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The birthday problem asks for the probability that at least two members of a group share a birthday. Use the explicit model of independent random variables taking $m$ labels with a uniform distribution on a finite set. Let $m\ge 1$ and $n\ge 0$ be integers. For calendar examples below, stipulate $m=365$ equally likely labels and omit leap days. For $0\le n\le m$, the probability of at least one matching pair isAt $n=0$, the empty product is one and $P_{0,m}=0$. At $n>m$, a repeat is certain by the pigeonhole principle. Count the complement to obtain the formula: the first label has $m$ choices, then a distinct second label has $m-1$, and so forth, out of $m^n$ equally likely label sequences.^[ref]

For the stipulated calendar, $P_{22,365}\approx 47.57\%$ and $P_{23,365}\approx 50.73\%$. Only 23 people suffice to put the probability above one half. There are $\left(\genfrac{}{}{0ex}{}{23}{2}\right)=253$ possible pairs, each with matching probability $1/365$. The many comparison opportunities explain why a small group can contain a repeat.

The distinction also matters for hash collisions. If independently generated outputs of a hash function are uniformly distributed over $m$ possible labels, the same counting formula governs repeats. The birthday square-root scale explains why looking for any repeated output requires far fewer labels than trying to match one predetermined output. The uniform-independent model is an assumption about those outputs; the birthday problem does not itself establish that a particular implementation meets it.

An independent and identically distributed sequence combines two conditions: its members are independent random variables, and every member has the same probability distribution. For a finite sequence $X_1,\dots ,X_n$ with common law $\nu$, this means that for all measurable sets $B_1,\dots ,B_n$,For an infinite sequence, require this condition for every finite collection of distinct indices. Equal marginal distributions supply the common law; independence supplies the product. Neither condition supplies the other. Grinstead and Snell call the combination an independent trials process.^[ref]

Choose a Bernoulli distribution with success probability $1/2$. Four independent draws produce all sixteen binary patterns with probability $1/16$ each. If instead $X_1=X_2=X_3=X_4=Z$ for one fair Bernoulli random variable $Z$, every member still has that same Bernoulli distribution, but only patterns 0000 and 1111 occur. Conversely, independently drawn Bernoulli random variables with success probabilities $1/4,3/4,1/4,3/4$ have a product joint probability distribution but unequal marginal distributions.

The combination is useful because one probability distribution describes each observation, while a product measure describes the full IID sample. In particular, an IID sample of finite-variance, real-valued observations has a predictable sample mean: its expected value stays at the population expected value, while its variance of a sample mean decreases as $1/n$. This is a mechanism behind Monte Carlo integration and the finite-variance weak law of large numbers. Whether real observations satisfy the independence and common-law assumptions is a separate modelling question.

Let $X_1,\dots ,X_n$ be an IID sample of real-valued random variables with finite expected value $\mu$ and finite variance ${\sigma }^2$, where $n\ge 1$. Its sample mean $\overline{X}=n^{-1}{\sum }_i{X}_i$ satisfiesLinearity of expectation gives the first identity. The variance of a sum and vanishing cross covariances give $\mathrm{Var}({\sum }_i{X}_i)=n{\sigma }^2$; scaling by $1/n$ scales the variance by $1/n^2$. Pairwise independent random variables with the same expected value and variance suffice for these identities, even without full independence. Grinstead and Snell state the corresponding finite-variance independent-trials result.^[ref]

The standard deviation of the sample mean is $\sigma /\sqrt{n}$. Four times as many observations halve this scale; they do not divide it by four. In the fair Bernoulli distribution example, $\sigma =1/2$, so sample sizes $4$ and $64$ give standard deviations $1/4$ and $1/16$, respectively. The common-coin averaging example shows why merely repeating observations with a common marginal law does not justify this calculation.

Chebyshev's inequality turns the variance identity into a probability bound: for $\epsilon >0$,This proves convergence in probability of the sample mean to $\mu$ in this finite-variance setting, a form of the weak law of large numbers. It is an upper bound, so values above one convey no improvement over the universal probability bound one. Cauchy sample mean stability lies outside these hypotheses.