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For independent random variables $X,Y$ with the same binomial distribution $\mathrm{Bin}(n,p)$, partition the tie event into the mutually exclusive events $X=Y=j$. Independence makes each probability the square of its binomial probability, proving the formula. At $p=1/2$, Vandermonde's identity gives $\mathrm{Pr}(X=Y)=4^{-n}\left(\genfrac{}{}{0ex}{}{2n}{n}\right)$. Symmetry under exchange of the two variables then gives $\mathrm{Pr}(X>Y)=\mathrm{Pr}(Y>X)=(1-\mathrm{Pr}(X=Y))/2$.

For independent aggregate Poisson distributions with means $n\lambda$ and $n\lambda e^{\beta }$, conditioning on the total eliminates $\lambda$. Independent Poisson conditioning gives the displayed binomial distribution. Its conditional likelihood is proportional to $e^{\beta Y_1}(1+e^{\beta }{)}^{-m}$, also the profile likelihood after maximizing over the unrestricted positive baseline rate. Unequal exposures replace the binomial success probability by $n_1{e}^{\beta }/(n_0+n_1{e}^{\beta })$.

Two independent boson species occupy uniform single-particle modes on $L$ sites, with fixed occupation numbers $n_A,n_B$. This differs from a fixed-number two-mode condensate whose particles all occupy one superposition of the two modes. Here $⟨a_i^{†}{a}_j⟩=n_A/L$, $⟨b_i^{†}{b}_j⟩=n_B/L$, and the local total number operator has variance $(n_A+n_B)L^{-1}(1-L^{-1})$. The latter follows from two independent binomial distributions with success probability $1/L$. This state satisfies a prescribed Schwinger boson constraint only in the mean unless the local variance vanishes.

The count $K_n$ has binomial distribution with mean $n/4$ and variance $3n/16$. The information content per symbol is an affine function of $K_n/n$, so its variance isChebyshev's inequality therefore gives the finite-sample lower boundThe bound is often loose, but it proves that every fixed positive tolerance eventually captures nearly all the probability. Direct summation of the binomial distribution's probability masses gives $P(T_{0.1}^{(640)})\approx 0.999779$. A finite sample does not need to have a nearly full typical set: the 40-symbol example above captures only about $0.639$.

For a binomial distribution count, compare a point null $\theta =1/2$ with the uniform alternative prior distribution on $(0,1)$. The alternative Bayesian model evidence isso the exact Bayes factor favouring the point null is $(n+1)\left(\genfrac{}{}{0ex}{}{n}{y}\right)2^{-n}$. With standardized deviation $z=2(y-n/2)/\sqrt{n}$, the local normal distribution approximation is $B_{01}\approx \sqrt{2n/\pi }{e}^{-z^2/2}$. At fixed $z$ the frequentist p-value remains approximately constant, while the factor grows like $\sqrt{n}$. This demonstrates the prior-width mechanism behind the Jeffreys-Lindley paradox for a Gaussian point null in a discrete sampling model.

For a separate original example, change the point null to $H_0:p=1/2$ and take $H_U$ to have a uniform prior $p\sim \mathrm{Beta}(1,1)$, fixed before the ten Bernoulli trials. Here $p$ is drawn once under $H_U$ and shared by all ten trials: they are independent conditional on $p$. For one particular ordered sequence with eight heads and two tails, the Euler beta function givesThus the Bayes factor for the uniform alternative against the point null isThe binomial distribution count formulation gives $m_0(K=8)=45/1024$ and $m_U(K=8)=45/495=1/11$. Their ratio is again $1024/495$; these count masses must not be combined with the ordered-sequence masses above. The general $B_{01}$ formula at the start of this article puts the point null in the numerator, so it equals $B_{0U}$ here.

