3 msΒ·
Let'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(contrac
by theli0nheart 5y ago
Let'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.