3 msΒ·
It's not clear what fallacy you're thinking about because the fallacy of the inverse is not it. From https://en.wikipedia.org/wiki/Confusion_of_the_inverse htt
by theli0nheart 5y ago
It's not clear what fallacy you're thinking about because the fallacy of the inverse is not it.
From https://en.wikipedia.org/wiki/Confusion_of_the_inverse https://en.wikipedia.org/wiki/Confusion_of_the_inverse:
> Confusion of the inverse, also called the conditional probability fallacy or the inverse fallacy, is a logical fallacy whereupon a conditional probability is equated with its inverse; that is, given two events A and B, the probability of A happening given that B has happened is assumed to be about the same as the probability of B given A, when there is actually no evidence for this assumption.[1][2] More formally, P(A|B) is assumed to be approximately equal to P(B|A).
- Closi 5y agoThe inverse of "any contract that has been hacked was insecure" is "any contract that hasn't been hacked must be secure". I think this is what OP meant. If something has been around for a long time it does probably mean it's less likely there is a really obvious security flaw, but it doesn't necessarily mean it is 'rock-solid' as plenty of things that have been seen to be 'rock-solid' in the past have turned out to be insecure.
- solomonb 5y agoyes this is what I was implying. I am curious how else the previous commenter would interpret my statement.
- theli0nheart 5y agoLet's formalize this to make it easier to discuss. π(π) = π(contract has been hacked) π(Β¬π) = π(contract has not been hacked) π(π) = π(contract is hack resistant) Relevant conditional probabilities: π(π|Β¬π) = π(contract is hack-resistant given that it has not been hacked) π(Β¬π|π) = π(contract has not been hacked given that it is hack-resistant) The fallacy of the inverse would be assuming that: > The probability that a contract is hack-resistant, given that it has not been hacked, is approximately equal to the probability that it has not been hacked, given that it's hack resistant. More succinctly: π(π|Β¬π) β π(Β¬π|π) In https://news.ycombinator.com/item?id=27666484 https://news.ycombinator.com/item?id=27666484, the fallacy of the inverse was presented as "those contracts not being hacked yet is no proof that they are resistant to hacks" or "NOT (NOT A implies B)", i.e.: Β¬(Β¬π β π) In summary: π(π|Β¬π) β π(Β¬π|π) => the fallacy of the inverse and Β¬(Β¬π β π) => statement in comment These two statements are fundamentally different. Note that the first statement is a comparison of probabilities, and the second is not. They're not the same. There might be another fallacy at play here, but it's not the fallacy of the inverse.
- Closi 5y agoTo be honest, I donβt fully follow the above, and Iβm not sure if this formalises the same problem. Copying the definition from Wikipedia: > If P, then Q. > Therefore, if not P, then not Q. In this case, we could plug in the below values: P = Contract Hacked Q = Insecure I.e. if the contract has been hacked the code was insecure. Therefore if the contract has not been hacked, it is not insecure.