My take at it is basically this: average over all possible distributions until you have further evidence. (Preferably, let other people play the game first to gather the evidence at no cost to myself.)
If someone tells me a coin has an unknown binomial distribution, and we really genuinely don’t know anything about this distribution (not even the distribution of possible distributions), I take the set of all possible distributions and assume they are all equally likely. Since they are symmetric, the average is a 50:50 fair coin.
In your example, you throw not just one coin, but a different one each time. Differently put, the sequence of coins is a random variable whose distribution we don’t know. I therefore begin by assuming it is the average over all possible distributions, and within that average distribution, the distribution of first coins is symmetric, so its average is 50:50, so for the first coin toss I’ll assume that it’s a 50:50.
Say the first coin toss now comes out tails. This provides me with a slight piece of evidence that your set of unfair coins may be biased towards containing more tail-preferring coins than head-preferring coins. Therefore, I will assume that the second coin is more likely to be a tail-preferring coin than a head-preferring one. I’d have to do the maths to find out exactly how much my ante would change.
With each new coin toss, I learn more about the likelihood of each distribution of coins within the set of all possible distributions.
I suspect that the level of indirection can actually be safely removed and we can regard the whole thing as a single random binary digit generator with a single distribution, about which we initially don’t know anything and gradually build up evidence. I further suspect that if the distribution of coins is uniformly distributed, and therefore symmetric, the sequence of coin tosses is equivalent to a single, fair coin. Once again, someone would have to do the maths to prove whether this is actually equivalent.
Preferably, let other people play the game first to gather the evidence at no cost to myself.
For the record, this is not permitted.
My take at it is basically this: average over all possible distributions
It’s easy to say this but I don’t think this works when you start doing the maths to get actual numbers out. Additionally, if you really take ALL possible distributions then you’re already in trouble, because some of them are pretty weird—e.g. the Cauchy distribution doesn’t have a mean or a variance.
distribution about which we initially don’t know anything and gradually build up evidence
I’d love to know if there are established formal approaches to this. The only parts of statistics that I’m familiar with assume known distributions and work from there. Anyone?
Odd. The printed book has another page and a half for that chapter, including the solution to Stage 4. (No surprises in the solution—same as stage 3 except you start with 40 fewer Green widgets.)
My take at it is basically this: average over all possible distributions until you have further evidence. (Preferably, let other people play the game first to gather the evidence at no cost to myself.)
If someone tells me a coin has an unknown binomial distribution, and we really genuinely don’t know anything about this distribution (not even the distribution of possible distributions), I take the set of all possible distributions and assume they are all equally likely. Since they are symmetric, the average is a 50:50 fair coin.
In your example, you throw not just one coin, but a different one each time. Differently put, the sequence of coins is a random variable whose distribution we don’t know. I therefore begin by assuming it is the average over all possible distributions, and within that average distribution, the distribution of first coins is symmetric, so its average is 50:50, so for the first coin toss I’ll assume that it’s a 50:50.
Say the first coin toss now comes out tails. This provides me with a slight piece of evidence that your set of unfair coins may be biased towards containing more tail-preferring coins than head-preferring coins. Therefore, I will assume that the second coin is more likely to be a tail-preferring coin than a head-preferring one. I’d have to do the maths to find out exactly how much my ante would change.
With each new coin toss, I learn more about the likelihood of each distribution of coins within the set of all possible distributions.
I suspect that the level of indirection can actually be safely removed and we can regard the whole thing as a single random binary digit generator with a single distribution, about which we initially don’t know anything and gradually build up evidence. I further suspect that if the distribution of coins is uniformly distributed, and therefore symmetric, the sequence of coin tosses is equivalent to a single, fair coin. Once again, someone would have to do the maths to prove whether this is actually equivalent.
For the record, this is not permitted.
It’s easy to say this but I don’t think this works when you start doing the maths to get actual numbers out. Additionally, if you really take ALL possible distributions then you’re already in trouble, because some of them are pretty weird—e.g. the Cauchy distribution doesn’t have a mean or a variance.
I’d love to know if there are established formal approaches to this. The only parts of statistics that I’m familiar with assume known distributions and work from there. Anyone?
If I’m not mistaken, the Cauchy distribution wouldn’t be included because it’s not supported on a bounded interval.
You should probably look at Jaynes’s book “Probability Theory: the Language of Science”. In particular, I think that the discussion there dealing with the Widget Problem and with Laplace’s Rule of Succession may be relevant to your question.
And just as it gets really interesting, that chapter ends. There is no solution provided for stage 4 :/
Odd. The printed book has another page and a half for that chapter, including the solution to Stage 4. (No surprises in the solution—same as stage 3 except you start with 40 fewer Green widgets.)