Probability updating question - 99.9999% chance of tails, heads on first flip

You are 99% confident of your results.

What does this mean?

If it's a probability, what gets the other 1%?

Here's the thing: in Bayesian updating, probability mass moves from one hypothesis to another. That means you need another hypothesis. For Bayes' theorem, you need three terms: the prior probability of your hypothesis (99%), the probability of seeing the experimental result assuming your hypothesis is true (0.0001%), and the probability of seeing the experimental result whether or not the hypothesis is true. You need another piece of information, and that will depend on your state of knowledge. For instance, the coin could be perfectly fair, slightly balanced in favor of tails (40:60), slightly balanced in favor of heads (60:40), 90:10 (either direction), etc. You need another hypothesis, or probably a bunch of them.

If you want to make the math simple, you can just consider two possibilities: your hypothesis, and the null hypothesis (the coin is perfectly fair, 50:50). If that's the case:

P(A|X) = P(A)P(X|A) / (P(X|A)P(A) + P(X|N)*P(N))

P(A|X) = (.99)*(.000001) / ((.99)*(.000001) + (.5)*(.01))

which works out to: P(A|X) = 0.000197961... (Nota bene: Check my math. I haven't slept much recently, and I could easily have made any number of mistakes)

Again, this is a really simplistic calculation, because there are so many other plausible hypotheses, and I haven't studied quite enough probability theory to do all the calculations, even if I had more information.

It seems some information is missing, so I'll try reformulating the problem my way:

A jar contains 1 fair coin (equal odds for heads and tails), and 99 unfair coins (99.9% chances of tails, 0.1% of heads). You pull out a coin, flip it, and it comes up tails. What are your expectations for the second flip?

P(fair & heads) = 0.01 * 0.5 = 0.005

P(unfair & heads) = 0.99 * 0.001 ~= 0.001

so P(heads on first throw) = 0.006

so P(fair | heads on first throw) = 5/6

so P(heads on second throw | heads on first throw) = 5/6 0.05 + 1/6 0.001 ~= 0.42

... at least, that's one way of interpreting "you are 99% confident in your results", I consider that the remaining 1% is "your analysis was completely wrong and the coin is just as likely to land on heads or tails". A more realistic situation would be one where your confidence would be distributed among possible coins in coinspace, something like "90% confident of less than 0.01% odds for heads, 99% confident in less than 1% odds for heads, 99.9% confident in less than 10% odds for heads, etc.".

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.

Seems a little confused. If 99% refers to a model of the coin, and the larger number equals the conditional probability of tails within that model, then I think P(model) suddenly drops to less than .02%. That assumes the other 1% goes to a uniform prior and we can treat the chance of heads if (not-model) as 50%. In this example I think I could have told you beforehand the model leaves out too much, because my sources say the outcome depends more on how you throw the coin and you won't ever get 99.9999% from this.

If you want to know the posterior probability of heads or tails look at the other comments.

Have you read this?