Common Geometric Model Errors: Indexing, Independence, and Fixed p

Why Geometric Calculations Go Wrong

A waiting-time problem can look geometric while failing to satisfy the geometric model. Sometimes the model is appropriate but the count is shifted: “success on trial 3” is not the same event as “three failures before success.” In other cases, the success chance changes from trial to trial, or the outcomes are dependent. In either situation, mechanically applying a geometric formula can give a plausible but incorrect answer.

The reliable approach is to define the waiting-time variable, state what is assumed about the trials, and then write the event in terms of the actual sequence of successes and failures. The distinction matters in applications such as repeated hypothetical claim reviews, inspections, or screening attempts. A geometric model is aligned with the SOA VEE Mathematical Statistics syllabus when its assumptions and indexing are handled correctly.

Definition: Let \(T\) be the number of the trial on which the first success occurs. In the standard geometric model, trials are independent and each has the same success probability \(p\), with \(0<p<1\). Thus \(T\) takes values \(1,2,3,\ldots\), and \(P(T=k)=(1-p)^{k-1}p\). If instead \(F\) counts the number of failures before the first success, then \(F=T-1\), so \(F\) takes values \(0,1,2,\ldots\).

First Check the Index

The word “waiting time” does not determine the variable’s support by itself. If the random variable counts trials up to and including the first success, its smallest possible value is 1. If it counts failures before the first success, its smallest possible value is 0. These are two ways to record the same sequence, but their values differ by one.

For example, if the first success occurs on trial 4, then there were 3 failures before it. The event \(T=4\) is the same event as \(F=3\), not \(F=4\). A useful translation is \(T=F+1\); use it before substituting into a probability formula.

Worked Example: Counting the Trial and the Failures

A hypothetical inspection process has independent trials, each with success probability \(p=0.4\) and failure probability \(1-p=0.6\). Let \(T\) be the trial of first success and \(F\) the number of failures before first success. Find (a) the probability that the first success is on trial 3, (b) the probability of exactly two failures before success, and (c) the probability of at least three failures before success.

Method and assumptions. The trials are independent and have the same success probability \(0.4\), so the geometric model applies. For (a), the sequence must be failure, failure, success:

$$ P(T=3)=(0.6)(0.6)(0.4)=0.1440. $$

For (b), exactly two failures before the success is the same event as success on trial 3, because \(F=T-1\). Therefore,

$$ P(F=2)=P(T=3)=0.1440. $$

For (c), at least three failures means the first three trials are all failures. The outcome of later trials is unrestricted:

P(F\geq 3)=P(T\geq 4)=(0.6)^3=0.2160.

Check. At least three failures is the complement of success within the first three trials. Thus \(P(T\leq 3)=1-0.2160=0.7840\), consistent with the survival probability above. The common indexing mistake is to calculate “success on trial 3” as \((0.6)^3(0.4)=0.0864\). That sequence has three failures followed by success: it is \(T=4\), or \(F=3\), not \(T=3\).

Why Independence and a Fixed Probability Matter

The standard geometric formula describes a particular sequence model, not every process that stops at its first success. It requires both independence and the same success probability on every trial. Independence means the result of one trial does not change the probabilities for another. A fixed probability means that each trial has success probability \(p\), regardless of its position.

To see why these conditions produce the formula, the event \(T=k\) requires failures on trials \(1\) through \(k-1\), followed by success on trial \(k\). Independence allows the probabilities of those outcomes to be multiplied. Equal success probabilities make each failure probability \(1-p\), and the final success probability is \(p\).

\begin{aligned} P(T=k) &=P(F_1,\ldots,F_{k-1},S_k)\\ &=P(F_1)\cdots P(F_{k-1})P(S_k)\\ &=(1-p)^{k-1}p. \end{aligned}

Here \(F_j\) denotes failure on trial \(j\), and \(S_j\) denotes success on trial \(j\). If trials are independent but their success probabilities are \(p_1,p_2,\ldots\), the corresponding exact-time probability is instead

P(T=k)=\left(\prod_{j=1}^{k-1}(1-p_j)\right)p_k.

If outcomes are dependent, multiplying unconditional probabilities is not generally valid. The chain rule still applies, but the factors must be conditional probabilities given the preceding outcomes. In particular, the chance of success after a run of failures may differ from the chance on the first trial.

Worked Example: Independent Trials With Changing Probabilities

A hypothetical review process has independent attempts, but the success probabilities are \(p_1=0.25\), \(p_2=0.50\), and \(p_3=0.75\) on the first three attempts. Let \(T\) be the attempt of first success. Find the probability that success first occurs on attempt 3 and the probability of success by the end of attempt 3.

Method and assumptions. The attempts are independent, but their success probabilities differ, so a single geometric parameter does not describe \(T\). For first success on attempt 3, the first attempt must fail, the second must fail, and the third must succeed:

$$ P(T=3)=(1-0.25)(1-0.50)(0.75) =(0.75)(0.50)(0.75)=0.28125. $$

Rounded to four decimal places, \(P(T=3)=0.2813\). For success by attempt 3, use the complement of failure on all three attempts:

\begin{aligned} P(T\leq 3) &=1-(1-0.25)(1-0.50)(1-0.75)\\ &=1-(0.75)(0.50)(0.25)\\ &=1-0.09375\\ &=0.90625=0.9063 \text{ to four decimal places.} \end{aligned}

Check. The probability of no success through attempt 3 is \(0.09375\), and \(0.90625+0.09375=1\). A common mistake is to average the probabilities, obtain \(p=0.50\), and then use a geometric formula: that would give \(P(T=3)=(0.50)^2(0.50)=0.1250\). Averaging does not reproduce the actual attempt-by-attempt probabilities; the correct exact-time probability is \(0.2813\).

