3 ms·

yes this is what I was implying. I am curious how else the previous commenter would interpret my statement.

theli0nheart 5y ago
Let's formalize this to make it easier to discuss. 𝐏(𝐀) = 𝐏(contract has been hacked) 𝐏(¬𝐀) = 𝐏(contract has not been hacked) 𝐏(𝐁) = 𝐏(contract is hack resistant) Relevant conditional probabilities: 𝐏(𝐁|¬𝐀) = 𝐏(contract is hack-resistant given that it has not been hacked) 𝐏(¬𝐀|𝐁) = 𝐏(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: 𝐏(𝐁|¬𝐀) ≅ 𝐏(¬𝐀|𝐁) 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.: ¬(¬𝐀 → 𝐁) In summary: 𝐏(𝐁|¬𝐀) ≅ 𝐏(¬𝐀|𝐁) => the fallacy of the inverse and ¬(¬𝐀 → 𝐁) => 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 ago
  To be honest, I don’t fully follow the above, and I’m 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.

Let's formalize this to make it easier to discuss. 𝐏(𝐀) = 𝐏(contract has been hacked) 𝐏(¬𝐀) = 𝐏(contract has not been hacked) 𝐏(𝐁) = 𝐏(contract is hack resistant) Relevant conditional probabilities: 𝐏(𝐁|¬𝐀) = 𝐏(contract is hack-resistant given that it has not been hacked) 𝐏(¬𝐀|𝐁) = 𝐏(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: 𝐏(𝐁|¬𝐀) ≅ 𝐏(¬𝐀|𝐁) 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.: ¬(¬𝐀 → 𝐁) In summary: 𝐏(𝐁|¬𝐀) ≅ 𝐏(¬𝐀|𝐁) => the fallacy of the inverse and ¬(¬𝐀 → 𝐁) => 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 ago
To be honest, I don’t fully follow the above, and I’m 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.