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DanielLC comments on Probability updating question - 99.9999% chance of tails, heads on first flip - Less Wrong Discussion
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Comments (16)
Do you mean that you're 99% confident in your reasoning that it comes up tails 99.9999% of the time? If so, you'd be much less than 99.9999% sure of heads in the first place.
You generally use beta distribution for coinflips. One beta distribution that would get your certainty is alpha = 999999, beta = 1. Landing on heads would get you a posterior of alpha = 999999, beta = 2, which would give you a certainty of about 99.9998% of landing on tails.
My problem is that your confidence isn't well specified. If you could give me a standard deviation, that would work better. Also, with something like this, a beta distribution isn't actually a very good prior. The most likely reason for it to land on heads is that you messed up, and the probability is more like 99.9%, which would be crazy unlikely under the prior I gave.
You can be 99.9999% sure of heads - and 99% confident of that - if you memorised your confidence - but then subsequently could not remember for sure if there were six "9"s - or maybe seven.
If there was a 99% chance that you remember correctly, and a 1% chance that there was an extra nine, you'd be slightly more than 99.9999% confident if heads.
I think the idea is to have a point mass at 0.999999 containing 0.99 of the prior probability. My intuition is that would behave more like a Beta(0,1) than a Beta(1,999999).
Beta(0,1) is an improper prior. Do you mean Beta(1,1), the uniform prior?
In that case, it's a silly prior. You can't be certain that it's exactly that probability.
If that's what you're using, you'd have a 50% chance of getting heads if you were wrong, and a 0.0001% chance if you were right, so:
P(pi = 0.000001) = 0.99
P(H) = P(H | pi = 0.000001)P(pi = 0.000001) + P(H | pi != 0.000001)P(pi != 0.000001)
= 0.0000010.99 + 0.50.01
= 0.00500099
~= 0.005
P(pi = 0.000001 | H) = P(H | pi = 0.000001)*P(pi = 0.000001)/P(H)
= 0.000001*0.99/0.005
= 0.000198
So there's a 0.02% chance that you were right about there being a 99.9999% chance of heads. Essentially, you can ignore that. You now have a 66.7% chance that the coin will land on heads, with much the same distribution as if you started with a uniform prior.
If you meant Beta(1,2) the answer is similar.
Edit: How do you make an asterisk show up, instead of italicizing?
I usually copy the ascii middle dot ( · ) from somewhere, or just make the multiplication implicit. It is kind of a pain.
I had in mind something like a mixture prior with 0.99*Beta(0,1) + 0.01*Beta(0.5,0.5). Yes, the Beta(0,1) component makes the prior improper. Once heads is observed the Beta(0,1) component is updated to Beta(1,1) and its mixture weight is also updated and basically becomes negligible such that the component effectively drops out of the mixture.
I'm treating the behavior of the above distribution as a guide for my intuition as to what will happen to the mixture weight when there is no Beta(0,1) but rather a point mass at 0.999999.
To get two asterisks in a line without italics, do \* this \*.
Use a backslash to escape it; for example: * (which is \ followed by the *)