The Law of Large Numbers

Frequencies and probabilities

Consider repeating an experiment $n$ times under identical conditions — say, flipping a coin or measuring the spin of a quantum particle prepared in the same state. Let $n_A$ be the number of trials in which event $A$ occurs. The sample frequency (also called the relative frequency) is the ratio

$f_n(A) = \frac{n_A}{n}.$

For a single trial $f_1(A)$ is either $0$ or $1$ . As $n$ grows, the sample frequency fluctuates less wildly and settles toward a stable value. The Law of Large Numbers (LLN) makes this precise.

Statement of the law

Let $X_1, X_2, \ldots$ be independent, identically distributed random variables each with expectation value $\mu = E[X]$ . Define the sample mean after $n$ trials as

$\bar{X}_n = \frac{X_1 + X_2 + \cdots + X_n}{n}.$

The weak law of large numbers states that for any tolerance $\varepsilon > 0$ ,

$P\!\left(|\bar{X}_n - \mu| > \varepsilon\right) \xrightarrow{n \to \infty} 0.$

In plain language: the sample mean converges in probability to the true mean as the number of trials grows without bound. The strong law strengthens this to almost-sure convergence, but for physics the weak form already captures everything we need.

Applied to an indicator variable (1 if $A$ occurs, 0 otherwise) the mean equals $P(A)$ , so the LLN says $f_n(A) \to P(A)$ . This is the bridge between the abstract definition of probability and what we actually observe in a laboratory.

A concrete derivation via Chebyshev

To see why the LLN holds, recall Chebyshev's inequality: for any random variable $Y$ with mean $\mu$ and finite variance $\sigma^2$ ,

$P(|Y - \mu| > \varepsilon) \leq \frac{\sigma^2}{\varepsilon^2}.$

Apply this to $\bar{X}_n$ . If the individual $X_i$ all have the same variance $\sigma^2$ , the variance of the sample mean is

$\operatorname{Var}(\bar{X}_n) = \frac{\sigma^2}{n},$

because independent trials contribute independently. Chebyshev then gives

$P(|\bar{X}_n - \mu| > \varepsilon) \leq \frac{\sigma^2}{n\,\varepsilon^2}.$

The right-hand side goes to zero as $n \to \infty$ for any fixed $\varepsilon$ and finite $\sigma^2$ . This is the core of the weak law.

Why this matters for quantum mechanics

In quantum mechanics, probability is not a frequency by definition — it enters through the Born rule: the probability of measuring outcome $k$ on a system in state $|\psi\rangle$ is $p_k = |\langle k | \psi \rangle|^2$ . The LLN is what connects that theoretical probability to what an experimentalist actually records in a table of shot outcomes.

Every quantum computing experiment reports measurement statistics — the fraction of shots landing in each basis state. Those fractions are sample frequencies. The LLN guarantees they converge to the true Born-rule probabilities as the shot count grows, which is why running more shots reduces statistical uncertainty in an experiment. Understanding this convergence is the foundation for reading any quantum measurement result.

Try it

This is a numerical exercise — your code should return a number, not a circuit. A fair coin is flipped $1000$ times and $520$ tosses land heads. Compute the sample frequency of heads.

Run your code to see the quantum state.