Stopping-time models

Derive a waiting-time expectation and see why conditional expectations need fresh reasoning.

14 minutes · advanced · Original Poker4Quant lesson

What you will learn

Solve simple independent waiting-time recurrences and a competing-event conditional problem.

Prerequisites: Expected value as a weighted average · Bayes with a frequency table

Count the current trial before a possible restart

For independent identical Bernoulli trials with success probability p>0, let T be the number of trials up to and including the first success. On the first trial, you always spend one step. With probability 1−p, failure returns you to the same waiting problem.

The recurrence E[T]=1+(1−p)E[T] gives E[T]=1/p. It relies on the same success probability and restart distribution after failure. Sampling cards without replacement generally violates that restart assumption.

Conditioning on the winner of a race can change the problem

Roll a fair die repeatedly until a 5 or 6 occurs. Each roll is decisive with probability 2/6, so the expected time to either is 3. Conditional on the decisive result being 6, that waiting-time distribution remains the same by symmetry of the two decisive labels.

Equivalently, P(T=t and winner=6)=(4/6)^(t−1)(1/6). Divide by P(winner=6)=1/2 to get the geometric mass with decisive probability 1/3. Do not use the unconditional time to first 6, which has mean 6, for this conditional event.

Worked example

A fair die is rolled until the first 6, but you are told that this 6 occurred before any 5. What is the expected number of rolls?

Treat 5 and 6 as competing decisive events; 1–4 are neutral restarts.
Conditioning on 6 winning the race leaves each waiting time weighted by 2×(4/6)^(t−1)/6.
That is a geometric distribution with parameter 1/3.
Its mean is 1/(1/3)=3.

The conditional expected count is 3 rolls, including the final 6. The conditioning excludes all paths where 5 appears first.

Hands-on: Build a stopping rule and play a competing-event race

Classify guided outcomes, inspect repeated paths and target-winner subsets, then distinguish independent restarts from a changing no-replacement population.

Reading stopping process…

Common mistakes

Using 1/p for a nonstationary no-replacement process.
Forgetting the final success trial in the count.
Reusing an unconditional expectation after conditioning on a race outcome.

Keep these ideas

Define exactly what stops the process.
A restart recurrence needs a justified reset.
Conditioning may alter the distribution, not just a final label.