They are not equivalent. If one green room copy is chosen at random, then the game will be played exactly once whether the coin flip resulted in heads or tails. But if every green room copy plays, the the game will be played 18 times if the coin came up heads and 2 times if the coin came up tails.
However, being chosen for the game (since the agent knows that in both cases exactly one copy will be chosen) also carries information the same way as being in the green room. Therefore, (by the same logic) it would imply an additional anthropic update: “Although I am in a green groom, the fact that I am chosen to play the game makes it much less probable that the coin is head.” So (by calculating the correct chances), he can deduce:
I am in a green room + I am chosen ⇒ P(head)=0.5
OTOH:
I am in a green room (not knowing whether chosen) ⇒ P(head)=0.9
[EDIT]: I just noted that you already argued the same way, I have plainly overlooked it.
They are not equivalent. If one green room copy is chosen at random, then the game will be played exactly once whether the coin flip resulted in heads or tails. But if every green room copy plays, the the game will be played 18 times if the coin came up heads and 2 times if the coin came up tails.
Good point.
However, being chosen for the game (since the agent knows that in both cases exactly one copy will be chosen) also carries information the same way as being in the green room. Therefore, (by the same logic) it would imply an additional anthropic update: “Although I am in a green groom, the fact that I am chosen to play the game makes it much less probable that the coin is head.” So (by calculating the correct chances), he can deduce:
I am in a green room + I am chosen ⇒ P(head)=0.5
OTOH:
I am in a green room (not knowing whether chosen) ⇒ P(head)=0.9
[EDIT]: I just noted that you already argued the same way, I have plainly overlooked it.