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VEE Mathematical Statistics · Geometric and Negative Binomial Models · Tutorial 120 of 880

How Success Probability Changes Waiting-Time Distributions

Learn which features of a waiting-time distribution move consistently as success becomes more likely—and which exact-time probabilities can move either way.

Intermediate 13 min read

What You'll Learn

  • Compare geometric and negative binomial waiting times under different success probabilities.
  • Calculate how means and variances change as the success probability increases.
  • Use a coupling argument to show why completion times become stochastically shorter.
  • Find and compare waiting-time quantiles, including geometric medians.
  • Distinguish monotone deadline probabilities from exact-completion probabilities.

Introduction

A larger chance of success on each trial should make a target easier to reach. For geometric and negative binomial waiting times, this intuition has precise consequences: completion times become stochastically shorter, their means decrease, and their quantiles cannot increase. But one tempting extension is false: the probability of completing on one particular trial does not have to move in the same direction. Keeping these distinctions clear is useful whenever a process consists of repeated independent attempts.

Throughout, assume trials are independent, each trial has the same success probability \(p\), and \(0<p<1\). The target is fixed while \(p\) changes. These assumptions matter: if the trial probabilities change over time or trials are dependent, the formulas and comparisons below need not apply.

Definition: Let \(T_r\) be the number of trials required to obtain \(r\) successes, where \(r\) is a fixed positive integer. Then \(T_1\) has the geometric distribution, and \(T_r\) has the negative binomial waiting-time distribution. The support is \(1,2,\ldots\) for \(T_1\), and \(r,r+1,\ldots\) for \(T_r\). For \(t\ge r\), $$ P(T_r=t)=\binom{t-1}{r-1}p^r(1-p)^{t-r}. $$

The final trial in \(T_r=t\) must be a success. Among the first \(t-1\) trials, exactly \(r-1\) must be successes; the binomial coefficient counts the possible arrangements. For \(r=1\), the formula reduces to \(P(T_1=t)=p(1-p)^{t-1}\), the geometric probability mass function.

What Increasing \(p\) Changes

For a fixed target \(r\), the mean and variance of the waiting time are

$$ E[T_r]=\frac{r}{p}, \qquad \operatorname{Var}(T_r)=\frac{r(1-p)}{p^2}. $$

Thus, increasing \(p\) decreases the mean. It also decreases the variance: \((1-p)/p^2=1/p^2-1/p\) has derivative \((p-2)/p^3\), which is negative for \(0<p<1\). These are statements about the expected waiting time and its spread, not a claim that every individual probability in the mass function decreases.

For a deadline of \(n\) trials, completion by that time means at least \(r\) successes among the first \(n\) trials. If \(B_n\) is the number of successes in those trials, then \(B_n\sim\operatorname{Bin}(n,p)\), and

$$ P(T_r\le n)=P(B_n\ge r) =\sum_{j=r}^{n}\binom{n}{j}p^j(1-p)^{n-j}. $$

When \(n<r\), this probability is zero. The deadline probability increases with \(p\), as the coupling argument below establishes. A waiting-time quantile also moves in a predictable direction. For \(0<\alpha<1\), define the \(\alpha\)-quantile by

$$ q_\alpha(p)=\min\{n\in\{r,r+1,\ldots\}:P(T_r\le n)\ge\alpha\}. $$

Because it is the first trial number at which the cumulative probability reaches \(\alpha\), this quantile cannot increase as \(p\) increases. It may remain the same across a range of probabilities because the trial count is integer-valued.

Why the Completion Time Becomes Shorter

A direct coupling makes the comparison precise. For each trial \(i\), take an independent random variable \(U_i\) uniformly distributed on \((0,1)\). At success probability \(p\), call trial \(i\) a success when \(U_i\le p\). This construction has the required independent Bernoulli trials for every fixed \(p\).

1
Use the same trial values.
Compare two processes with probabilities \(p_1<p_2\), using the same sequence \(U_1,U_2,\ldots\) for both.
2
Compare successes trial by trial.
If \(U_i\le p_1\), then \(U_i\le p_2\) as well. Every success under \(p_1\) is also a success under \(p_2\).
3
Compare completion times.
The process with probability \(p_2\) has reached any fixed target \(r\) no later than the process with probability \(p_1\). Hence \(T_r(p_2)\le T_r(p_1)\) for every outcome in this construction.

In particular, \(P(T_r(p_2)\le n)\ge P(T_r(p_1)\le n)\) for every deadline \(n\). This is stochastic ordering: a larger success probability gives a waiting time that is stochastically smaller. It explains both the increase in deadline probabilities and the nonincrease in waiting-time quantiles.

