Sampling a discrete uniform distribution ($X=x_{1+\lfloor nU\rfloor }$)

Let the numerical support of a probability distribution be $x_1<\cdots <x_n$. Its cumulative distribution function counts how many values have been passed:It is zero below $x_1$, equals $j/n$ on $x_j\le t<x_{j+1}$ for $1\le j<n$, and is one at and above $x_n$. Every jump is $1/n$, even when the horizontal gaps differ.

For inverse transform sampling, take an ideal continuous uniform distribution for $U$ on $[0,1)$. Partition that interval into $n$ intervals of width $1/n$ and outputEach interval has probability $1/n$, proving the desired discrete uniform distribution. Using the floor function, the index is $j=1+\lfloor nU\rfloor$; for consecutive integer values the output is $X=a+\lfloor nU\rfloor$. With the usual quantile function convention $Q(u)=\inf \{t:F_X(t)\ge u\}$, one instead writes $Q(u)=x_{\lceil nu\rceil }$ for $0<u\le 1$, using the ceiling function. The endpoint conventions differ only at a finite set of points with probability zero under the ideal continuous uniform distribution.

For $S=\{1,2,4,8\}$, the four intervals $[0,1/4)$, $[1/4,1/2)$, $[1/2,3/4)$, and $[3/4,1)$ select $1,2,4,8$, respectively. Thus $U=3/10$ selects $2$, while $U=9/10$ selects $8$. The upper half of the interval selects values at least four, recovering probability $1/2$. Equal widths refer to the sampling probability coordinate, not to distances between output values. This is an exact mathematical construction assuming an ideal continuous uniform distribution; a finite computer random-number source needs its own justification of equal output chances.