When can a binomial with n = 10,000 and p = 0.0003 be approximated by a Poisson distribution?

## The result The **Poisson limit theorem**, sometimes called the law of rare events, says that if you let n → ∞ and p → 0 while holding np = λ fixed, then `C(n, k) p^k (1-p)^(n-k) → e^(-λ) λ^k / k!` for every fixed k. In words: many trials, each very unlikely to succeed, produce a count that behaves like a Poisson with rate equal to the expected number of successes. ## Applying it to the example With n = 10,000 emails and a bounce probability p = 0.0003 per email, independently: `λ = np = 10,000 × 0.0003 = 3`. Compare the exact binomial with Poisson(3): | k | Binomial(10000, 0.0003) | Poisson(3) | |---|---|---| | 0 | 0.04976 | 0.04979 | | 1 | 0.14934 | 0.14936 | Both agree to three decimal places, and the agreement holds across the rest of the range. In practice the Poisson answer *is* the binomial answer here. ## Why it works — the variance view The cleanest intuition is dispersion. A binomial has mean np and variance np(1-p); a Poisson has mean λ and variance λ. Set λ = np and the two means already match exactly — no approximation involved. Only the variances differ, by the factor (1-p). When p = 0.0003 that factor is 0.9997, so the discount is negligible and the shapes coincide. When p = 0.4 the factor is 0.6 and the binomial is far tighter than a Poisson with the same mean, so the approximation is worthless however large n is. **Large n alone is never the condition; small p is.** ## Rules of thumb Commonly quoted working conditions are n ≥ 100, p ≤ 0.01, and np no more than about 10. These are guidelines, not theorems — the honest formulation is that accuracy improves as p shrinks with λ held fixed. A quick self-check is to compare the two variances: if np(1-p) and np differ by less than a percent or so, the substitution is safe. ## What you gain Three things. First, arithmetic: the exact binomial needs `C(10000, k)`, an enormous number multiplied by an underflowing power, whereas the Poisson needs only λ. Second, parsimony: you specify one number instead of two, and you often only *know* one number — the observed average bounce count — without knowing whether it came from 10,000 sends at 0.0003 or 30,000 at 0.0001, which the model cannot distinguish. Third, additivity: independent Poisson counts add, so campaigns can be pooled by summing rates. ## What can go wrong **Dependence.** The theorem assumes independent trials with a common p. Real bounces are correlated: one bad domain, one blocklisting event, one malformed batch produces many bounces at once. Clustered failures give counts with variance far above λ, so the Poisson model's tail is too thin and a genuinely bad send looks impossibly extreme. **A varying p.** If the per-email probability differs across segments — a fresh list at 0.0001, a stale one at 0.002 — the pooled count is a mixture and again more dispersed than Poisson(np̄). The fix is to model segments separately rather than to pool. **Large p.** Already covered: the approximation degrades in a way that *systematically overstates* variability, making the model look conservative when it is simply wrong. ## Worked contrast Take n = 20, p = 0.3. Then np = 6, which looks like a fine λ, but the binomial variance is 20 × 0.3 × 0.7 = 4.2 versus the Poisson's 6. The binomial also cannot exceed 20, while the Poisson assigns positive probability to 30 or 50. Same mean, materially different distribution. Contrast with n = 20,000, p = 0.0003: λ = 6 again, binomial variance 5.9982, and the two are indistinguishable in practice. ## Common errors Setting λ = p rather than np; justifying the approximation by large n alone; assuming the mean is approximate when in fact the means match exactly and only the variances differ; and forgetting that the Poisson has no upper bound while a binomial count can never exceed n — which is harmless when λ ≪ n and absurd when it is not.