6 ms·
Sorry, I should be clearer: I was calculating based on the general mortality rate. The accident rate is much lower (0.000391). I agree that either measure is
by codex 13y ago
Sorry, I should be clearer: I was calculating based on the general mortality rate. The accident rate is much lower (0.000391). I agree that either measure is skewed, though; this is just a rough calculation. The annual death rate for 35-44 from all causes (see http://www.data360.org/dsg.aspx?Data_Set_Group_Id=587 http://www.data360.org/dsg.aspx?Data_Set_Group_Id=587) is 0.002, which puts the odds of a speaker death within one month of the conference at 15% over ten years. Poorer, but then again, this is for only one specific security conference. If you assume five security conferences a year of all sorts, it's back to 50% over ten years. I think it's a bit higher as these speakers live sedentary lifestyles, even for the U.S.
- tomjen3 13y agoYeah but that is for all causes and that includes things like long term cancer and other deceases that don't really strike suddenly.
- conroe64 13y agoMakes sense. Another thought: Barnaby Jack was one of the top speakers and was to speak on a very controversial subject. I would guess that out of the 5000 speeches that were presented in your scenario only a few of them, maybe 2%, would contain information controversial enough that foul play would appear as a reasonable scenario to an outside observer (and this is being generous). Let F = foul play occurred in order to disrupt a conference, D = death of speaker one month before conference Let P(D) = (.02 deaths per year for 25-35 y.o) / (12 months in a year) = 0.0017 Let P(F|D) = .001 (assuming 1 in 1000 chance foul play was involved given a death of a speaker at a conference) P(F) = P(F|^D) * P(^D) + P(F|D) * P(D) = 0 * 0.98 + 0.001 * .02 = 0.00002 Let P(D|F) = 1 (chance of death if foul play is involved, assumed 100%) So Bayes theorem gives us: P(F|D) = P(D|F)*P(F)/P(D) = 1 * 0.00002 / 0.0017 = 0.018 (chance of foul play for a single speech given the speaker died 1 month before) Let P(C) = 0.02 (probability of a controversial speech) P(D) = .0017 (from above) Let P(C&D|F) = .5 (assuming there is a 50% chance the speech was controversial given foul play did occur, and death always occurs from foul play) P(C&D) = .02 * 0.0017 = 0.000034 P(F) = 0.00002 (from above) P(F|C&D) = P(C&D|F) * P(F) / P(C&D) = 0.5 * 0.00002 / .000034 = around a 30% foul play was involved in Barnaby Jack's death There are a lot of assumptions here that could adjust the final figure up or down, but if I did my math right, foul play does seem a reasonable scenario, (but not a foregone conclusion). edit: removed line "P(F|D) = 0.00058 (from above)" as pointed out by user 0003. End result didn't change, though.
- 0003 13y agoP(F|D) = 0.00058 (from above) Can you explain this line?
- pkinsky 13y agoLet P(F|D) = .001 (assuming 1 in 1000 chance foul play was involved given a death of a speaker at a conference) P(F|D) = P(D|F)P(F)/P(D) = 1 0.00002 / 0.0017 = 0.018 (chance of foul play for a single speech given the speaker died 1 month before) Is this a typo? I don't understand how you're finding two different values for P(F|D).
- acqq 13y agoLet P(F|D) = .001 (assuming 1 in 1000 chance foul play was involved given a death of a speaker at a conference) Am I right that here you claim that every 1000th death of a speaker is by foul play? I'd expect such number of deaths to be more orders of magnitude less common.
- diminoten 13y agoAnd this is why Bayesian Decision Theory is utter bullshit. Do you seriously think you can, with any real degree of accuracy, predict how likely it was that foul play was involved?