The Discrete Uniform Model on a Finite Set

Why the Set of Possible Values Matters

A discrete uniform model assigns equal probability to each value in a specified finite set. The set need not contain every integer between its smallest and largest members. That distinction matters: a value can lie numerically between two possible values and still be impossible under the model. To calculate probabilities correctly, first identify the values the random variable can actually take.

The model is appropriate when the random variable’s distinct possible values are equally likely. A fair selection mechanism can support that assumption, but it does not guarantee it automatically. If several equally likely underlying outcomes produce the same value of the random variable, that value receives more probability than the others.

Definition: Let $S=\{x_1,x_2,\ldots,x_m\}$ be a nonempty finite set of $m$ distinct values. A random variable $X$ has a discrete uniform distribution on $S$ if $$ P(X=x)= \begin{cases} \frac{1}{m}, & x\in S,\\ 0, & x\notin S. \end{cases} $$ The set $S$ is the support of $X$: it consists exactly of the values with positive probability. The integer $m$ is the number of distinct values in that support.

The support is not the same as the entire interval from the least to the greatest support value. For example, if $S=\{-5,-1,2,9,15,22\}$, then 4 lies between the least and greatest values but is not in the support. Under the stated model, $P(X=4)=0$. In contrast, each of the six listed values has probability $1/6$.

When $X$ is uniform on $S$, the probability of an event involving $X$ can be found by counting which support values satisfy the event. Values that satisfy the event’s numerical description but do not belong to $S$ contribute no probability.

Result: If $X$ is uniform on a finite support $S$ with $m$ values, and $A$ is any set of values, then $$ P(X\in A)=\frac{|A\cap S|}{m}, $$ where $|A\cap S|$ is the number of distinct support values that are also in $A$.

Why Counting Support Values Gives the Probability

This result follows directly from the probability mass function. The event $X\in A$ occurs when $X=x$ for at least one value $x$ in $A\cap S$. Those possible values are mutually exclusive: $X$ cannot equal two different numbers in the same trial. Each contributes probability $1/m$, so adding their probabilities gives the event probability.

P(X\in A) =\sum_{x\in A\cap S}P(X=x) =\sum_{x\in A\cap S}\frac{1}{m} =\frac{|A\cap S|}{m}.

If $A\cap S$ is empty, the sum has no positive terms and the event has probability 0. If every support value is in $A$, the event has probability $m/m=1$. This counting method depends on the uniform assumption; if the support values have unequal probabilities, their probabilities must instead be added individually.

Worked Example: An Irregular Support With Gaps

A hypothetical monitoring device reports one of the six readings $-5,-1,2,9,15,22$, each with equal probability. Let $X$ be the reported reading. Find $P(-2\leq X\leq 10)$, $P(X=4)$, and $P(X\in\{-5,4,15\})$. State the support and check that the model is valid.

Method and model. The six listed readings are distinct and are assumed equally likely. Thus $X$ is discrete uniform on $S=\{-5,-1,2,9,15,22\}$, with $m=6$. Each support value has probability $1/6$, and every value outside $S$ has probability 0.

First event. The support values satisfying $-2\leq X\leq 10$ are $-1,2,9$. There are three such values, so $$ P(-2\leq X\leq 10)=\frac{3}{6}=\frac{1}{2}=0.5000. $$

Second event. The value 4 is not in $S$. Therefore $$ P(X=4)=0. $$ Being between the smallest and largest readings does not make 4 attainable.

Third event. Of the values in $\{-5,4,15\}$, the support values are $-5$ and $15$; 4 is not attainable. Therefore $$ P(X\in\{-5,4,15\})=\frac{2}{6}=\frac{1}{3}=0.3333. $$

Check. There are six distinct support values, each assigned probability $1/6$, so the total probability is $6(1/6)=1$. The event counts also use only values in the support.

Common mistake avoided. Counting every integer from $-5$ through $22$ would wrongly treat unsupported readings as possible. The model lists six possible readings, not all the integers in that interval.

Equal Outcomes Do Not Always Mean Equal Values

A probability model may begin with equally likely outcomes of an experiment and then define a random variable from those outcomes. If two or more outcomes give the same value of the variable, their probabilities combine. Consequently, the distinct values of the variable need not be equally likely, even if the underlying outcomes are.

For example, suppose a fair device has five equally likely settings and a rule assigns a numerical value to each setting. If two settings both produce the value 1, then $P(X=1)$ is the probability of either setting, not the probability of just one setting. The support still lists 1 only once: a set contains distinct values, not repeated copies.

Worked Example: A Transformation Produces Unequal Probabilities

A fair five-position device has positions $a,b,c,d,e$, each with probability $1/5$. A recorded score $X$ is assigned as follows: $a\mapsto -2$, $b\mapsto 1$, $c\mapsto 1$, $d\mapsto 4$, and $e\mapsto 7$. Find the support and probability mass function of $X$, then calculate $P(X\geq 4)$ and $P(X=1)$. Is $X$ discrete uniform on its support?

Method and model. Start from the five equally likely device positions. The score is determined by the stated mapping; it is not assigned independently as a uniform choice from the distinct scores.

Support. The distinct scores are $S=\{-2,1,4,7\}$, so the support has four values. The value 1 occurs at two positions, while each of the other scores occurs at one position.

Probability mass function. Each position has probability $1/5$. Thus $$ P(X=-2)=\frac{1}{5},\qquad P(X=1)=\frac{1}{5}+\frac{1}{5}=\frac{2}{5},\qquad P(X=4)=\frac{1}{5},\qquad P(X=7)=\frac{1}{5}. $$ The probabilities sum to $1/5+2/5+1/5+1/5=1$.

