Maybe I’m just really tired, but I seem to have grown a blind spot hiding a logical step that must be present in the argument given for SIA. It doesn’t seem to be arguing for the SIA at all, just for the right way of detecting a blue door independent of the number of observers.
Consider this variation: there are 150 rooms, 149 of them blue and 1 red. In the blue rooms, 49 cats and 99 human clones are created; in the red room, a human clone is created. The experiment then proceeds in the usual way (flipping the coin and killing inhabitants of rooms of a certain color).
The humans will still give a .99 probability of being behind a blue door, and 99 out of 100 equally-probable potential humans will be right. Therefore you are more likely to inhabit a universe shared by an equal number of humans and cats, than a universe containing only humans (the Feline Indication Argument).
If you are told that you are in that situation, then you would assign a probability of 50⁄51 of being behind a blue door, and a 1⁄51 probability of being behind a red door, because you would not assign any probability to the possibility of being one of the cats. So you will not give a probability of .99 in this case.
Maybe I’m just really tired, but I seem to have grown a blind spot hiding a logical step that must be present in the argument given for SIA. It doesn’t seem to be arguing for the SIA at all, just for the right way of detecting a blue door independent of the number of observers.
Consider this variation: there are 150 rooms, 149 of them blue and 1 red. In the blue rooms, 49 cats and 99 human clones are created; in the red room, a human clone is created. The experiment then proceeds in the usual way (flipping the coin and killing inhabitants of rooms of a certain color).
The humans will still give a .99 probability of being behind a blue door, and 99 out of 100 equally-probable potential humans will be right. Therefore you are more likely to inhabit a universe shared by an equal number of humans and cats, than a universe containing only humans (the Feline Indication Argument).
If you are told that you are in that situation, then you would assign a probability of 50⁄51 of being behind a blue door, and a 1⁄51 probability of being behind a red door, because you would not assign any probability to the possibility of being one of the cats. So you will not give a probability of .99 in this case.
Fixed, thanks. (I didn’t notice at first that I quoted the .99 number.)