If you are in a green room and someone asks you if you will bet that a head was flipped, you should say “yes”.
However, if that same person asks you if they should bet that heads was flipped, you should answer no if you ascertain that they asked you on the precondition that you were in a green room.
the probability of heads | you are in green room = 90%
the probability of you betting on heads | you are green room = 100% = no information about the coin flip
Your first claim needs qualifications: You should only bet if you’re being drawn randomly from everyone. If it is known that one random person in a green room will be asked to bet, then if you wake up in a green room and are asked to bet you should refuse.
P(Heads | you are in a green room) = 0.9
P(Being asked | Heads and Green) = 1⁄18, P(Being asked | Tails and Green) = 1⁄2
Hence P(Heads | you are asked in a green room) = 0.5
Of course the OP doesn’t choose a random individual to ask, or even a random individual in a green room. The OP asks all people in green rooms in this world.
If there is confusion about when your decision algorithm “chooses”, then TDT/UDT can try to make the latter two cases equivalent, by thinking about the “other choices I force”. Of course the fact that this asserts some variety of choice for a special individual and not for others, when the situation is symmetric, suggests something is being missed.
What is being missed, to my mind, is a distinction between the distribution of (random individuals | data is observed), and the distribution of (random worlds | data is observed).
In the OP, the latter distribution isn’t altered by the update as the observed data occurs somewhere with probability 1 in both cases. The former is because it cares about the number of copies in the two cases.
Yes, exactly.
If you are in a green room and someone asks you if you will bet that a head was flipped, you should say “yes”.
However, if that same person asks you if they should bet that heads was flipped, you should answer no if you ascertain that they asked you on the precondition that you were in a green room.
the probability of heads | you are in green room = 90%
the probability of you betting on heads | you are green room = 100% = no information about the coin flip
Your first claim needs qualifications: You should only bet if you’re being drawn randomly from everyone. If it is known that one random person in a green room will be asked to bet, then if you wake up in a green room and are asked to bet you should refuse.
P(Heads | you are in a green room) = 0.9 P(Being asked | Heads and Green) = 1⁄18, P(Being asked | Tails and Green) = 1⁄2 Hence P(Heads | you are asked in a green room) = 0.5
Of course the OP doesn’t choose a random individual to ask, or even a random individual in a green room. The OP asks all people in green rooms in this world.
If there is confusion about when your decision algorithm “chooses”, then TDT/UDT can try to make the latter two cases equivalent, by thinking about the “other choices I force”. Of course the fact that this asserts some variety of choice for a special individual and not for others, when the situation is symmetric, suggests something is being missed.
What is being missed, to my mind, is a distinction between the distribution of (random individuals | data is observed), and the distribution of (random worlds | data is observed).
In the OP, the latter distribution isn’t altered by the update as the observed data occurs somewhere with probability 1 in both cases. The former is because it cares about the number of copies in the two cases.