious if your resulting probability were a sum of very many independent, similarly probable alternatives (such sums do look normal even if the individual alternatives aren’t).
I’d say your preference can’t possibly influence the probability of this event. To clear up the air, can you explain how does taking into account your preference influence the estimate? Better, how does the estimate break up on the different defeaters (events making the positive outcome impossible)?
Sorry, I should have been more clear: my preferences influence the possible interpretations of the word ‘save’. I wouldn’t consider surviving indefinitely but without my preferences being systematically fulfilled ‘saved’, for instance; more like damned.