Since the only thing that matters (to your expected value) is the overall probability of heads v. tails, the problem is a simple one of parameter estimation. You start with a distribution over p (determined by this thing called the principle of maximum entropy, probably just uniform in this case) and then update it with new evidence (for example, P(p=0.01) drops almost to 0 as soon as you see a head). After 100 heads you get a very spiky function at 1. And yes this is optimal.
For any finite amount of data you won’t perfectly break even using a bayesian method, but it’s better than all the alternatives, as long as you don’t leave out some data.
For any finite amount of data you won’t perfectly break even using a bayesian method, but it’s better than all the alternatives, as long as you don’t leave out some data.
What!? Are you saying that you can predict in advance that you’ll lose money? Surely that can’t happen, because you get to choose how much you want to pay, so you can always just pay less. No?
For simplicity, let’s imagine betting on a single coin with P(heads) = 0.9. You say “how much will you pay to win $1 if it lands heads?” and I say “50 cents,” because at the start I am ignorant. You flip it and it lands heads. I just made 40 cents relative to the equilibrium value.
So it’s not predictably losing money. It’s predictably being wrong in an unpredictable direction.
I read this to say that you can’t calculate a value that is guaranteed to break even in the long term, because there isn’t enough information to do this. (which I tend to agree with)
Perhaps he means that you break even once opportunity costs are taken into account, that is you won’t win as much money as you theoretically could have won.
Yup. Except maybe with a little more confidence that Bayes’ rule applies here in the specific way of altering the probability distribution over p at each point.
Since the only thing that matters (to your expected value) is the overall probability of heads v. tails, the problem is a simple one of parameter estimation. You start with a distribution over p (determined by this thing called the principle of maximum entropy, probably just uniform in this case) and then update it with new evidence (for example, P(p=0.01) drops almost to 0 as soon as you see a head). After 100 heads you get a very spiky function at 1. And yes this is optimal.
For any finite amount of data you won’t perfectly break even using a bayesian method, but it’s better than all the alternatives, as long as you don’t leave out some data.
What!? Are you saying that you can predict in advance that you’ll lose money? Surely that can’t happen, because you get to choose how much you want to pay, so you can always just pay less. No?
For simplicity, let’s imagine betting on a single coin with P(heads) = 0.9. You say “how much will you pay to win $1 if it lands heads?” and I say “50 cents,” because at the start I am ignorant. You flip it and it lands heads. I just made 40 cents relative to the equilibrium value.
So it’s not predictably losing money. It’s predictably being wrong in an unpredictable direction.
I read this to say that you can’t calculate a value that is guaranteed to break even in the long term, because there isn’t enough information to do this. (which I tend to agree with)
Perhaps he means that you break even once opportunity costs are taken into account, that is you won’t win as much money as you theoretically could have won.
That sounds pretty much the same as what I said above.
Yup. Except maybe with a little more confidence that Bayes’ rule applies here in the specific way of altering the probability distribution over p at each point.