My calculation was simplistic, you’re right, but still useful for arriving at a conservative estimate. To add to the nitpickyness we should mention that the probability that someone is completely well calibrated on something like cryonics is almost surely 0. The 10% estimate should instead be for the chance that the person’s estimate is well calibrated or underconfident.
ETA: applying the same conservative method to the lottery odds would lead you to be even less willing to buy a ticket.
The point I’m making is that there is an additional parameter in the equation: your probability that cryonics is possible independent of that source. This needn’t be epsilon any more than the expectation of the lottery value need be epsilon.
My calculation was simplistic, you’re right, but still useful for arriving at a conservative estimate. To add to the nitpickyness we should mention that the probability that someone is completely well calibrated on something like cryonics is almost surely 0. The 10% estimate should instead be for the chance that the person’s estimate is well calibrated or underconfident.
ETA: applying the same conservative method to the lottery odds would lead you to be even less willing to buy a ticket.
The point I’m making is that there is an additional parameter in the equation: your probability that cryonics is possible independent of that source. This needn’t be epsilon any more than the expectation of the lottery value need be epsilon.
I agree. That’s why I said so previously =P