BTT Mathematics / Primary Mathematics Learning Hub / Random Walks
A random walk is a sequence of moves in which the next step is chosen according to stated probabilities. A Markov chain is a state-based model in which the probability of the next state depends on the current state and the transition rule used.
These ideas turn probability into motion. Instead of asking only “What is the chance of heads?”, we can ask “Where might the walker be after three steps?” or “How does the distribution across states change after one transition?”
This guide is Primary Mathematics enrichment. It connects Probability, Chance and Fairness, Simulation and Random Trials, Route Networks and Algorithms and Decision Trees.
Formal matrix methods are optional later-study material. Every exercise here can be solved by probability trees, tables or repeated state updates.
States and transitions · One-dimensional walks · Probability distributions · Absorbing states · Markov property · 24 questions · Worked answers
1. A transition rule tells us how probability moves between states
Suppose a system has two states A and B. From A, the next state is A with probability 1/2 and B with probability 1/2. From B, the next state is A with probability 1/4 and B with probability 3/4.
Each state’s outgoing probabilities must add to one because exactly one next state occurs.
Worked example A: One step from A
Starting at A, after one step the probabilities are P(A)=1/2, P(B)=1/2.
Worked example B: One step from B
Starting at B, after one step P(A)=1/4, P(B)=3/4.
Worked example C: Impossible transition
If a state sends probability 0 to another state, that move is impossible in one step under the stated rule.
Worked example D: Deterministic transition
A transition with probability1 is certain. Random-walk models may contain both random and certain moves.
2. A simple random walk moves left or right
Consider positions on a number line. At each step, move +1 with probability1/2 or −1 with probability1/2.
Worked example E: One step from zero
After one step, the walker is at −1 or +1, each with probability 1/2.
Worked example F: Two steps
The four equally likely step sequences are LL, LR, RL, RR. Final positions are −2,0,0,+2. Therefore:
P(−2)=1/4, P(0)=1/2, P(+2)=1/4.
Worked example G: Three steps
There are8 equally likely sequences. Final positions −3,−1,+1,+3 have counts1,3,3,1, so probabilities are 1/8,3/8,3/8,1/8.
Worked example H: Parity restriction
After an even number of ±1 steps from0, the final position is even. After an odd number, it is odd. This is a structural impossibility rule, not a low-probability statement.
3. A probability distribution records all state probabilities
A distribution assigns a probability to every possible state, with total probability1.
Worked example I: Two-state update
Suppose the current distribution is P(A)=1/2 and P(B)=1/2. Using the transition rule above:
New P(A)=(1/2)(1/2)+(1/2)(1/4)=1/4+1/8=3/8.
New P(B)=(1/2)(1/2)+(1/2)(3/4)=1/4+3/8=5/8.
Worked example J: Probability conservation
3/8+5/8=1. A transition redistributes probability; it does not create or destroy total probability.
Worked example K: Expected position after one fair step
From0, positions −1 and+1 each have probability1/2. Expected position=(−1)(1/2)+(1)(1/2)=0.
Worked example L: Expected position after two fair steps
Distribution is −2 with1/4, 0 with1/2, +2 with1/4. Expected position=(−2)(1/4)+0+(2)(1/4)=0.
4. An absorbing state cannot be left
A state S is absorbing if P(S→S)=1. Once entered, the chain stays there forever.
Worked example M: Game finish state
If “Finished” transitions to itself with probability1, reaching Finished ends the active part of the process.
Worked example N: Walk between 0 and 3
Positions0 and3 are absorbing. From positions1 or2, move left/right equally. Starting at1, one step gives probability1/2 of absorption at0 and1/2 of moving to2.
Worked example O: Two steps from1
Paths: 1→0 stays absorbed; or1→2 then2→1 or3. After two steps:
P(0)=1/2, P(1)=1/4, P(3)=1/4.
Worked example P: Absorption probability is cumulative
Once probability enters an absorbing state, it remains there in later distributions.
5. The Markov property uses the current state, not the full route
In a Markov chain, the transition probabilities for the next step depend on the present state under the model. Two different histories that arrive at the same state use the same next-step rule.
Worked example Q: History does not change the stated rule
If every visit to state B uses P(B→A)=1/4 and P(B→B)=3/4, then arriving at B after one step or after ten steps gives the same immediate transition probabilities.
Worked example R: A non-Markov rule
If the next move depends on “how many times B has been visited before,” then the state label B alone is not enough. The state description must be expanded to include that information before the process becomes Markov.
Worked example S: Transition tree versus state model
A probability tree emphasizes complete paths. A Markov diagram emphasizes current states and reusable transition rules. Both can represent the same small process.
Worked example T: Long-run caution
Some finite Markov chains settle toward stable proportions, some cycle, and some are absorbed. The transition structure determines the behavior; no one long-run pattern should be assumed without analysis.
Extension: Biased Walks, Pascal-Triangle Structure and Expected Drift
The fair random walk is only the first model. Suppose a learner moves +1 with probability p and −1 with probability 1−p. The walk now has a preferred direction when p is not 1/2. This is a useful extension for strong Primary learners because the arithmetic remains accessible while the reasoning becomes richer.
