5 ms·
That’s not necessarily accurate if you are already operating in probability per year in this case. It just means 7.75 occurrences per time.
by robbintt 6y ago
That’s not necessarily accurate if you are already operating in probability per year in this case. It just means 7.75 occurrences per time.
- rubidium 6y agoNope. Bad maths. Average of 7.75 per year !== >100% of it happening at least once in one year. There’s still some chance it doesn’t happen in one year.
- deleted 6y ago[deleted]
- OJFord 6y agoYou can have an expected value E[X] = 7.75 You cannot have a probability P[X ≥ 1] >1. There is no 'probability per year' (if it were meaningful year would have to be <1 for a >1 result anyway) - that time frame is built into 'X', it's 'inside' the probability.
- free_rms 6y agoBut OP didn't use either of those terms, they used "775% chance per year", which you can interpret colloquially as a speaker of English. It's been clear throughout this thread what was meant.
- a1369209993 6y ago> "chance per year", which you can interpret Probably something like reciprocal half-life; 775% / yr is about 1 / 1.5 month ie, happens about once every month and a half.
- drdeca 6y agoIf they are independent, then there should be a log-probability of it not happening per year though, right? Like, if the chance for each year is the same and they are independent, then the log of the probability that it doesn’t happen in a given year, If you multiply that that by the number of years, that will give the log of the probability that it doesn’t happen in any of the years. This seems like it could serve many of the purposes that people are intuitively looking for when they think of “probability per year”. And if the different years aren’t independent, uh, maybe some terms could be added to adjust for that?
- alentist 6y agoThat would be the expected value E(X), not the probability P(X > 0). The two are not conflatable.