= Memorylessness of the geometric distribution
{title2=$\mathbb P(T>m+r\mid T>m)=\mathbb P(T>r)$}
For the positive-integer <geometric distribution> with success <probability> $0<p\leq1$, the displayed identity holds for nonnegative <integers> $m,r$ whenever the conditioning <event> has positive <probability>. The first $m$ attempts have failed; future <independent> attempts use the same success <probability>. Equivalently, divide the tail $(1-p)^{m+r}$ by $(1-p)^m$. For a zero-based failures count $F=T-1$, use $\mathbb P(F\geq m+r\mid F\geq m)=\mathbb P(F\geq r)$. Constant success <probability> and <independence> are essential: accumulated wear or learning would require a different model.
The full remaining-count law is unchanged, not only one tail <probability>. With $0<p<1$, after $m$ failures set $R=T-m$ on the <event> $T>m$. For $r=1,2,\ldots$, its <probability mass function> under this <conditional probability> is
$$
\mathbb P(R=r\mid T>m)=\frac{q^{m+r-1}p}{q^m}=q^{r-1}p.
$$
Here <independence> makes the future trials unaffected by the observed failures, and the unchanged success <probability> makes their law identical to the original one. Consequently $\mathbb E[T-m\mid T>m]=1/p$, while $\mathbb E[T\mid T>m]=m+1/p$. For the constructed $p=0.3$ mechanism, five failures leave a <mean> of $10/3$ more attempts, making the conditional <mean> total $25/3$; the five already spent are still counted.
\Image[/geometric-distribution-conditional-residual.png]
{title=Conditioning on five failures, renormalizing, and resetting the count gives the original geometric law}
The same mechanism has constant <discrete hazard>:
$$
\mathbb P(T=k\mid T\geq k)=\frac{q^{k-1}p}{q^{k-1}}=p,\qquad k\geq1.
$$
Thus each reached attempt has the same chance of success. This differs from the decreasing unconditional <probability> $\mathbb P(T=k)$: later success requires surviving earlier attempts. At $p=1$, only conditioning on an <event> of positive <probability> is allowed: $T>m$ permits only $m=0$, and $T\geq k$ only $k=1$.
Among proper laws on the positive <integers>, <memorylessness of the geometric distribution> also characterizes the <geometric distribution>. Write $S(n)=\mathbb P(T>n)$. The identity gives $S(m+r)=S(m)S(r)$; if $S(m)=0$, both sides vanish anyway. Since $S(0)=1$, <induction> yields $S(n)=S(1)^n$. Properness excludes $S(1)=1$, so setting $p=1-S(1)>0$ recovers the <geometric distribution>, including $p=1$.<Random: Probability, Mathematical Statistics, Stochastic Processes by Kyle Siegrist/The Geometric Distribution>[Random: Probability, Mathematical Statistics, Stochastic Processes by Kyle Siegrist; Basic Theory, Distribution Functions and the Memoryless Property: conditional remaining-count identity and its converse]{ref}
The retry example stipulates <independent> <Bernoulli trials> with a known constant $p$; it is not evidence that a real retry system behaves this way. Shared changing conditions or adapting attempts can alter the next success <probability>. Even constant individual success <probabilities> alone do not imply <independence>.
For an original contrasting model, uniformly shuffle one success card and three failure cards, then draw by <sampling without replacement>. Its success position $U$ has the <discrete uniform distribution> on $\{1,2,3,4\}$. Initially the success <probability> is $1/4$; after one failure it is $1/3$. In particular,
$$
\mathbb P(U>2\mid U>1)=\frac{2/4}{3/4}=\frac23
\ne\frac34=\mathbb P(U>1).
$$
This shows concretely why <sampling without replacement need not be independent> and why the remaining wait can change. Every draw still has unconditional success <probability> $1/4$ when all four positions are considered; the relevant <discrete hazard> changes after failures.
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