Comment author:Cyan2
12 March 2008 05:46:32PM
2 points
[-]

If I'm being asked to accept or reject a number meant to correspond to the calculated or measured likelihood of heads coming up, and I trust the information about it being biased, then the only correct move is to reject the 0.5 probability.

Alas, no. Here's the deal: implicit in all the coin toss toy problems is the idea that the observations may be modeled as exchangeable. It really really helps to have a grasp on what the math looks like when we assume exchangeability.

In models where (infinite) exchangeability is assumed, the concept of long-run frequency can be sensibly defined. (Long-run frequency may or may not be a cogent concept in models without exchangeability.) The probability of heads in any one toss is the expectation of a probability density function (pdf) which encodes our knowledge about the long run frequency. (Roughly. There are some technical conditions for the existence of a pdf that I'm ignoring.)

Conrad, your idea that 0.5 is not an allowable probability is almost correct. In fact, the correct expression of this idea is that the pdf of the long-run frequency must be equal to zero at 0.5. But! -- its values in the neighborhood of 0.5 are not constrained, so the pdf may have a removable singularity.

Suppose our information about bias in favour of heads is equivalent to our information about bias in favour of tail. Our pdf for the long-run frequency will be symmetrical about 0.5 and its expectation (which is the probability in any single toss) must also be 0.5. It is quite possible for an expectation to take a value which has zero probability density. We can refuse to believe that the long-run frequency will converge to exactly 0.5 while simultaneously holding a probability of 0.5 for any specific single toss in isolation.

## Comments (190)

OldAlas, no. Here's the deal: implicit in all the coin toss toy problems is the idea that the observations may be modeled as exchangeable. It really really helps to have a grasp on what the math looks like when we assume exchangeability.

In models where (infinite) exchangeability is assumed, the concept of long-run frequency can be sensibly defined. (Long-run frequency may or may not be a cogent concept in models without exchangeability.) The probability of heads in any one toss is the expectation of a probability density function (pdf) which encodes our knowledge about the long run frequency. (Roughly. There are some technical conditions for the existence of a pdf that I'm ignoring.)

Conrad, your idea that 0.5 is not an allowable probability is almost correct. In fact, the correct expression of this idea is that the pdf of the long-run frequency must be equal to zero at 0.5. But! -- its values in the neighborhood of 0.5 are not constrained, so the pdf may have a removable singularity.

Suppose our information about bias in favour of heads is equivalent to our information about bias in favour of tail. Our pdf for the long-run frequency will be symmetrical about 0.5 and its expectation (which is the probability in any single toss) must also be 0.5. It is quite possible for an expectation to take a value which has zero probability density. We can refuse to believe that the long-run frequency will converge to exactly 0.5 while simultaneously holding a probability of 0.5 for any specific single toss in isolation.