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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)
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 *)