That’s possible, but I am not sure how I am fighting it in this case. Leave Omega in place- why do we assume equal probability of omega guessing incorrectly or correctly, when the hypothetical states he has guessed correctly each previous time? If we are not assuming that, why does cdc treat each option as equal, and then proceed to open two boxes?
I realize that decision theory is about a general approach to solving problems- my question is, why are we not including the probability based on past performance in our general approach to solving problems, or if we are, why are we not doing so in this case?
That’s possible, but I am not sure how I am fighting it in this case. Leave Omega in place- why do we assume equal probability of omega guessing incorrectly or correctly, when the hypothetical states he has guessed correctly each previous time? If we are not assuming that, why does cdc treat each option as equal, and then proceed to open two boxes?
I realize that decision theory is about a general approach to solving problems- my question is, why are we not including the probability based on past performance in our general approach to solving problems, or if we are, why are we not doing so in this case?