In contrast, maximum likelihood estimation gives $\hat{p}=0.8$: differentiating the log-likelihood yields $8/p-2/(1-p)=0$, and the endpoints have zero likelihood. Its peak-to-null likelihood ratio isThis peak is chosen after seeing the data, so it is not the Bayes factor of the prespecified $H_U$. The Bayesian model evidence for $H_U$ averages $L(p)$ over its entire uniform prior; it is smaller than the peak. It is also smaller than $L(0.7)=0.7^{8}0.3^2\approx 0.00518832$. Against the same $p=0.5$ null, a point alternative fixed at $p=0.7$ would give a Bayes factor about $5.31284$, greater than $2.06869$. An integrated alternative need not outperform a point alternative.

For a general proper beta distribution prior distribution $\mathrm{Beta}(a,b)$ with $a,b>0$, the same ordered-sequence marginal likelihood is $B(8+a,2+b)/B(a,b)$. This follows by substituting the normalized prior density into the integral, and shows exactly where the parameter prior distribution changes the Bayes factor. Holding the parameter prior distributions fixed while changing model prior odds leaves that Bayes factor unchanged.

Place $n$ labelled balls independently into $m$ labelled bins, each with probability $1/m$. The vector of bin counts has a multinomial distribution, and each count separately has the binomial distribution $\mathrm{Bin}(n,1/m)$. Distinct bin counts have covariance $-n/m^2$, so they are dependent for $n>0$ and $m>1$. Counts of bins of a specified size are sums of indicator variables; their expectations follow from the linearity of expectation, without any independence between bins.

Let $n$ be even and let $X_1,\dots ,X_n$ be independent Bernoulli variables with parameter $1/2$, arranged on a circle. The probability that some consecutive half-circle $I$ has $|{\sum }_{i\in I}{X}_i-{\sum }_{i\notin I}{X}_i|\ge w$ is at most $2n\exp (-2w^2/n)$. For a fixed $I$, the variable $Z_I={\sum }_{i\in I}{X}_i+{\sum }_{i\notin I}(1-X_i)$ has the binomial distribution $\mathrm{Bin}(n,1/2)$, and the signed imbalance is $Z_I-n/2$. Apply Hoeffding inequality to this single binomial variable, then take a union bound over the at most $n$ intervals. Consequently $w=c\sqrt{n\log n}$ suffices for uniform balance with probability tending to one whenever $c>1/\sqrt{2}$.

For $n\ge 2$, choose $n-2$ independent random variables, each with the uniform distribution on a finite set $\{1,\dots ,n\}$, and decode their values as a Prüfer sequence. Every sequence has probability $n^{-(n-2)}$, and the bijection gives each labelled tree the same probability. This produces a uniform spanning tree of the complete graph without repeatedly proposing sets of edges and rejecting those that fail to be trees. It supplies random inputs for testing implementations of breadth-first search and a tractable model of random connections. ^[ref]

This distribution is uniform over fixed-label trees, rather than over unlabelled trees. With four labels, the star graph shape has probability $4/16=1/4$ and the path graph shape has probability $12/16=3/4$. Also, for a fixed label $v$, the Prüfer degree formula givesHere the binomial distribution counts the independent code positions equal to $v$. The expected value agrees with the average vertex degree from the degree-sum formula. Individual labels have the same distribution by symmetry, though their vertex degrees are constrained together by their fixed sum. When $n=1$, simply return the single isolated vertex.

For independent and identically distributed random variables $U_1,\dots ,U_n$ with the continuous uniform distribution on $(0,1)$, the $r$th order statistic has Beta distributionIts probability density function, mean, and variance areThe probability density function is zero outside $[0,1]$; values assigned to its endpoints do not change any probability. Peter Kempthorne's MIT lecture notes derive the uniform order statistic law and these moments. ^[ref].

