Geometric Waiting Times and Support Conventions

Two Ways to Count the Same Wait

Suppose independent trials are repeated, each with success probability $p$, until the first success occurs. The waiting time is easy to describe but easy to count in two different ways: one can count the trial on which success occurs, or count only the failures that came before it. The resulting variables describe the same event in different units, but their supports begin at different integers.

For example, if the first trial is a success, the first-success trial number is 1, while the number of failures before success is 0. This one-trial difference is the central support convention to keep in view. In this tutorial, $p$ always denotes the probability of success on any one trial, and $1-p$ denotes the probability of failure.

Definition: Let $Z_1,Z_2,\ldots$ be independent Bernoulli trials with $P(Z_i=1)=p$ and $P(Z_i=0)=1-p$, where $Z_i=1$ represents success. The first-success trial number is $$ T=\min\{i\geq 1: Z_i=1\}. $$ The number of failures before the first success is $$ K=T-1. $$ When no success ever occurs, set $T=\infty$ and $K=\infty$. For $0<p<1$, these are geometric waiting-time variables under the two respective conventions.

The subscript starts at 1 because the first trial is trial 1, not trial 0. The failure count starts at 0 because it counts failures already observed before success: if success happens immediately, there have been no failures. Some sources use the letter $X$ for either convention. A symbol alone therefore does not identify the convention; the definition and support must be checked.

Supports and Parameter Range

For a discrete random variable, its support is the set of values that occur with positive probability. When $0<p<1$, every finite waiting time is possible: the first success might occur on any positive-integer trial. Equivalently, any nonnegative integer number of failures might precede it. Thus the two supports are shifted by one.

Theorem: If $0<p<1$, then $$ \mathcal{S}_T=\{1,2,3,\ldots\}, \qquad \mathcal{S}_K=\{0,1,2,3,\ldots\}. $$ The variables are related pointwise by $K=T-1$, so the support of $K$ is obtained by subtracting 1 from every value in the support of $T$.

To justify that these are the full supports, first consider any integer $t\geq 1$. The event $T=t$ requires failures on trials $1,\ldots,t-1$, followed by success on trial $t$. By independence, its probability is $(1-p)^{t-1}p$, which is positive because $0<p<1$. No value less than 1 can be a trial number. Thus $T$ has exactly the stated support. Since $K=T-1$, each integer $k\geq 0$ corresponds to $T=k+1$, and has positive probability; no negative failure count is possible. This proves the theorem. $\square$

The restriction $0<p<1$ gives the nondegenerate, infinite supports above. At $p=1$, success occurs on the first trial with certainty, so $T=1$ and $K=0$ with probability 1; each support is a singleton. At $p=0$, no success occurs, so under the extended definitions $T=K=\infty$ with probability 1. Many definitions of a geometric distribution restrict the parameter to $0<p\leq 1$, thereby excluding this no-success case. In the usual nondegenerate geometric model, $0<p<1$ is assumed.

A useful check is to ask what happens on the earliest possible success. If the model says “the number of trials until the first success,” the minimum is 1. If it says “the number of failures before the first success,” the minimum is 0. A claimed support beginning at 0 for the former, or at 1 for the latter, signals a convention mismatch.

Worked Example: Naming the Variable and Its Support

A quality-control system tests one item at a time. Each item independently passes with probability $p=0.92$. Let $T$ be the test number of the first item that passes. State the convention, support, and corresponding failure-count variable.

The counted quantity is the trial number: if the first tested item passes, then $T=1$. Since $0<0.92<1$, any positive-integer trial number is possible, so $$ \mathcal{S}_T=\{1,2,3,\ldots\}. $$

If instead the question records the number $K$ of failed items before the first pass, then $K=T-1$, and $$ \mathcal{S}_K=\{0,1,2,3,\ldots\}. $$ For instance, a pass on test 4 corresponds to $T=4$ and $K=3$, not $K=4$. The pass probability is unchanged by the choice of convention; only the recorded count changes.

Translate Values and Events Carefully

The pointwise identity $K=T-1$ gives a direct translation for every observed wait. If the first success occurs on trial $t$, exactly $t-1$ trials have failed beforehand. Conversely, if $k$ failures occur before success, success is on trial $k+1$. This correspondence is one-to-one: no value in either convention is left unmatched.

