Makes 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 spe

Makes 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.