Residual Waiting Times After a Run of Failures

Introduction

A sequence of failures can feel as if success is overdue. In a model where each trial has the same success probability and trials are independent, however, the failures already observed do not change the distribution of the wait ahead. The remaining wait is measured from the next trial, not from the beginning of the sequence.

This distinction matters in applications such as hypothetical quality checks, repeated claims screening, or a sequence of independent equipment tests. After a run of failures, a question about the next few trials is conditional on what has already happened. A question about the trial number of the first success instead includes the elapsed trials. Both can be answered with the geometric model, but they are not the same quantity.

Definition: Suppose trials are independent, each with success probability \(p\), where \(0<p\leq 1\). Let \(T\) be the trial number of the first success. When \(p>0\), \(T\) is finite with probability 1 and has support \(\{1,2,3,\ldots\}\). Write \(q=1-p\). After \(r\) failures, define the remaining waiting time as \(W_r=T-r\), conditional on \(T>r\). It counts the trials beginning with the next trial and ending with the first subsequent success.

The restriction \(p>0\) is essential for this finite-valued geometric model. If \(p=0\), success never occurs, so \(T\) is not a positive-integer-valued random variable and its finite expectation formulas do not apply. For \(0<p<1\), the event \(T>r\) has positive probability for every finite nonnegative integer \(r\). If \(p=1\), then \(T=1\) with certainty: conditioning on \(r\) failures is possible only for \(r=0\), since one or more failures cannot occur.

Elapsed Trials and Remaining Trials

The event \(T>r\) means that the first \(r\) trials were all failures. Thus \(r\) is the number of elapsed trials, while \(W_r\) counts only the trials still to come. For example, after six failures, the next trial is the first remaining trial. If success occurs immediately, then \(W_6=1\), even though the success occurs on trial \(T=7\) of the full sequence.

For \(0<p<1\), the conditional probability that the first success occurs exactly \(k\) trials from now is

P(W_r=k\mid T>r) =\frac{P(T=r+k)}{P(T>r)} =\frac{q^{r+k-1}p}{q^r} =q^{k-1}p,\qquad k=1,2,3,\ldots.

The result does not depend on \(r\). The remaining wait has the same geometric distribution as the original wait. Equivalently, the probability of no success in the next \(m\) trials, given the first \(r\) were failures, is \(q^m\); the probability of at least one success in those next \(m\) trials is \(1-q^m\).

Why the Memoryless Formula Holds

The reason is that the probability of a particular future pattern factors into the probability of the observed past pattern and the probability of the future pattern. Independence allows the observed failures to cancel when we condition on them.

Write the conditioning event.
After \(r\) failures, the event is \(T>r\), with probability \(q^r\) when \(0<p<1\).

Specify a residual wait.
For the first success to be exactly \(k\) trials from now, there must be \(k-1\) further failures followed by a success. Together with the first \(r\) failures, this has probability \(q^{r+k-1}p\).

Condition and cancel the past.
Divide the joint probability by \(P(T>r)=q^r\). The result is \(q^{k-1}p\), which is the original geometric probability for a wait of \(k\) trials.

This argument also gives a useful conditional tail identity. For any nonnegative integer \(m\), \(T>r+m\) means there have been no successes in either the first \(r\) trials or the next \(m\) trials. Therefore

P(T>r+m\mid T>r) =\frac{q^{r+m}}{q^r} =q^m,\qquad 0<p<1.

In particular, “no success in the next \(m\) trials” is not the same as “success within the next \(m\) trials.” These are complementary events, with probabilities \(q^m\) and \(1-q^m\), respectively.

Expected Residual Waiting Time

Because \(W_r\) has the same distribution as \(T\), its mean is \(1/p\). This is the expected number of additional trials until success, regardless of how many failures have already occurred. The expected absolute trial number is different: \(T=r+W_r\) under the condition \(T>r\), so its conditional mean is \(r+1/p\).

