6 ms·

Isn't it sad that since the NSA revelations anything is really "on the table" for our imaginations? Nothing seems too far-fetched these days...

by vxxzy 13y ago

frou_dh 13y ago
I know it's a hot topic, but the Snowden leak is hardly the genesis of distrust.
- jetti 13y ago
  For me, it is more the case of Michael Hastings[1] that would bring foul play from a government agent rather than the Snowden case. [1]http://www.foxnews.com/us/2013/06/24/journalist-michael-hastings-sends-chilling-e-mail-to-colleagues-before-death/ http://www.foxnews.com/us/2013/06/24/journalist-michael-hast...
  - 00rion 13y ago
    Indeed. I can't believe more people don't know about this.
    - lawnchair_larry 13y ago
      I can't believe you crazies found HN.
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.
nikster 13y ago
It is a little sad. Those days where one could rule out foul play by the US government in a case like this - those days are most certainly over. That said in this case we'll have to trust his family.

Isn't it sad that since the NSA revelations anything is really "on the table" for our imaginations? Nothing seems too far-fetched these days...

by vxxzy 13y ago

Isn't it sad that since the NSA revelations anything is really "on the table" for our imaginations? Nothing seems too far-fetched these days...

frou_dh 13y ago

I know it's a hot topic, but the Snowden leak is hardly the genesis of distrust.

jetti 13y ago
For me, it is more the case of Michael Hastings[1] that would bring foul play from a government agent rather than the Snowden case. [1]http://www.foxnews.com/us/2013/06/24/journalist-michael-hastings-sends-chilling-e-mail-to-colleagues-before-death/ http://www.foxnews.com/us/2013/06/24/journalist-michael-hast...
- 00rion 13y ago
  Indeed. I can't believe more people don't know about this.
  - lawnchair_larry 13y ago
    I can't believe you crazies found HN.

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.

nikster 13y ago

It is a little sad. Those days where one could rule out foul play by the US government in a case like this - those days are most certainly over. That said in this case we'll have to trust his family.