4 ms·

The U.S. mortality rate is 0.008 per year. That means that in any gathering of 100 people (roughly the number of speakers at Black Hat), there is a 55% chance

by codex 13y ago

The U.S. mortality rate is 0.008 per year. That means that in any gathering of 100 people (roughly the number of speakers at Black Hat), there is a 55% chance that at least one will die in the year before the event and a 6.5% chance they would die one month before the event. Over ten years of conferences, the odds that someone would die within one month of speaking is 49%. The under-40 crowd doesn't really appreciate this since death predominantly affects older people.

conroe64 13y ago
That doesn't consider that the age and economic status of conference speakers isn't representative of the U.S. as a whole. Not that it's impossible for it to be an accident, but the odds are certainly less than 49%, given that the crowd is mostly urban professionals.
- 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 ago
    Yeah 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 ago
    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.
    - 0003 13y ago
      P(F|D) = 0.00058 (from above) Can you explain this line?
      - marshray 13y ago
        http://en.wikipedia.org/wiki/Bayes%27_theorem http://en.wikipedia.org/wiki/Bayes%27_theorem
      - conroe64 13y ago
        Sorry, that line was erroneous, but didn't affect the rest of the calculation as far as I can tell. I have removed it.
samstave 13y ago
So... if I am asked to talk at BlackHat, then there is a 55% chance I'll die! ;-) (JK I know thats not what you meant...)
- cgio 13y ago
  There is 100% chance we will all die, whether invited to talk at BH or not.
  - PeterisP 13y ago
    Historically 100% have died, but it's not necessary to be that way - consider it a bug to be fixed.
joe_the_user 13y ago
The use of statistics and such can't really take away from this being a suspicious seeming coincidence. It can, however, show that coincidences are relatively common and so do not automatically imply a hidden hand. However, given that recent revelations have made a hidden hand a less-than-extraordinary event, we can get all Bayesian and say a rise in the prior has increased posterior probability here.

by codex 13y ago

conroe64 13y ago

That doesn't consider that the age and economic status of conference speakers isn't representative of the U.S. as a whole. Not that it's impossible for it to be an accident, but the odds are certainly less than 49%, given that the crowd is mostly urban professionals.

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 ago
  Yeah 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 ago
  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.
  - 0003 13y ago
    P(F|D) = 0.00058 (from above) Can you explain this line?
    - marshray 13y ago
      http://en.wikipedia.org/wiki/Bayes%27_theorem http://en.wikipedia.org/wiki/Bayes%27_theorem
    - conroe64 13y ago
      Sorry, that line was erroneous, but didn't affect the rest of the calculation as far as I can tell. I have removed it.

samstave 13y ago

So... if I am asked to talk at BlackHat, then there is a 55% chance I'll die! ;-) (JK I know thats not what you meant...)

cgio 13y ago
There is 100% chance we will all die, whether invited to talk at BH or not.
- PeterisP 13y ago
  Historically 100% have died, but it's not necessary to be that way - consider it a bug to be fixed.

joe_the_user 13y ago

The use of statistics and such can't really take away from this being a suspicious seeming coincidence. It can, however, show that coincidences are relatively common and so do not automatically imply a hidden hand. However, given that recent revelations have made a hidden hand a less-than-extraordinary event, we can get all Bayesian and say a rise in the prior has increased posterior probability here.