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.
𝐏(𝐀) = 𝐏(contract has been hacked)
𝐏(¬𝐀) = 𝐏(contract has not been hacked)
𝐏(𝐁) = 𝐏(contract is hack resistant)
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 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)
¬(¬𝐀 → 𝐁)
𝐏(𝐁|¬𝐀) ≅ 𝐏(¬𝐀|𝐁) => the fallacy of the inverse
and
¬(¬𝐀 → 𝐁) => statement in comment
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.
> 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.
From 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).