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Manfred comments on Bayesian probability as an approximate theory of uncertainty? - Less Wrong Discussion
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Comments (35)
Let's figure it out! Suppose the payoffs are as used here.
Now, the correct answer is to go straight with p=2/3. How is that figured out? Because it maximizes the expected value given by p · (1-p) · 4 + p · p.
Since the probabilities are dependent on your strategy, we will leave them as equations - if you go straight with probability p, then P(first crossing) is 1/(1+p), and P(second crossing) is p/(1+p). So the question is, what is the correct mixed strategy to take, in terms of these probabilities and the utilities?
Well, there's a rather dumb way: you just extract p from the probabilities and plug it into the expected value.
That is, you look at the ratio P(second crossing)/P(first crossing), and then choose a strategy such that that ratio maximizes the expected utility equation. That's the function.
I see. For any math problem where I can figure out the right answer, I can write a function that receives the phase of the moon as argument and returns the right answer. Okay, I guess you're right, I do want the function to look like expected utility maximization based on the supplied probabilities, rather than some random formula.
Hardly random - it makes perfect sense that the ratio of probabilities is equal to the parameter of the strategy, and so any maximization of the parameter can be rewritten as a maximization of the ratio of probabilities.