4 ms·
The problem with a "unit" of risk like this is that probabilities don't simply sum like this. Assuming the probability of dying in each activity is independent
by swolchok 10y ago
The problem with a "unit" of risk like this is that probabilities don't simply sum like this.
Assuming the probability of dying in each activity is independent, planning to do a 7000-micromort activity A and then (after appropriate recovery time) an 8000-micromort activity B is not an overall activity with 15000 micromorts of risk. (The math is close enough when talking about single-digit micromorts, but it doesn't work in general.)
Pr(dying when doing A followed by B) = Pr({doing A and dying} or {doing A and surviving and then doing B and dying}) = Pr(doing A and dying) + PR(doing A and surviving and then doing B and dying) = 7000 / 1000000 + (1 - 7000 / 1000000) * 8000 / 1000000 = 14944 / 1000000
The larger the risks or the longer the series of events, the less accurate the assumption of linearity. This also doesn't take into account the fact that dying in extreme activities is not independent; by failing to die in A, you've demonstrated some small amount of survival fitness.
- swolchok 10y agoIf the micromort was sold as an expected number of deaths, then I wouldn't have as much of a problem with it. (My major problem is that you can't die more than once, so probability seems to be more of a concern than expected number.) The article calls it "a one-in-a-million chance of dying", though. For example, playing Russian Roulette once is about 166667 micromorts. Playing it 6 times is 1 mort of risk, but the probability of dying after 6 plays is not 1. (It's 1 - (5/6)6, or about 2/3.)
- AstralStorm 10y agoWelcome to the world of bayesian statistics. Frequentist approach like this micromort unit ignores conditional probability.