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.
You’re assigning 0% probability to (cryonics_working|estimate_miscalibrated). Therefore you should buy the lottery ticket.
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