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).
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).