Since you will make the same decision both times, the only coherent state is α=1/(p+1).
I’m curious how you arrived at this. Shouldn’t it be α = (1/2)p + (1 - p) = 1 - p/2? (The other implies that you are a thirder in the Sleeping Beauty Problem—didn’t Nick Bostrum have the last word on that one?) The payoff becomes α[p^2+4p(1-p)] + (1-α)[p+4(1-p)] = (1 - p/2)(4p − 3p^2) + (p/2)(4 − 3p) = 6p - (13/2)p^2 + (3/2)p^3, which has a (local) maximum around p = 0.577.
alpha = 1/(p+1) because the driver is at Y p times for every 1 time the driver is at X; so times the driver is at X / (times the driver is at X or Y) = 1 / (p+1).
The problem with pengvado’s calculation isn’t double counting. It purports to give an expected payoff when made at X, which doesn’t count the expected payoff at Y. The problem is that it doesn’t really give an expected payoff. alpha purports to be the probability that you are at X; yet the calculation must be made at X, not at Y (where alpha will clearly be wrong). This means we can’t speak of a “probability of being at X”; alpha simply is 1 if we use this equation and believe it gives us an expected value.
Or look at it this way: Before you introduce alpha into the equation, you can solve it and get the actual optimal value for p. Once you introduce alpha into the equation, you guarantee that the driver will have false beliefs some of the time, because alpha = 1/(p+1), and so the driver can’t have the correct alpha both at X and at Y. You have added a source of error, and will not find the optimal solution.
If you want to find the value of p that leads to the optimal decision you need to look at the impact on expected value of choosing one p or another, not just consider expected value at the end. Currently, it maximises expectations, not value created, with situations where you pass through X and Y being double counted.
I’m a “who’s offering the bet”er on Sleeping Beauty (which Bostrom has said is consistent with, though not identical to, his own model). And in this case I specified bets offered and paid separately at each intersection, which corresponds to the thirder conclusion.
I’m curious how you arrived at this. Shouldn’t it be α = (1/2)p + (1 - p) = 1 - p/2? (The other implies that you are a thirder in the Sleeping Beauty Problem—didn’t Nick Bostrum have the last word on that one?) The payoff becomes α[p^2+4p(1-p)] + (1-α)[p+4(1-p)] = (1 - p/2)(4p − 3p^2) + (p/2)(4 − 3p) = 6p - (13/2)p^2 + (3/2)p^3, which has a (local) maximum around p = 0.577.
The conclusion remains the same, of course.
alpha = 1/(p+1) because the driver is at Y p times for every 1 time the driver is at X; so times the driver is at X / (times the driver is at X or Y) = 1 / (p+1).
The problem with pengvado’s calculation isn’t double counting. It purports to give an expected payoff when made at X, which doesn’t count the expected payoff at Y. The problem is that it doesn’t really give an expected payoff. alpha purports to be the probability that you are at X; yet the calculation must be made at X, not at Y (where alpha will clearly be wrong). This means we can’t speak of a “probability of being at X”; alpha simply is 1 if we use this equation and believe it gives us an expected value.
Or look at it this way: Before you introduce alpha into the equation, you can solve it and get the actual optimal value for p. Once you introduce alpha into the equation, you guarantee that the driver will have false beliefs some of the time, because alpha = 1/(p+1), and so the driver can’t have the correct alpha both at X and at Y. You have added a source of error, and will not find the optimal solution.
If you want to find the value of p that leads to the optimal decision you need to look at the impact on expected value of choosing one p or another, not just consider expected value at the end. Currently, it maximises expectations, not value created, with situations where you pass through X and Y being double counted.
I’m a “who’s offering the bet”er on Sleeping Beauty (which Bostrom has said is consistent with, though not identical to, his own model). And in this case I specified bets offered and paid separately at each intersection, which corresponds to the thirder conclusion.