Probability course
Probability course
Module 7 of 8

Waiting times & processes

Use geometric trials, memorylessness, and small state models.

20 min2 worked examples3 concept checks

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Learning objectives
  • State which convention a geometric variable uses.
  • Use the memoryless property only when assumptions hold.
  • Build a recurrence when a pattern requires multiple states.

Concepts

First success after repeated attempts

Suppose trials are independent with fixed success probability p>0, and T counts trials through the first success. Then P(T=k)=(1−p)k−1pP(T=k)=(1-p)^{k-1}p for k≥1, and E[T]=1/p. If instead you count failures before success, that variable is T−1 and its mean is (1−p)/p. Naming the convention prevents an off-by-one error.

Past failures do not make success due

For geometric T, P(T>m+n∣T>m)=P(T>n)P(T>m+n|T>m)=P(T>n). Given m failures, the remaining wait follows the same model because future trials remain independent with the same p. This does not apply to sampling without replacement, a changing success chance, or a mechanism that learns from failures.

A pattern needs a state

Waiting for consecutive successes is different from waiting for a single success. Track the useful suffix of the history: for two heads in a row, either you have no trailing head or one trailing head. Write 'one trial plus the expected remaining work' for each state, then solve the simultaneous equations.

Examples

Worked example 1

A 20% success chance

Independent trials succeed with probability 0.2. T counts attempts through the first success. Find E[T] and P(T>3).

  1. Use the trial-count convention

    The successful trial is included, so T starts at 1 and E[T]=1/0.2=5.

  2. Translate the tail event

    T>3 means that all of the first three attempts fail. The fourth attempt need not be the success.

  3. Multiply independent failures

    P(T>3)=0.8³=0.512. A mean wait of 5 does not make the wait exactly five trials.
Worked example 2

Two heads in a row

Toss an independent fair coin until HH first appears. What is the expected number of tosses from the start?

  1. Name the two unfinished states

    Let E₀ be the remaining mean wait with no trailing head, and E₁ with one trailing head. The completed HH state has zero remaining wait.

  2. Write first-step equations

    From no trailing head: E₀=1+(1/2)E₀+(1/2)E₁. From one head: E₁=1+(1/2)E₀, since a head finishes and a tail resets the useful suffix.

  3. Solve and interpret

    The first equation gives E₀=2+E₁. Substitute the second to get E₀=3+E₀/2, hence E₀=6. Treating disjoint pairs as single trials would answer a different stopping rule.

Concept checks

Choose an answer, then check it. You can retry; your latest checked answer determines completion.

0 of 3 concept checks passed
Concept check 1Each independent attempt succeeds with probability 1/5. What is the expected number of failures before the first success?
Concept check 2A fixed-p independent trial process has failed ten times. What is the chance the next trial succeeds?
Concept check 3While waiting for two consecutive heads, the current last toss is H. A tail arrives. What state should a first-step recurrence use next?

Key results

  • Say whether you count trials or failures.
  • A geometric tail means several failures in a row.
  • Memorylessness comes from the model, not from intuition about luck.
  • For patterns, model the useful history as states.

Practice problems

Solve related problems or try a timed probability drill.

Practice problemsHints and worked solutions included.Timed probability drillShort questions on this module’s topic.
Next in the course

Quantify what you do not know

Connect sample size, uncertainty, and honest conclusions.

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