4 ms·

From the article: > NSA itself planted it surreptitiously. > Apple, complicit with the NSA, added it. Neither seems very likely given how visible the goto is

From the article: > NSA itself planted it surreptitiously. > Apple, complicit with the NSA, added it. Neither seems very likely given how visible the goto is. Something just a little more subtle like a semicolon at the end of the if() clause might look better. Of course given how glaring it is, it could be a case of plausible deniability: would we do something to stupid, so unsubtle.

ggreer 13y ago
"A sneaky bug? Only the NSA could have pulled that off! An obvious bug? Only the NSA would be so brazen in trying to throw us off their track!" If the witch had led an evil and improper life, she was guilty; if she had led a good and proper life, this too was a proof, for witches dissemble and try to appear especially virtuous. After the woman was put in prison: if she was afraid, this proved her guilt; if she was not afraid, this proved her guilt, for witches characteristically pretend innocence and wear a bold front. Or on hearing of a denunciation of witchcraft against her, she might seek flight or remain; if she ran, that proved her guilt; if she remained, the devil had detained her so she could not get away.[1] If you argue that an obvious bug is evidence of NSA involvement, then you must also believe a subtle bug would be evidence against NSA involvement. You can't have it both ways. 1. From Conservation of Expected Evidence http://lesswrong.com/lw/ii/conservation_of_expected_evidence/ http://lesswrong.com/lw/ii/conservation_of_expected_evidence...
- terminus 13y ago
  > If you argue that an obvious bug is evidence of NSA involvement, then you must also believe a subtle bug would be evidence against NSA involvement. You can't have it both ways. Thanks for taking the argument to a mathematical plane. My reasoning was the following: P(is-involved) = P(is-involved|subtle)P(subtle) + P(is-involved|subtle)P(~subtle) Now, we know that 0 < P(is-involved) < 1. Assuming that there have been past instances of their involvement in both subtle and unsubtle bugs, I think my argument where I can attach a probability of their involvement in both the case was fair. Or do you think I missed something. Note: I think I got it. Essentially your point is that given that I am using the unsubtle-ty of the bug to argue both ways, implies that this variable provides us no useful information and can be removed from the discussion. Thanks for pointing out my muddy reasoning.
  - nicky0 13y ago
    Shouldn't it be P(is-involved) = P(is-involved|subtle)P(subtle) + P(is-involved|~subtle)P(~subtle)

ggreer 13y ago

"A sneaky bug? Only the NSA could have pulled that off! An obvious bug? Only the NSA would be so brazen in trying to throw us off their track!" If the witch had led an evil and improper life, she was guilty; if she had led a good and proper life, this too was a proof, for witches dissemble and try to appear especially virtuous. After the woman was put in prison: if she was afraid, this proved her guilt; if she was not afraid, this proved her guilt, for witches characteristically pretend innocence and wear a bold front. Or on hearing of a denunciation of witchcraft against her, she might seek flight or remain; if she ran, that proved her guilt; if she remained, the devil had detained her so she could not get away.[1] If you argue that an obvious bug is evidence of NSA involvement, then you must also believe a subtle bug would be evidence against NSA involvement. You can't have it both ways. 1. From Conservation of Expected Evidence http://lesswrong.com/lw/ii/conservation_of_expected_evidence/ http://lesswrong.com/lw/ii/conservation_of_expected_evidence...

terminus 13y ago
> If you argue that an obvious bug is evidence of NSA involvement, then you must also believe a subtle bug would be evidence against NSA involvement. You can't have it both ways. Thanks for taking the argument to a mathematical plane. My reasoning was the following: P(is-involved) = P(is-involved|subtle)P(subtle) + P(is-involved|subtle)P(~subtle) Now, we know that 0 < P(is-involved) < 1. Assuming that there have been past instances of their involvement in both subtle and unsubtle bugs, I think my argument where I can attach a probability of their involvement in both the case was fair. Or do you think I missed something. Note: I think I got it. Essentially your point is that given that I am using the unsubtle-ty of the bug to argue both ways, implies that this variable provides us no useful information and can be removed from the discussion. Thanks for pointing out my muddy reasoning.
- nicky0 13y ago
  Shouldn't it be P(is-involved) = P(is-involved|subtle)P(subtle) + P(is-involved|~subtle)P(~subtle)