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Consider successive independent Bernoulli trials with the same success probability $0<p\le 1$. A geometric distribution models waiting for their first success. Two conventions coexist, so always say what is counted. Let $T$ count all trials up to and including the successful trial, and let $F=T-1$ count preceding failures. Their probability mass functions areThe factors describe failures followed by one success; independence permits multiplication. Summing the geometric series verifies that each law has total mass one. Their expected values differ by one, while shifting by a constant leaves the variance unchanged:The positive convention and these moments are also given by Penn State STAT 414.^[ref]

For a constructed retry mechanism with independent attempts and $p=0.3$, success on attempt four has probability $\mathbb{P}(T=4)=0.7^3(0.3)=0.1029$. More than four attempts are needed precisely when the first four fail: $\mathbb{P}(T>4)=0.7^4=0.2401$. The expected values are $\mathbb{E}T=10/3$ and $\mathbb{E}F=7/3$, and the common variance is $70/9\approx 7.7778$. The large spread explains why the average waiting count is not a deadline.

The two probability mass functions have identical heights with different labels: $T=4$ and $F=3$ describe the same event. For $0<p<1$, consecutive masses have ratio $1-p$, explaining the name geometric distribution. The support of a probability distribution is the positive integers for $T$ and the nonnegative integers for $F$. At $p=1$, it collapses to $T=1$ and $F=0$, both with variance zero; interpret an empty product, including $(1-p{)}^0$, as one. These counts concern attempts; converting them into elapsed time requires specifying the time spent per attempt.

The survival function makes the counting convention especially visible. Put $q=1-p$. For nonnegative integers $n$, and for $k\ge 1$,The event $T>n$ requires $n$ failures, whereas $T\ge n$ requires only $n-1$ when $n\ge 1$. For the constructed $p=0.3$ mechanism, $\mathbb{P}(T\ge 4)=0.343$ differs from $\mathbb{P}(T>4)=0.2401$. The geometric success deadline uses the latter tail to control the chance of still waiting.

The striking memorylessness of the geometric distribution says that, for nonnegative integers $m,r$ with $\mathbb{P}(T>m)>0$,After five failures, the conditional probability of requiring more than two additional attempts is still $0.7^2=0.49$. The analogous zero-based identity uses $\mathbb{P}(F\ge m+r\mid F\ge m)=\mathbb{P}(F\ge r)$; keeping the inequalities straight avoids an off-by-one error. This property follows from fresh independent trials with unchanged $p$; learning or changing conditions can destroy it. At $p=0$, success never occurs, so there is no proper finite waiting-count law.