Equal Marginal Probabilities Do Not Prove Independence

Even when every trial has the same marginal probability of success, the trials may be dependent. A marginal probability such as \(P(S_2)=0.5\) describes trial 2 without conditioning on trial 1. Independence requires, among other equivalent conditions, that learning the outcome of trial 1 does not change the probability of success on trial 2. Thus \(P(S_2\mid S_1)\) must equal \(P(S_2)\) when those conditional probabilities are defined.

Worked Example: Same Marginals, Dependent Trials

A hypothetical sequence has \(P(S_1)=0.50\). On each later trial, the probability of success is \(0.75\) if the immediately preceding trial was a success, and \(0.25\) if the immediately preceding trial was a failure. Find (a) the probability that the first success occurs on trial 3, (b) the probability that no success occurs in the first three trials, and (c) the marginal probability of success on trial 2. Decide whether the standard geometric model applies.

Method and assumptions. The stated conditional probabilities describe dependence on the preceding result. For first success on trial 3, trial 1 must fail, trial 2 must fail given that failure, and trial 3 must succeed given the preceding failure:

\begin{aligned} P(T=3) &=P(F_1)P(F_2\mid F_1)P(S_3\mid F_2)\\ &=(0.50)(0.75)(0.25)\\ &=0.09375=0.0938 \text{ to four decimal places.} \end{aligned}

For no success in the first three trials, the third trial must also fail given that trial 2 failed:

\begin{aligned} P(T>3) &=P(F_1)P(F_2\mid F_1)P(F_3\mid F_2)\\ &=(0.50)(0.75)(0.75)\\ &=0.28125=0.2813 \text{ to four decimal places.} \end{aligned}

The marginal probability of success on trial 2 is found by conditioning on trial 1:

\begin{aligned} P(S_2) &=P(S_2\mid S_1)P(S_1)+P(S_2\mid F_1)P(F_1)\\ &=(0.75)(0.50)+(0.25)(0.50)\\ &=0.50. \end{aligned}

Although trial 1 and trial 2 each have marginal success probability \(0.50\), they are not independent: \(P(S_2\mid S_1)=0.75\ne P(S_2)=0.50\). The standard geometric model therefore does not apply. In particular, using \(p=0.50\) would predict \(P(T=3)=(0.50)^2(0.50)=0.1250\), rather than the correct \(0.0938\). As a check, success by trial 3 has probability \(1-0.28125=0.71875\), which is the complement of the computed three-failure probability.

A Practical Model-Checking Sequence

Before using a geometric probability, write down what one trial is and what counts as success. Then check the model conditions in order. This prevents an indexing convention from obscuring a more fundamental problem with the assumptions.

Define the waiting-time variable.
State whether it counts trials through success (\(T=1,2,\ldots\)) or failures before success (\(F=0,1,\ldots\)).

Check the success probability.
Confirm that the success probability is the same on every trial. If it changes with the trial number, use the specified probabilities rather than a single geometric parameter.

Check independence.
Ask whether prior outcomes change the probabilities for later trials. If they do, use conditional probabilities along the relevant sequence.

Write the event as outcomes.
For first success on trial \(k\), write \(k-1\) failures followed by a success. For no success through trial \(k\), write \(k\) failures.

Check the complement and index.
Compare an at-most probability with the complement of the corresponding all-failure event, and verify that the number of preceding failures is one less than the trial number of success.

Common Exam Traps

Using the trial count as the failure count. First success on trial \(k\) means \(k-1\) preceding failures, not \(k\).
Using one \(p\) when the probabilities change. The standard formula \((1-p)^{k-1}p\) requires the same \(p\) on every trial; an average probability is not a substitute.
Assuming equal marginals imply independence. The probabilities \(P(S_j)\) can all be equal while conditional probabilities still depend on earlier outcomes.
Multiplying unconditional probabilities in a dependent sequence. For a specified outcome path, use the relevant conditional probabilities, such as \(P(F_2\mid F_1)\), not automatically \(P(F_2)\).
Forgetting the event after the stopping point. “Success on trial 3” specifies failure, failure, success. “At least three failures before success” specifies three failures, with later outcomes unrestricted.

Key takeaway: A geometric waiting time counts trials to first success under independent, identically distributed Bernoulli trials with fixed success probability \(p\). Confirm the support and those assumptions before using the formula; otherwise calculate from the actual sequence probabilities.

Check Your Understanding

For each problem, identify the relevant trial sequence before calculating.

Independent trials have fixed success probability \(p=0.30\). Let \(T\) be the trial of first success. Find the probability that the first success is on trial 4. (Answer: \(0.1029\))
Independent trials have fixed success probability \(p=0.20\). Let \(F\) count failures before first success. Find \(P(F=2)\), and express the event using \(T\). (Answer: \(P(F=2)=P(T=3)=0.1280\))
Three independent attempts have success probabilities \(0.10\), \(0.40\), and \(0.50\), respectively. Find the probability of first success on attempt 3 and the probability of success by attempt 3. (Answer: \(P(T=3)=0.2700\); \(P(T\leq 3)=0.7300\))
A dependent sequence has \(P(S_1)=0.40\), \(P(S_2\mid F_1)=0.50\), and \(P(S_3\mid F_2)=0.20\). Find the probability of no success in the first three trials. (Answer: \(0.2400\))
Every trial in a sequence has marginal success probability \(0.60\), but \(P(S_2\mid S_1)=0.80\). Can the standard geometric model be justified from these facts? (Answer: No; the conditional probability differs from the marginal probability, so independence is not established and in fact fails.)

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