For a geometric waiting time, the quantile can also be calculated directly. Since \(P(T_1>n)=(1-p)^n\), the smallest integer \(n\) with \(P(T_1\le n)\ge\alpha\) is

$$ q_\alpha(p)= \left\lceil\frac{\log(1-\alpha)}{\log(1-p)}\right\rceil. $$

Both logarithms are negative, so their ratio is positive. The ceiling is essential because the waiting time must be an integer number of trials.

Worked Comparisons

Worked Example: A Geometric Wait at Two Success Probabilities

A laboratory procedure is repeated independently until its first successful result. Compare \(p=0.20\) with \(p=0.40\). For each model, find the probability that the first success occurs exactly on trial 3, the probability it occurs by trial 4, the mean waiting time, and the median.

Method. Use the geometric mass function for exact completion, the geometric cumulative probability for completion by a deadline, and the quantile formula for the median. The common mistake to avoid is assuming that the exact-trial probability and the by-deadline probability must change in the same way.

At \(p=0.20\), the probability of first success on trial 3 is

$$P(T_1=3)=(0.20)(0.80)^2=0.20(0.64)=0.1280.$$

The probability of success by trial 4 is

$$P(T_1\le4)=1-(0.80)^4=1-0.4096=0.5904.$$

The mean is \(1/p=1/0.20=5\) trials. The median is

$$q_{0.5}=\left\lceil\frac{\log(0.5)}{\log(0.80)}\right\rceil =\lceil3.1063\ldots\rceil=4.$$

Checking the boundary, \(P(T_1\le3)=1-0.80^3=0.4880<0.5\), whereas \(P(T_1\le4)=0.5904\ge0.5\).

At \(p=0.40\), the corresponding exact-trial probability is

$$P(T_1=3)=(0.40)(0.60)^2=0.40(0.36)=0.1440.$$

The probability of success by trial 4 is \(1-(0.60)^4=1-0.1296=0.8704\), and the mean is \(1/0.40=2.5\) trials. The median is

$$q_{0.5}=\left\lceil\frac{\log(0.5)}{\log(0.60)}\right\rceil =\lceil1.3569\ldots\rceil=2.$$

Indeed, \(P(T_1\le1)=0.40<0.5\), while \(P(T_1\le2)=1-0.60^2=0.64\ge0.5\). Increasing \(p\) raises the probability of completion by trial 4 and lowers both the mean and median. Here it also raises the probability of completion exactly on trial 3; that last change is not guaranteed in general.

Worked Example: A Three-Success Target and a Deadline

A quality-control sequence continues until three acceptable units have been found. Compare success probabilities \(p=0.25\) and \(p=0.50\). Calculate the mean and variance, the probability of completing by trial 8, and the median waiting time for each probability.

Method. Apply the negative binomial moment formulas. For the deadline probability and median, count successes in the stated number of trials and use the binomial distribution. The common trap is to confuse the negative binomial waiting time with the number of successes by the deadline.

For \(p=0.25\), the mean and variance are

$$E[T_3]=\frac{3}{0.25}=12,\qquad \operatorname{Var}(T_3)=\frac{3(1-0.25)}{(0.25)^2} =\frac{2.25}{0.0625}=36.$$

To finish by trial 8 requires at least three successes in eight trials. Subtract the probabilities of zero, one, or two successes:

$$ \begin{aligned} P(T_3\le8) &=1-\left[(0.75)^8+8(0.25)(0.75)^7+\binom82(0.25)^2(0.75)^6\right]\\ &=1-(0.1001129+0.2669678+0.3114624)\\ &=0.3214569\approx0.3215. \end{aligned} $$

To find the median, at trial 10 the probability of having at least three successes is \(1-(0.0563135+0.1877117+0.2815676)=0.4744072<0.5\). At trial 11 it is \(1-(0.0422351+0.1548622+0.2581036)=0.5447991\ge0.5\). Thus the median is 11 trials.

For \(p=0.50\), the mean and variance are

$$E[T_3]=\frac{3}{0.50}=6,\qquad \operatorname{Var}(T_3)=\frac{3(1-0.50)}{(0.50)^2} =\frac{1.5}{0.25}=6.$$

By trial 8, the probability of at least three successes is

$$ P(T_3\le8)=1-\frac{\binom80+\binom81+\binom82}{2^8} =1-\frac{1+8+28}{256} =\frac{219}{256}=0.85546875\approx0.8555. $$

For the median, \(P(T_3\le4)=\binom43/16+\binom44/16=5/16=0.3125\), while \(P(T_3\le5)=(10+5+1)/32=0.5\), using the probabilities of three, four, or five successes. Therefore the median is 5 trials. The larger success probability lowers the mean from 12 to 6, lowers the variance from 36 to 6, raises the deadline probability, and lowers the median from 11 to 5.

