Uncertainties in the probabilities can’t be absorbed into a single, more conservative probability? If I’m 10% sure that someone’s estimate that cryonics has a 50% likelihood of working is well calibrated, isn’t that the same as being 5% sure that cryonics is likely to work?
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.
Uncertainties in the probabilities can’t be absorbed into a single, more conservative probability? If I’m 10% sure that someone’s estimate that cryonics has a 50% likelihood of working is well calibrated, isn’t that the same as being 5% sure that cryonics is likely to work?
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
Potentially more—perhaps their process for calibration is poor, but the answer coincidentally happens to be right.