Memorylessness of the Geometric Distribution

Why Earlier Failures Do Not Change the Model

Suppose a process is repeated until the first success: a device is tested repeatedly, for example, or a customer is contacted until one accepts an offer. After several failures, it is natural to wonder whether success is now “due.” In a geometric waiting-time model, it is not. The chance of success on each future trial remains the same, and the distribution of the additional wait is unchanged.

This feature is called memorylessness. It is not a general fact about waiting times. It follows from the specific assumptions of the geometric model: trials are independent, and each trial has the same probability of success.

Model and definition: Suppose trials are independent, each with success probability $p$, where $0<p<1$. Let $T$ be the number of trials up to and including the first success, so $T$ takes values $1,2,3,\ldots$. Write $q=1-p$ for the failure probability. The geometric waiting time is memoryless if, for every pair of nonnegative integers $m$ and $k$, $$ P(T>m+k\mid T>m)=P(T>k). $$ Here $T>m$ means that the first $m$ trials were all failures.

The event $T>m+k$ says there were no successes in the first $m+k$ trials. Given that $T>m$, it asks for $k$ more consecutive failures. The equality says that the chance of those additional failures is just what it would have been at the start. Equivalently, the number of additional trials until success, conditional on $T>m$, has the same distribution as $T$.

Deriving the Memoryless Property

In the geometric model, the probability of no success in the first $j$ trials is $q^j$. This follows because each of the $j$ trials fails with probability $q$, and the trials are independent. In particular, $P(T>j)=q^j$.

Write the conditional probability.
Since $T>m+k$ implies $T>m$, the definition of conditional probability gives $P(T>m+k\mid T>m)=P(T>m+k)/P(T>m)$.

Substitute the geometric tail probabilities.
The numerator is $q^{m+k}$, and the denominator is $q^m$. Since $q>0$, the denominator is positive.

Simplify.
Dividing gives $q^{m+k}/q^m=q^k$, which is $P(T>k)$. Thus the conditional tail probability equals the original tail probability.

P(T>m+k\mid T>m) =\frac{P(T>m+k)}{P(T>m)} =\frac{q^{m+k}}{q^m} =q^k =P(T>k).

The same conclusion holds for the entire remaining-wait distribution, not only for probabilities of waiting longer than $k$ trials. Let $R=T-m$ be the number of additional trials until the first success, given that the first $m$ trials failed. For any integer $r\geq 1$,

\begin{aligned} P(R=r\mid T>m) &=P(T=m+r\mid T>m)\\ &=\frac{P(T=m+r)}{P(T>m)}\\ &=\frac{q^{m+r-1}p}{q^m}\\ &=q^{r-1}p. \end{aligned}

The formula $q^{r-1}p$ is the geometric probability of a first success on trial $r$: the first $r-1$ trials fail and the next succeeds. Therefore, conditional on any specified number of initial failures, the additional wait $R$ is geometric with the same success probability $p$. This is the distributional meaning of memorylessness.

Worked Examples

Worked Example: Success Within a Few More Trials

Imagine a hypothetical independent contact process in which each contact has a $20\%$ chance of success. Given four unsuccessful contacts, find the probability of at least one success in the next three contacts. Check the model conditions: the scenario specifies independent trials and a constant success probability $p=0.20$, so a geometric model applies. The failure probability is $q=1-0.20=0.80$.

It is easiest to find the complement, that all three additional contacts fail. By memorylessness, or directly from independence, the conditional probability of three more failures is

P(T>4+3\mid T>4) =\frac{0.80^7}{0.80^4} =0.80^3 =0.512.

Thus the probability of at least one success in those three contacts is

$$ 1-0.512=0.488. $$

Given the four earlier failures, the chance of at least one success in the next three contacts is $0.488$, or $48.8\%$. The four failures do not increase that probability; it is the same as the probability of success within the first three contacts.

Worked Example: First Success on the Second Additional Trial

Suppose each trial in a hypothetical inspection process independently finds a defect with probability $p=0.30$. After five trials with no defect found, calculate the probability that the first defect is found on the second additional trial. The trials meet the geometric assumptions because their success probability is constant and they are independent. Here $q=0.70$.

The first additional trial must fail, and the second additional trial must succeed. The conditional probability is

P(R=2\mid T>5)=q^{2-1}p=(0.70)(0.30)=0.210.

The same result follows by forming a ratio using the original waiting time. First, $T=7$ requires six failures followed by a success, so $P(T=7)=0.70^6(0.30)$. Also, $P(T>5)=0.70^5$. Therefore,

P(T=7\mid T>5) =\frac{0.70^6(0.30)}{0.70^5} =(0.70)(0.30) =0.210.

After five failures, the probability that the first success occurs on the second additional trial is $0.210$. It equals the probability that the first success occurs on trial 2 from the beginning.

Worked Example: Checking When Memorylessness Fails

Consider a hypothetical process where the first trial succeeds with probability $0.20$, but every later trial succeeds with probability $0.50$. Assume the trials are independent. This does not satisfy the constant-probability condition for a geometric model. Compare the chance of success on the next trial at the start with the chance after the first trial fails.

At the start, the probability of success on the first trial is $0.20$. Conditional on that trial failing, the next trial has success probability $0.50$, by the stated trial probabilities and independence. These differ:

P(\text{success on trial 1})=0.20 \qquad\text{but}\qquad P(\text{success on trial 2}\mid\text{trial 1 fails})=0.50.

For another comparison, the probability of no success in the first two trials is $(0.80)(0.50)=0.40$, while the probability of failure on the first trial is $0.80$. Hence

P(T>2\mid T>1)=\frac{0.40}{0.80}=0.50, \qquad P(T>1)=0.80.

The two probabilities are not equal, so the waiting time in this process is not memoryless. Independence alone is not enough: the success probability must also remain constant across trials.

What Memorylessness Does—and Does Not—Say

Memorylessness concerns the probability distribution of the additional wait, conditional on having reached a point with no success. It does not say that the total number of trials already spent disappears, or that success becomes more likely after failures. If $m$ trials have failed, the total wait is at least $m+1$, but the conditional distribution of the additional wait is still geometric with parameter $p$.

The result is useful whenever a process really does consist of repeated, independent trials with an unchanged success probability. It allows conditional questions after a run of failures to be answered using the same geometric probabilities as at the beginning. It also provides a check on model choice: if the probability changes with time, past outcomes reveal relevant information about the future, or trials are dependent, the memoryless conclusion need not hold.

Common Mistake

A common mistake is to argue, “We have already had many failures, so success is more likely on the next trial.” That reasoning treats the sequence as though it must compensate for its past. Under the geometric model, it does not: independence and the constant value of $p$ imply that the next trial still has success probability $p$. Conversely, it is also a mistake to apply memorylessness just because a process involves repeated attempts. First verify that the attempts are independent and have the same success probability.

Key takeaway: For a geometric waiting time $T$, conditioning on $m$ initial failures leaves the additional wait with the same geometric distribution: $P(T-m=r\mid T>m)=q^{r-1}p$ for $r\geq1$. This property follows from independent trials with a constant success probability.

Check Your Understanding

Answer each question using the geometric model where its assumptions are stated.

State the two key trial assumptions needed for a geometric waiting time to be memoryless.
If $p=0.10$, what is the probability of no success in the next two trials, given that the first eight trials failed?
For $p=0.25$, find the probability that the first success occurs on the third additional trial after $m$ failures.
Explain why an increasing success probability from trial to trial can invalidate memorylessness, even if trials are independent.
In your own words, explain why previous failures do not make success “due” under the geometric model.