Worked Example: Why an Exact-Trial Probability Can Decrease

A system is monitored until two successful checks have occurred. Compare \(p=0.40\) with \(p=0.60\). Find the probability of completion exactly on trial 5, the mean and variance, and the median for each model.

Method. Evaluate the negative binomial mass function at \(t=5\), then compare moments and medians. The common mistake is to infer that every exact-completion probability increases when \(p\) increases.

For \(r=2\), completion exactly on trial 5 requires one success in the first four trials and a success on trial 5:

$$P(T_2=5)=\binom41p^2(1-p)^3.$$

At \(p=0.40\), this is \(4(0.16)(0.216)=0.13824\). At \(p=0.60\), it is \(4(0.36)(0.064)=0.09216\). Thus the exact-trial probability decreases, despite the higher success probability.

At \(p=0.40\), the mean is \(2/0.40=5\) and the variance is \(2(0.60)/(0.40)^2=1.20/0.16=7.5\). For the median, \(P(T_2\le3)=P(\operatorname{Bin}(3,0.40)\ge2)=3(0.16)(0.60)+0.064=0.352<0.5\). At trial 4, the probability is \(1-(0.60)^4-4(0.40)(0.60)^3=1-0.1296-0.3456=0.5248\). The median is 4.

At \(p=0.60\), the mean is \(2/0.60=3.3333\ldots\) and the variance is \(2(0.40)/(0.60)^2=0.80/0.36=2.2222\ldots\). At trial 2, completion probability is \(0.60^2=0.36<0.5\). At trial 3, it is \(3(0.60)^2(0.40)+(0.60)^3=0.432+0.216=0.648\). The median is 3. The moments and median become smaller, but the mass at exactly trial 5 also becomes smaller. This is consistent: the overall distribution shifts earlier, while probability is redistributed among particular trial numbers.

Common Mistake: Treating Every Mass as Monotone

The coupling argument proves that cumulative probabilities \(P(T_r\le n)\) increase with \(p\), and the mean and variance formulas show that both moments decrease. It does not prove that \(P(T_r=t)\) increases for every \(t\), or that it decreases for every \(t\). The probability at a particular trial reflects a balance between having enough successes by that time and not having completed earlier.

For example, the incorrect inference “\(p\) rises from \(0.40\) to \(0.60\), so \(P(T_2=5)\) must rise” would overlook the three required failures among the first four trials. The correct calculation gives \(0.13824\) at \(p=0.40\) and \(0.09216\) at \(p=0.60\). Always identify whether the requested event is exact completion, completion by a deadline, or a quantile; these are different features of the distribution.

Key takeaway: For a fixed target and independent trials with constant success probability, increasing \(p\) makes the geometric or negative binomial waiting time stochastically shorter. Its mean, variance, and quantiles decrease or stay unchanged, while deadline probabilities increase. Exact-trial probabilities need not follow either monotone pattern.

Check Your Understanding

For each problem, assume independent trials with the stated constant success probability. Check whether your result distinguishes exact completion from completion by a deadline.

  1. A device is tested until its first successful calibration. Compare \(p=0.10\) and \(p=0.20\). Find the mean waiting time and the probability of success by trial 3 for each. (Answer: Means 10 and 5 trials; deadline probabilities 0.2710 and 0.4880.)
  2. A geometric waiting time has success probability \(0.30\). Find its median, and compare it with the median when \(p=0.50\). (Answer: The medians are 2 trials and 1 trial, respectively.)
  3. A process requires two successes. Compare the mean and variance at \(p=0.20\) and \(p=0.40\). (Answer: At \(p=0.20\), mean \(10\), variance \(40\); at \(p=0.40\), mean \(5\), variance \(7.5\).)
  4. A process requires three successes. Find \(P(T_3=4)\) at \(p=0.30\) and \(p=0.60\). Does the exact-trial probability increase? (Answer: The probabilities are \(3(0.30)^3(0.70)=0.0567\) and \(3(0.60)^3(0.40)=0.2592\); it increases in this comparison.)
  5. A process requires two successes. Find the median waiting time at \(p=0.25\) and \(p=0.50\). For \(p=0.25\), verify the cumulative probability on either side of the median. (Answer: The medians are 7 trials and 3 trials, respectively; at \(p=0.25\), \(P(T_2\le6)=0.4661<0.5\) and \(P(T_2\le7)=0.5551\ge0.5\).)

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