doing math

courses › Beyond College

Probability theory

level 59 course

Long-run behaviour, random walks, and clever expectations.

Pen and paper is fine · no calculator needed why?

Learn first (about 3 minutes)

Opens at level 59.

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Builds on: Expected value & variance (not open yet) · Eigenvalues (not open yet)

the lesson

The idea, the techniques and a tip for each skill, right here. The Learn page adds worked examples for every skill and untimed practice.

Read the lesson · about 3 minutes

The idea

Each problem here has a short exact answer. The tools are balance, fairness, symmetry and adding up waits.

Symmetry: n independent uniform random points dropped on the line from 0 to 1 cut it into n + 1 gaps of equal average size. So the largest point averages n/(n + 1), and the smallest 1/(n + 1).

Techniques

Balance the flows

Two states, with fixed chances of switching each step.

  1. Let p be the long-run share of time in A, so 1 − p is the share in B.
  2. Balance: p × (chance A to B) equals (1 − p) × (chance B to A). Solve for p.
worked example

Example: A process has states A and B. Each step it moves A → B with probability 0.2 and B → A with probability 0.6, and otherwise stays put. What long-run fraction of the time is it in A? Give an exact fraction.

  1. 0.2p equals 0.6 × (1 − p), so 0.8p equals 0.6.
  2. p = 0.6 ÷ 0.8 = 3/4.

Answer: 3/4

A fair game keeps the average

Betting $1 at a time until you reach a target or go broke.

  1. Fair game: you end with the target N or nothing, so N × (chance of reaching N) equals your start.
  2. Uneven odds, win chance p, loss chance q, start i: the chance is ((q/p)ⁱ − 1) ÷ ((q/p)ᴺ − 1).
worked example

Example: You bet $1 at a time on fair coin flips, starting with $4, and stop at $10 or at $0. What is the probability you reach $10? Give an exact fraction.

  1. Your expected money stays $4, so 10 × P equals 4.
  2. P = 4/10 = 2/5.

Answer: 2/5

Add up the waits

Expected tries to see every one of n equally likely types.

  1. With k types missing, each try finds a new one with chance k/n: n/k tries on average.
  2. Add the waits from the number missing now down to 1, then round.
worked example

Example: A spinner has 4 equal sections. How many spins do you expect to need before it has landed on every section at least once? Round to 1 decimal place.

  1. Waits with 4, 3, 2, then 1 missing: 4/4 + 4/3 + 4/2 + 4/1.
  2. That is 25/3 ≈ 8.33, which rounds to 8.3.

Answer: 8.3

Tips by skill

  • TipLong-run share: Balance the flows. The share of time in a state is the chance of moving into it, divided by the sum of both chances.
  • TipGambler’s ruin: Fair game: the chance of reaching the target is start ÷ target. Uneven odds: ((q/p)ⁱ − 1) ÷ ((q/p)ᴺ − 1).
  • TipExpected maximum: n draws cut the line from 0 to 1 into n + 1 gaps of equal average size. Count gaps to the point you want.
  • TipCollect them all: With k types still missing, the next new one takes n/k tries on average. Add those waits, then round.

Watch out for

  • Swapping the two chances. The share of time in A has the chance of moving into A on top.
  • Assuming a fair game gives even odds of reaching the target. From $3 with a $10 target, it is 3/10.
  • Giving 1/2, the average of one draw, for the largest of several draws.

skills · practice stats

From rounds of this course only: box, review and test-out answers are left out. Once a skill has 40 tries, it compares your first 20 tries with your last 20.

rest ladder

Win 3 of your last 4 rounds and the course rests. A win is 90% right, within 2× the round's par. Pass the review when it comes back and the next rest is longer.

  1. 1 day
  2. 3 days
  3. 7 days
  4. 14 days
  5. 30 days
  6. 60 days
  7. mastered · every 90 days

your rounds

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