E[W_r\mid T>r]=\frac{1}{p}, \qquad E[T\mid T>r]=r+\frac{1}{p}, \qquad 0<p<1.

A third quantity is sometimes useful: the number \(F_r=W_r-1\) of additional failures before the first subsequent success. Since one remaining trial is the success trial itself, \(F_r\) has support \(\{0,1,2,\ldots\}\) and

E[F_r\mid T>r] =E[W_r-1\mid T>r] =\frac{1}{p}-1 =\frac{q}{p}.

For the boundary case \(p=1\), after zero failures there is one remaining trial with certainty: \(W_0=1\), and there are zero additional failures. The conditional formulas above involving \(T>r\) are not used for \(r\geq1\), because that conditioning event has probability zero.

Worked Applications

Worked Example: Probability of Success Within a Fixed Residual Window

A hypothetical inspection process consists of independent checks, each passing with probability \(p=0.20\). The first six checks have failed. Find (a) the probability that the first pass is on the second check from now, (b) the probability of at least one pass in the next four checks, and (c) the expected number of additional checks until the first pass.

This is a geometric residual-wait problem with \(q=1-p=1-0.20=0.80\). The six failures are elapsed checks; they do not count as part of any “next” checks. No interest rate or mortality assumption applies: this is a hypothetical Bernoulli-trial model, not a financial or life-contingency calculation.

(a) Exactly two additional checks. The next check must fail and the one after it must pass:

P(W_6=2\mid T>6)=q^{2-1}p=(0.80)(0.20)=0.1600.

(b) At least one pass in the next four checks. First calculate the probability that all four checks fail, then take the complement:

P(\text{at least one pass in next 4}\mid T>6) =1-q^4 =1-(0.80)^4 =1-0.4096 =0.5904.

(c) Expected additional checks. The conditional residual wait has a geometric distribution with success probability \(0.20\), so

E[W_6\mid T>6]=\frac{1}{0.20}=5\text{ checks}.

As a check on the counting, the expected number of additional failures is \(q/p=0.80/0.20=4\); adding the success check gives five additional checks. A common mistake is to count the six failed checks as part of the next-four window. The conditioning describes the past, while the window begins at check seven.

Worked Example: A Wait With a Required Initial String of Failures

A hypothetical sequence of independent screening decisions has success probability \(p=0.30\) per decision. Eight decisions have failed. Find the probability that the next two decisions fail and the third succeeds, the probability of success within the next three decisions, and the expected decision number of the first success in the full sequence.

Here \(q=1-0.30=0.70\), and the event describing the eight earlier failures is \(T>8\). No interest rate or mortality assumption applies; the model is Bernoulli trials with fixed success probability \(0.30\).

Next two fail, then success. This is an exact residual waiting time of three decisions: two failures followed by a success.

P(W_8=3\mid T>8) =q^2p =(0.70)^2(0.30) =(0.49)(0.30) =0.1470.

Success within the next three decisions. The complement is three consecutive additional failures:

P(W_8\leq3\mid T>8) =1-q^3 =1-(0.70)^3 =1-0.343 =0.6570.

Expected absolute decision number. The expected additional wait is \(1/p=1/0.30=10/3\) decisions. The first eight failures have already occurred, so the expected full-sequence trial number is

E[T\mid T>8] =8+\frac{10}{3} =\frac{34}{3} =11.333\ldots\text{ decisions}.

The exact value is \(34/3\); reported to three decimal places, it is 11.333 decisions. A frequent error is to report \(10/3\) as the expected absolute decision number. That is the expected number of decisions still to come, not the trial index measured from the start.

Worked Example: Minimum Wait, Exact Wait, and Additional Failures

A hypothetical device produces independent test outcomes, with a successful outcome on each test with probability \(p=0.40\). Ten tests have failed. Find (a) the probability that at least four further tests are needed, (b) the probability that success occurs exactly on the fourth further test, and (c) the expected number of further failures before success.