Probability of a score at least 4. The qualifying support values are 4 and 7, each with probability $1/5$. Therefore $$ P(X\geq 4)=P(X=4)+P(X=7)=\frac{1}{5}+\frac{1}{5}=\frac{2}{5}=0.4000. $$

Probability of a score of 1. Two of the five positions produce that score, giving $$ P(X=1)=\frac{2}{5}=0.4000. $$

Conclusion and check. $X$ is not uniform on $\{-2,1,4,7\}$, because the value 1 has probability $2/5$, while each other support value has probability $1/5$. The four probabilities total 1, and the larger probability for 1 matches the two device positions that generate it.

Common mistake avoided. Assigning probability $1/4$ to each distinct score would incorrectly assume the scores themselves were selected uniformly. The stated fair mechanism selects positions uniformly, and two positions produce the same score.

Events and Conditional Probabilities on a Finite Support

For a uniform variable on $S$, an event is evaluated by finding the support values that meet its condition. A condition may combine several restrictions, such as “even and greater than 10.” A conditional probability uses the same idea after restricting attention to values consistent with the condition. The conditioning event must have positive probability.

Conditional-probability rule for a finite uniform support: If $X$ is uniform on $S$, and $B$ is an event with $B\cap S\neq\varnothing$, then $$ P(X\in A\mid X\in B) =\frac{|A\cap B\cap S|}{|B\cap S|}. $$ The numerator counts support values satisfying both conditions; the denominator counts support values satisfying the condition after the vertical bar.

Worked Example: A Conditional Probability With Several Restrictions

A hypothetical system selects one of the eight codes $0,3,8,11,20,26,35,41$, each with equal probability. Let $X$ be the selected code. Find (a) the probability that the code is even, (b) the probability that it exceeds 10, and (c) the probability that it is even given that it exceeds 10. Also identify an integer between the smallest and largest support values that cannot be selected.

Method and model. The support is $S=\{0,3,8,11,20,26,35,41\}$, with $m=8$. The selection is uniform over these eight distinct codes, so each has probability $1/8$.

(a) Even code. The even support values are $0,8,20,26$, giving four qualifying values: $$ P(X\text{ is even})=\frac{4}{8}=\frac{1}{2}=0.5000. $$

(b) Code exceeding 10. The qualifying values are $11,20,26,35,41$, giving five: $$ P(X>10)=\frac{5}{8}=0.6250. $$

(c) Even given greater than 10. Among $11,20,26,35,41$, the even values are $20$ and $26$. There are two qualifying values in the condition set of five, so $$ P(X\text{ is even}\mid X>10)=\frac{2}{5}=0.4000. $$ Equivalently, $P(X\text{ is even and }X>10)=2/8$, and dividing by $P(X>10)=5/8$ gives $(2/8)/(5/8)=2/5$.

Impossible value and check. The integer 15 lies between 0 and 41 but is not in $S$, so $P(X=15)=0$. The counts are consistent: there are eight support values in total, five above 10, and two that are both even and above 10.

Common mistake avoided. The conditional denominator is five, not eight: once we know $X>10$, only the five codes satisfying that condition remain possible.

Modeling Checks and Exam Traps

Before using a discrete uniform model, verify that the stated values are both exhaustive and equally likely. “One value is selected” is not enough information unless the selection rule or model specifies the probabilities. Also check whether the listed entries are distinct values or repeated outcomes of a mechanism. These distinctions determine the support and the probability mass function.

Do not fill gaps automatically. A support such as $\{2,7,13\}$ does not include 3 through 6 or 8 through 12.
Do not count a value twice in a support set. If a mechanism has repeated outcomes with the same numerical value, combine their probabilities for that value.
Do not use the counting formula for a nonuniform variable. For unequal probabilities, add the probability mass of each qualifying support value instead of dividing the number of values by the support size.
Check conditional denominators. A conditional probability counts only the support values allowed by the conditioning event.

A useful first step is to write the support explicitly, then mark which members satisfy each event. This prevents an interval description from being mistaken for a complete list of possible values. It also makes clear when an event has probability zero or one.

Key takeaway: A finite discrete uniform distribution gives the same probability to every distinct value in its support and probability zero to every value outside it. For any event, count only the support values that satisfy the event; if the values are not equally likely, use their actual probabilities instead.

Check Your Understanding

For each problem, first identify the stated support and use only attainable values in your calculation.

A random variable is uniform on $S=\{-6,0,5,12,19\}$. Find $P(-1\leq X\leq 15)$ and $P(X=8)$. (Answer: $3/5=0.6000$; $0$.)
A fair four-outcome device gives numerical values $2,2,6,10$, one value for each outcome. Find the support and the probability of each support value. Is the resulting variable uniform on its support? (Answer: Support $\{2,6,10\}$; probabilities $1/2,1/4,1/4$, respectively; no.)
A code is selected uniformly from $\{1,4,9,16,25,36\}$. Find the probability that the code is a perfect square greater than 8. (Answer: $4/6=2/3=0.6667$.)
A variable is uniform on $S=\{-3,2,6,11,18,24\}$. Find $P(X\geq 6\mid X\text{ is even})$. (Answer: $3/4=0.7500$.)
A fair five-outcome mechanism produces the values $0,3,3,8,12$. Find the probability of a value at least 8, and explain why dividing the number of qualifying distinct values by the number of distinct support values is not the correct calculation. (Answer: $3/5=0.6000$; the outcomes are equally likely but the distinct values are not, since 3 is produced twice.)

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