tl;dr No. The probability is approximately 10⁻²⁸⁵³ — effectively zero.
- Network hashrate: H
- Miner's hashrate: h = H / 2016
- Miner's win probability per block: p = 1/2016 ≈ 0.0496%
- Blocks per difficulty epoch: n = 2016
- Goal: mine 55% of blocks → X ≥ 1109 blocks (55% × 2016 = 1108.8, rounded up)
Each of the 2016 blocks is won independently with probability p.
X ~ Binomial(n = 2016, p = 1/2016)
Expected blocks mined: λ = n × p = 2016 × (1/2016) = 1
When n is large and p is tiny, Binomial(n, p) ≈ Poisson(λ).
Since λ = np = 1:
P(X = k) = e⁻¹ / k!
This is exact to many decimal places for our parameters.
To reach 55% the miner needs at least 1109 blocks — which is 1108 standard deviations above their expected value of 1.
P(X ≥ 1109) = Σ[k=1109 → ∞] e⁻¹ / k!
For a Poisson variable with mean λ, and any target t > λ:
P(X ≥ t) ≤ (e·λ / t)ᵗ · e⁻λ
Plug in t = 1109, λ = 1:
log P ≤ 1109 · ln(e / 1109) − 1
= 1109 · (1 − ln 1109) − 1
= 1109 · (1 − 7.0112) − 1
= 1109 · (−6.0112) − 1
≈ −6,666 (nats)
≈ −2,895 (base 10)
P(X ≥ 1109) ≤ 10⁻²⁸⁹⁵
This is an upper bound — the true value is smaller still.
Compare to the 51% threshold (≥1009 blocks): that was 10⁻²⁵⁹².
Raising the bar to 55% makes it ~10³⁰³ times less likely — or about the square of all atoms in the universe, less probable again.
| Thing | Count |
|---|---|
| Atoms in the observable universe | ~10⁸⁰ |
| Planck-time intervals since the Big Bang | ~10⁶¹ |
| Attempts needed to expect one success | 10²⁸⁹⁵ |
The gap between 10⁸⁰ (atoms) and 10²⁸⁹⁵ (attempts needed) is itself 10²⁸¹⁵ — a number with 2,815 digits.
This event cannot happen.