Set \(q=1-0.40=0.60\). “At least four further tests are needed” means the first three further tests all fail; success on the fourth is allowed. No interest rate or mortality assumption applies; the stated probability model is independent Bernoulli trials with \(p=0.40\).

(a) At least four further tests. The condition is \(W_{10}\geq4\), which requires three failures at the start of the residual sequence:

P(W_{10}\geq4\mid T>10) =q^3 =(0.60)^3 =0.2160.

(b) Exactly four further tests. The first three must fail and the fourth must succeed:

P(W_{10}=4\mid T>10) =q^3p =(0.60)^3(0.40) =(0.216)(0.40) =0.0864.

(c) Further failures before success. The expected number of additional failures excludes the eventual success test:

E[F_{10}\mid T>10] =\frac{q}{p} =\frac{0.60}{0.40} =1.5\text{ failures}.

The expected number of further tests, including the success test, is \(1/p=2.5\). This checks the distinction: 1.5 expected failures plus one success test equals 2.5 expected tests. The common mistake is to use \(1/p\) as the expected number of failures; it counts trials, including the success.

Common Errors and Practical Checks

Residual-wait questions often combine conditioning, complements, and trial indexing. Before calculating, identify the origin of the count: does it begin at the first trial, or at the next trial after the observed failures? Then translate the event into the number of additional failures and whether a success trial is included.

Adding elapsed failures to a residual wait. After \(r\) failures, \(W_r\) counts from the next trial. Add \(r\) only when the question asks for the absolute trial index \(T\).
Confusing failures with trials. A wait of \(k\) trials consists of \(k-1\) failures followed by one success. The expected additional failures are \(q/p\); the expected additional trials are \(1/p\).
Using the wrong complement. “Success within the next \(m\)” has probability \(1-q^m\). “No success in the next \(m\)” has probability \(q^m\).
Conditioning on an impossible run. If \(p=1\), a run of one or more failures has probability zero, so a conditional probability given that run is undefined. For a finite geometric waiting time, require \(p>0\).

Key takeaway: Given \(r\) failures in independent trials with fixed success probability \(p>0\), the expected wait from the next trial is \(1/p\), not \(r+1/p\). The latter is the expected absolute trial number of the first success, conditional on those \(r\) failures.

The memoryless property is useful whenever the model genuinely has independent trials with an unchanged success probability. If the probability changes over time, or outcomes are dependent, the cancellation in the conditional-probability calculation no longer follows from this geometric model. The assumption is therefore part of the calculation, not merely background wording.

Check Your Understanding

For each problem, distinguish the observed failures from the residual sequence. Unless stated otherwise, trials are independent with the specified fixed success probability.

A hypothetical calibration procedure succeeds on each attempt with probability \(0.25\). After seven failures, find the probability of success within the next three attempts, the expected number of further attempts, and the expected number of further failures. (Answer: probability \(0.5781\); expected further attempts \(4\); expected further failures \(3\).)
A screening test has success probability \(0.10\) per independent test. After twelve failures, find the probability that the next five tests all fail and the probability of success within those five tests. Also give the expected further tests. (Answer: \(0.5905\), \(0.4095\), and \(10\) tests.)
For independent trials with \(p=0.60\), four failures have occurred. Find the probability that at least three further trials are needed, the probability success occurs exactly on the third further trial, and the expected further wait in trials. (Answer: \(0.1600\), \(0.0960\), and \(1.6667\) trials.)
A hypothetical review process has \(p=0.20\) per independent review. Nine reviews have failed. Find the probability of success within the next three reviews and the expected absolute review number of the first success. (Answer: \(0.4880\) and review number \(14\).)
Independent audit attempts succeed with probability \(p=0.35\). After six failures, find the probability that the next two attempts fail and the third succeeds, the expected number of further failures, and the expected number of further attempts. (Answer: \(0.1479\), \(1.8571\) failures, and \(2.8571\) attempts.)

What You'll Learn