Given each challenge is sampled at random (i.e. all challenges are independent) among N indexes, the probability that a single challenge hit a node that the prover can prove is (1-\alpha)
This means that for K challenges, the probability that the prover who lost an \alpha fraction of the data is not caught is
$p = (1-\alpha)^K$
As one can observe, given we are considering the percentage of storage lost (i.e. \alpha), the soundness error only depends on \alpha and K, not N.
Of course, the absolute value of data lost depends on ProofSet size, but this is not considered in the security analysis.
Assuming one proof per day, K challenges each, the soundness error decreases over time as
$$ p_{\text{day } T} = (1-\alpha)^{K\cdot T} $$
If we call \epsilon the "probability of evasion”, we have that
| α (fraction of data lost) | Per-day evasion ( (1−α)^5 ) | Per-day detection | 30-day evasion ( (1−α)^(150) ) | 30-day detection |
|---|---|---|---|---|
| 1% | 0.95099 | 4.901 % | 0.22145 | 77.855 % |
| 5% | 0.77378 | 22.622 % | 0.00046 | 99.954 % |
| 20% | 0.32768 | 67.232 % | 2.91 × 10⁻¹⁵ | ≈ 100 % |