Result: For each integer $t\geq 1$, the event that the first success is on trial $t$ is exactly the event that there are $t-1$ failures before the first success: $$ \{T=t\}=\{K=t-1\}. $$ More generally, for any integer $m\geq 1$, $$ \{T\leq m\}=\{K\leq m-1\}, \qquad \{T>m\}=\{K\geq m\}. $$

For the first equality, $T\leq m$ means success occurs on one of trials $1,\ldots,m$. Since $K=T-1$, this happens exactly when $K\leq m-1$. Negating both events gives $T>m$ exactly when $K>m-1$. Because $K$ is integer-valued, $K>m-1$ is the same as $K\geq m$. The same translation works for probabilities: equal events have equal probabilities. $\square$

These translations are especially useful for phrases such as “by the fourth trial” or “after three failures.” “By the fourth trial” corresponds to $T\leq 4$, or $K\leq 3$. “More than four trials are needed” corresponds to $T>4$, or $K\geq 4$. Writing the inequality before substituting numbers helps keep the boundary correct.

Worked Example: Translating a Deadline

A laboratory repeats a test independently until it obtains a positive result. Let $T$ be the trial of the first positive result, and let $K$ count negative results before that positive result. Express the event “the first positive result occurs within the first five tests” using both variables.

“Within the first five tests” includes tests 1, 2, 3, 4, and 5, so with the trial-number convention the event is $T\leq 5$. The corresponding failure count is one less than the successful trial number. Therefore the same event is $$ \{T\leq 5\}=\{K\leq 4\}. $$

The endpoint checks the translation: a positive result on test 5 means four preceding negative results, so it is included by both inequalities. A positive result on test 6 means five preceding negatives, so it is excluded by both. The phrase “within five tests” does not mean $K\leq 5$; that would also include a positive result on test 6.

Worked Example: Converting a Recorded Count

A call center records the number of unanswered calls before the first answered call. For one customer, the recorded count is $K=2$. Convert this observation to the trial-number convention, and identify the support of each variable if the probability of an answered call on each call is $p=0.30$, independently.

Two unanswered calls are followed by the first answered call, so that answer is on call $$ T=K+1=2+1=3. $$ Because $0<0.30<1$, all nonnegative integers are possible for $K$, while all positive integers are possible for $T$: $$ \mathcal{S}_K=\{0,1,2,\ldots\}, \qquad \mathcal{S}_T=\{1,2,3,\ldots\}. $$

If the first call had been answered, the recorded failure count would have been $K=0$, but the trial number would still be $T=1$. This boundary case confirms why adding one is necessary when converting failures to the trial of success.

Why the Convention Matters

Both variables describe a wait in the same repeated-trial model, but numerical statements about them are not interchangeable without translation. A threshold, a reported observation, or a formula must be interpreted with its support. This is particularly important when comparing answers from different sources: one may call the trial-number version “geometric,” while another uses that name for the failures-before-success version. Their random variables differ by one even though they use the same success probability.

The support also provides an immediate diagnostic. A probability assigned to $T=0$ is incompatible with a trial number beginning at 1. A probability assigned to $K=-1$ is incompatible with a count of failures. And if $p=1$, treating every positive integer as a possible first-success trial would ignore the degenerate boundary case.

Common Mistakes

Starting the trial number at zero. The first trial is numbered 1, so the first-success trial number cannot be 0.
Counting the success as a failure. The failures-before-success variable counts only the unsuccessful trials preceding the first success; immediate success gives $K=0$.
Using the same support for both conventions. The trial count starts at 1 and the failure count starts at 0; their supports differ by a shift of one.
Translating a threshold without checking equality. $T\leq m$ corresponds to $K\leq m-1$, not $K\leq m$. Test the boundary case of success exactly on trial $m$.
Assuming the name alone settles the convention. State what is counted and list the support whenever ambiguity is possible.

Key takeaway: Define the variable by what it counts. If $T$ is the trial number of the first success and $K$ is the number of failures before it, then $K=T-1$; for $0<p<1$, their supports are respectively the positive integers and the nonnegative integers.

Check Your Understanding

For each question, distinguish the trial-number convention from the failures-before-success convention.

A first success occurs on trial 7. What value does each convention record?
For $0<p<1$, state the support of a variable counting failures before the first success. Why is 0 included?
Rewrite the event $T\leq 6$ using the number $K$ of failures before the first success.
If $p=1$, what are the values of $T$ and $K$, and what are their supports?
A report says “the number of trials required is 0.” Explain why this cannot be a first-success trial number, and give a related count for which 0 is possible.