This seems sort of similar to the famous “Quantum suicide” concept, although less elegant :-)
We can view Time 2 as the result of an experiment where 8 copies of you start the game, are assigned to two groups W and L (4 copies in each), then a coin is tossed, and if it comes up heads, than we kill 75% of the L group. After such an experiment, you definitely should expect yourself being in the W group given that you’re alive with a 80% probability. Time 3 (not shown in your figure) corresponds to a second round of the experiment, when no coin is tossed, but simply 75% of the W group are exterminated. So if you’re doing a single suicide, expect to be in W with 80% probability, but if you’re doing a double-suicide experiment, then expect a 50% probability of being in W. If before exterminating the W group you dump all their memories into the one surviving copy, then expect a 50% chance of being in L and a 50% chance of being in W, but with four times as much memories of having been in W after the first stage of the experiment. Finally, you can also easily calculate all probabilities if you find yourself in group W after the first suicide, but are unsure as to which version of the game you’re playing. I think that’s what Bostrom’s answer corresponds to.
The suicide experiment is designed more clearly, and it helps here. What would change if you had an exactly 100% probability of winning (the original Quantum suicide)? What if vice versa? And if you have a non-zero probability of either outcome, what if you just view it as the appropriately weighted sum of the two extremes?
This seems sort of similar to the famous “Quantum suicide” concept, although less elegant :-)
We can view Time 2 as the result of an experiment where 8 copies of you start the game, are assigned to two groups W and L (4 copies in each), then a coin is tossed, and if it comes up heads, than we kill 75% of the L group. After such an experiment, you definitely should expect yourself being in the W group given that you’re alive with a 80% probability. Time 3 (not shown in your figure) corresponds to a second round of the experiment, when no coin is tossed, but simply 75% of the W group are exterminated. So if you’re doing a single suicide, expect to be in W with 80% probability, but if you’re doing a double-suicide experiment, then expect a 50% probability of being in W. If before exterminating the W group you dump all their memories into the one surviving copy, then expect a 50% chance of being in L and a 50% chance of being in W, but with four times as much memories of having been in W after the first stage of the experiment. Finally, you can also easily calculate all probabilities if you find yourself in group W after the first suicide, but are unsure as to which version of the game you’re playing. I think that’s what Bostrom’s answer corresponds to.
The suicide experiment is designed more clearly, and it helps here. What would change if you had an exactly 100% probability of winning (the original Quantum suicide)? What if vice versa? And if you have a non-zero probability of either outcome, what if you just view it as the appropriately weighted sum of the two extremes?