I believe that the argument for why you are probably on a type (3) island is analogous to that of why the great filter probably lies in our future because both come about from updating “on your own existence by weighting possible worlds as more likely the more observers they contain.”
Katja wrote:
For instance if you were born of an experiment where the flip of a fair coin determined whether one (tails) or two (heads) people were created, and all you know is that and that you exist, SIA says heads was twice as likely as tails.
The coin example Katja does use this reasoning. The islands example doesn’t have to.
Take
Islands that are easy to reach and are not prone to disaster.
Islands that are hard to reach and are not prone to disaster.
Islands that are easy to reach and are prone to disaster.
Islands that are hard to reach and are prone to disaster.
Assign prior odds of 1⁄3 to reaching each of the two “easy to reach” cases, and 1⁄6 to reaching the two “hard to reach” cases. Note I’m putting “reaching an island” in my priors, and observations about how many other people are on that island into my posteriors. You should get the same answer if you assign equal priors and mash everything into your posteriors.
Suppose “prone to disaster” means “probability 1⁄2 that there was a disaster on this island just before we arrived”, and “not prone to disaster” means “there was never a disaster on this island”.
Suppose “easy to reach” means “0 to 7 other tribes have reached this island, with each number having equal probability.”
Suppose “hard to reach” means “0 to 1 other tribes have reached this island, with each number having equal probability.”
Assume only disasters eliminate tribes.
Observation B is: No other tribes on island.
Let A denote one of the 4 possible islands. P(A|B) = P(A,B) / P(B) = P(B|A)P(A)/P(B)
We’ll compute just P(B|A)P(A) for each A, as P(B) is the same for all of them.
P(1|B) ~ P(B|1)P(1) = 1⁄8 x 1⁄3 = 1⁄24
P(2|B) ~ P(B|2)P(2) = 1⁄2 x 1⁄6 = 1⁄12
P(3|B) ~ P(B|3)P(3) = [1/2 + 1⁄2 x 1⁄8] x 1⁄3 = 9⁄48 = 3⁄16
P(4|B) ~ P(B|4)P(4) = [1/2 x 1 + 1⁄2 x 1⁄2] x 1⁄6 = 3⁄4 x 1⁄6 = 1⁄8
This shows that island type 3 is the most likely; yet we never assumed that worlds with more people are more likely.
OP wrote:
Katja wrote:
The coin example Katja does use this reasoning. The islands example doesn’t have to.
Take
Islands that are easy to reach and are not prone to disaster.
Islands that are hard to reach and are not prone to disaster.
Islands that are easy to reach and are prone to disaster.
Islands that are hard to reach and are prone to disaster.
Assign prior odds of 1⁄3 to reaching each of the two “easy to reach” cases, and 1⁄6 to reaching the two “hard to reach” cases. Note I’m putting “reaching an island” in my priors, and observations about how many other people are on that island into my posteriors. You should get the same answer if you assign equal priors and mash everything into your posteriors.
Suppose “prone to disaster” means “probability 1⁄2 that there was a disaster on this island just before we arrived”, and “not prone to disaster” means “there was never a disaster on this island”.
Suppose “easy to reach” means “0 to 7 other tribes have reached this island, with each number having equal probability.” Suppose “hard to reach” means “0 to 1 other tribes have reached this island, with each number having equal probability.”
Assume only disasters eliminate tribes. Observation B is: No other tribes on island.
Let A denote one of the 4 possible islands. P(A|B) = P(A,B) / P(B) = P(B|A)P(A)/P(B)
We’ll compute just P(B|A)P(A) for each A, as P(B) is the same for all of them.
P(1|B) ~ P(B|1)P(1) = 1⁄8 x 1⁄3 = 1⁄24
P(2|B) ~ P(B|2)P(2) = 1⁄2 x 1⁄6 = 1⁄12
P(3|B) ~ P(B|3)P(3) = [1/2 + 1⁄2 x 1⁄8] x 1⁄3 = 9⁄48 = 3⁄16
P(4|B) ~ P(B|4)P(4) = [1/2 x 1 + 1⁄2 x 1⁄2] x 1⁄6 = 3⁄4 x 1⁄6 = 1⁄8
This shows that island type 3 is the most likely; yet we never assumed that worlds with more people are more likely.