Worked extension A: Expected change in one step
If p=0.6, the expected change in one step is (0.6)(+1)+(0.4)(−1)=0.2. Over many independent steps, the expected position therefore drifts to the right. After n steps from 0, the expected position is n(2p−1). This does not predict the exact path; it describes the centre of the distribution across many repeated walks.
Worked extension B: Route probability and combinations
To finish at +2 after four ±1 steps, the learner needs three right steps and one left step. There are four possible orders for that combination. In a fair walk, each route has probability (1/2)4, so the total probability is 4/16=1/4. In a biased walk with right-step probability p, each such route has probability p3(1−p), so the total is 4p3(1−p).
Why Pascal’s triangle appears
The number of routes to each reachable position follows the same combination structure that produces Pascal’s triangle. This gives learners a bridge between repeated coin tosses, combinations and state probabilities. The triangle is not a separate trick: it records how many different histories lead to each state.
From routes to transition probabilities
A transition model asks a different question: once the system is in a particular state, what are the probabilities of its possible next states? A weather toy model might say that a Dry state moves to Dry with probability 0.7 and Wet with probability 0.3, while Wet moves to Dry with probability 0.4 and Wet with probability 0.6. The learner can update a whole probability distribution one step at a time without listing every possible history.
This is the conceptual doorway to a Markov chain: the next-step rule is organised by the current state. At this level, the goal is not matrix algebra. The goal is to see the compression: a large tree of possible histories can often be replaced by a small state model when the rule genuinely depends only on the present state.
Teaching boundary
Use this extension only after the learner can distinguish a route from a state, can check that total probability is 1, and can explain why different histories may arrive at the same position. The enrichment should deepen probability structure rather than turn Primary Mathematics into premature notation.
6. Practice: 24 original questions
Questions 1–8: Simple random walks
1. From0, what are the possible positions after one fair ±1 step?
2. Give their probabilities.
3. After two fair steps, what is P(0)?
4. After two steps, what is P(+2)?
5. After three steps, what is P(+1)?
6. After three steps, what is P(+3)?
7. Can a fair ±1 walk starting at0 be at position1 after two steps?
8. Explain using parity.
Questions 9–16: State distributions
9. From A, with A→A=1/2 and A→B=1/2, give the one-step distribution.
10. From B, with B→A=1/4 and B→B=3/4, give the one-step distribution.
11. Starting with half the probability in A and half in B, find the next P(A).
12. Find the next P(B).
13. Check that the two answers sum to1.
14. Find the expected position after one fair step from0.
15. Find the expected position after two fair steps.
16. What property must every row of outgoing transition probabilities satisfy?
Questions 17–24: Absorption and Markov reasoning
17. What is an absorbing state?
18. In the 0–3 absorbing walk, starting at1, what is P(0) after one step?
19. What is P(2) after one step?
20. After two steps, what is P(3)?
21. After two steps, what is P(1)?
22. State the Markov property in words.
23. If the next rule depends on the number of previous visits, is the current state label alone sufficient?
24. Why does probability already in an absorbing state never decrease?
7. Worked answers
Answers 1–8
1. −1 and +1.
2. 1/2 each.
3. 1/2. Paths LR and RL.
4. 1/4.
5. 3/8.
6. 1/8.
7. No.
8. Two ±1 steps change position by an even total, so the final position must be even.
Answers 9–16
9. A:1/2, B:1/2.
10. A:1/4, B:3/4.
11. 3/8.
12. 5/8.
13. 1.
14. 0.
15. 0.
16. They must sum to1.
Answers 17–24
17. A state that transitions to itself with probability1.
18. 1/2.
19. 1/2.
20. 1/4.
21. 1/4.
22. The next-step distribution depends on the present state under the stated model, not on the full route used to reach it.
23. No. The state description would need to include the extra history information.
24. By definition the state sends all of its probability back to itself.
8. Teaching and transfer
If a learner mixes paths and states, use two views of the same example: a probability tree for routes and a state table for where probability is located after each step.
When probabilities do not sum to one
Stop before continuing. Outgoing transitions from one state and the full state distribution after an update must both sum to one.
When “random” is mistaken for “anything can happen”
Randomness operates inside structural constraints. Parity may make a position impossible even though individual steps are random.
When Markov means “memoryless” too vaguely
Say exactly what the state contains. A process is Markov only relative to a state description that contains enough information to determine the next-step probabilities.
When expected value is confused with a likely outcome
The expected position may be0 even when probability of being exactly at0 is zero, as after one fair ±1 step.
Connect to simulation
The same process can be simulated repeatedly to compare experimental frequencies with the exact transition calculations in Simulation, Experimental Probability and Random Trials.
Continue through this enrichment collection
For assignment constraints, use Graph Matchings, Bipartite Assignment and Pairing Problems. For topological diagrams, use Knot Theory, Links, Crossings and Simple Invariants. For counting decompositions, use Integer Partitions, Ferrers Diagrams and Counting Structures.
Return to the BTT Primary Mathematics Learning Hub.
Original enrichment guide with 24 original practice questions and separate worked answers. Transition rules and boundary conditions are stated locally for every finite example.