For $n=5$, the minimum, sample median, and sample maximum correspond to $r=1,3,5$. Their probability density functions are $5(1-t{)}^4$, $30t^2(1-t{)}^2$, and $5t^4$, respectively. The sample median therefore has Beta distribution $\mathrm{Beta}(3,3)$, mean $1/2$, and variance $1/28\approx 0.0357143$. Its cumulative distribution function isFor instance, $\mathbb{P}(U_{(3)}\le 1/4)=53/512\approx 0.103516$. The mean of the minimum is $1/6$, and the mean of the sample maximum is $5/6$. The fixed observed sample median $0.4$ in the order statistic example is one possible outcome; $1/2$ is the average over repeated statistical samples from this uniform distribution.

Each curve has area one. The extreme order statistics favor the ends, while the sample median favors the center. The endpoint density height $5$ is legitimate: the probability density function gives area per unit length, rather than point probability. In particular, a continuous uniform distribution has zero probability of producing an endpoint.

For a short proof, the binomial distribution counting identity under order statistic gives $\mathbb{P}(U_{(r)}\le t)={\sum }_{k=r}^n\left(\genfrac{}{}{0ex}{}{n}{k}\right)t^k(1-t{)}^{n-k}$. Taking its derivative cancels consecutive terms, leaving the displayed probability density function. Recognize its normalizing factorial coefficient as that of the Beta distribution, then substitute $\alpha =r$, $\beta =n+1-r$ into the Beta distribution mean and variance formulas. Although the original observations are independent, their order statistics generally are dependent; see the positive covariance of two uniform order statistics.

A uniform prior on a finite-volume parameter region assigns constant probability density there and zero outside. On $[0,1]$ it is $\mathrm{Beta}(1,1)$, giving a beta distribution posterior after binomial distribution data. Uniformity depends on the coordinate: a nonlinear reparametrization introduces a Jacobian and generally does not preserve a uniform density.

Start with the empty graph and reveal the edges of the complete graph in a uniformly random order. After $m$ steps its edge set is a uniform $m$-element subset. Independent uniform edge labels couple this process with all binomial random graphs: $G(n,p)=G_{n,M(p)}$, where $M(p)$ has a binomial distribution with parameters $\left(\genfrac{}{}{0ex}{}{n}{2}\right),p$. For a monotone graph property, concentration of $M(p)$ transfers a threshold estimate between edge probability and edge count.

Let $X_1,X_2,\dots$ be real independent and identically distributed random variables with $\mathbb{E}|X_1|<\infty$ and expected value $\mu =\mathbb{E}[X_1]$. The weak law of large numbers states that their sample mean has convergence in probability to $\mu$: for every fixed $\epsilon >0$,Thus a large sample is unlikely to give an average with a specified fixed-size error. The strong law of large numbers makes the stronger statement of almost sure convergence along an entire sequence. Under the displayed assumptions both laws apply; their names distinguish the conclusions, not two competing predictions.

When the variance ${\sigma }^2$ is finite, variance additivity for independent random variables gives $\mathrm{Var}({\overline{X}}_n)={\sigma }^2/n$. The Chebyshev inequality immediately supplies a useful quantitative form and a brief proof:For independent trials with a Bernoulli distribution of parameter $p=0.2$, the expected value is $0.2$ and the variance is $0.16$. At $n=1000$, the chance of an error of at least $0.1$ is therefore at most $0.016$. The exact binomial distribution gives a much smaller probability, as the figure under the strong law of large numbers shows. The bound is conservative, but it is valid for every sample size under these assumptions and does not require a normal approximation.

The weak law of large numbers addresses a fixed tolerance $\epsilon$. The central limit theorem instead resolves errors on the shrinking scale $\sigma /\sqrt{n}$ and gives their approximate normal distribution. In particular, the weak law of large numbers alone supplies neither a bell-shaped error law nor a universal rate of approach. For example, a Pareto distribution with minimum one and shape $3/2$ has expected value three but infinite variance: both the weak law of large numbers and strong law of large numbers apply, while the finite-variance statement of the central limit theorem does not.