Annotated exemplar · AI HL
How long does a game of Snakes and Ladders last on my grandmother's board?
Exemplar written by IB Math Revision for teaching — do not submit or copy (academic misconduct). The data and the student voice are illustrative. Schools check coursework with similarity software such as Turnitin.
An AI HL exploration that models a family board as a 101-state Markov chain, finds the full distribution of game length from matrix powers, proves the tail-sum formula for the mean, catches its own off-by-one error with a simulation, and identifies which snake lengthens the game most.
How this exemplar would be marked
| Criterion | Mark | Why |
|---|---|---|
| A · Presentation | 4/4 | Coherent, well organised and concise: rules stated first, two worked rows then technology, conclusion answers both parts of the aim. |
| B · Mathematical communication | 3/4 | Clear representations and a stated matrix convention, but the transition matrix is called T in one section and P in another, and some probabilities are quoted without stating the event precisely. |
| C · Personal engagement | 3/3 | The family board, the student's own rules, real games recorded and a design goal for the grandmother make this unmistakably the student's own. |
| D · Reflection | 3/3 | Critical: a disagreement with simulation is investigated and resolved, the method's range of validity is evaluated, and rule changes are quantified. |
| E · Use of mathematics | 5/6 | Relevant, correct AI HL mathematics (transition matrices, absorbing states, matrix powers, a justified tail-sum formula, independence for two players). Held at 5 because the long-run behaviour (why F(n) → 1, and the truncation at n = 600) is asserted rather than justified. |
| Total | 18/20 | A very strong AI HL exploration; consistent notation and a sentence of rigour about convergence would make it a 20. |
Marks are our judgement of this teaching exemplar against the current criteria, explained criterion by criterion. They are not IB moderation results.
Excerpts with examiner annotations
Free sections are shown below with comments; the rest is in the full exemplar, available in the protected viewer with the IA package or a Pro plan.
1. Introduction
Every Christmas my family plays Snakes and Ladders on a board my grandmother painted when my mother was a child. It is not the same as the shop-bought boards: it has eight ladders and eight snakes, including a snake from 87 all the way down to 24 that my cousins call “the monster”. Games on it seem to go on for ever, and last year one lasted so long that we gave up. I wanted to know whether that was bad luck or whether this board really produces long games.
The aim of this exploration is to find the probability distribution of the length of a game on my grandmother's board — including the expected and median number of turns — and to find which snake makes the biggest difference to how long the game takes.
Snakes and Ladders involves no decisions at all: where you move depends only on where you are and the roll of the die. That is exactly the situation a Markov chain describes, so I decided to model the game with a transition matrix, which we had just met in class, and check the results by simulation and by playing.
C A personal, specific context (the hand-painted board, “the monster”) that leads to a clear question only this student could ask.
A Precise aim with two clearly answerable parts.
E The choice of a Markov chain is justified by the structure of the game — the maths is chosen because it fits.
2. The board and the rules
Table 1 shows the board. I used our family rules: each player starts off the board (square 0), rolls one fair six-sided die per turn, and moves forward that many squares; landing on the bottom of a ladder moves you to its top, and landing on the head of a snake moves you to its tail. To finish you must land exactly on 100; if a roll would take you past 100, you do not move. There is no extra turn for a six.
| Ladders | Snakes |
|---|---|
| 3 → 22 | 17 → 4 |
| 8 → 30 | 47 → 26 |
| 28 → 84 | 62 → 19 |
| 36 → 44 | 64 → 60 |
| 51 → 67 | 87 → 24 |
| 58 → 77 | 93 → 73 |
| 71 → 91 | 95 → 75 |
| 80 → 99 | 98 → 78 |
I made three assumptions: the die is fair, each roll is independent of the previous ones, and I consider one player on their own (in a real game the players do not affect each other, so a game with several players ends when the first of them finishes; I return to this in Section 8).
A The exact rules are stated before modelling — essential, because the answer depends on them.
B Clear table; assumptions listed explicitly.
3. Building the transition matrix
I defined the states as the square a player is on after a turn, 0 to 100, so there are 101 states. The transition matrix \(\mathbf{T}\) is a \(101 \times 101\) matrix in which the entry in row \(i\) and column \(j\) is the probability of moving from square \(i\) to square \(j\) in one turn. Each row adds up to 1. (I use row vectors, so a distribution \(\mathbf{s}_n\) after \(n\) turns satisfies \(\mathbf{s}_{n+1} = \mathbf{s}_n\mathbf{T}\).)
Worked example — row 0. From square 0 the rolls 1 to 6 reach squares 1 to 6. Square 3 is the bottom of a ladder to 22, so rolling a 3 takes the player to 22. So row 0 has probability \(\tfrac{1}{6}\) in each of the columns 1, 2, 4, 5, 6, 22, and 0 everywhere else.
Worked example — row 97 (the exact-finish rule). From 97, rolling 1 lands on 98, which is a snake to 78; rolling 2 lands on 99; rolling 3 lands on 100; rolling 4, 5 or 6 would overshoot, so the player stays on 97. So row 97 has \(\tfrac{1}{6}\) in column 78, \(\tfrac{3}{6}\) in column 97, \(\tfrac{1}{6}\) in column 99, \(\tfrac{1}{6}\) in column 100.
Squares at the bottom of a ladder or the head of a snake can never be occupied at the end of a turn, so their rows do not matter; I gave them a 1 on the diagonal so that every row still sums to 1. Square 100 is an absorbing state: once there, the player stays there, so row 100 has a 1 in column 100. Building 101 rows by hand would be repetitive and error-prone, so after doing these two rows by hand I built the matrix with a short spreadsheet formula (Appendix 1) and checked it by confirming that every row sums to 1 and by comparing rows 0, 45 and 97 with my hand calculations.
E Two representative rows are worked in full, including the tricky exact-finish rule, and the rest is generated with technology that is then checked — understanding demonstrated without repetition.
A Concise: the student explains why they did not write out 101 rows.
B The matrix convention (row vectors) is stated; notation mostly consistent, but the matrix is later also called P in places (see Section 4) — a small inconsistency.
4. The distribution of the game length
In the full exemplar (about 2.5 pages). Matrix powers give P(finished by turn n); the median, mode and the long right tail. Open in the protected viewer
5. Where are the players during the game?
In the full exemplar (about 1.5 pages). State vectors after 5, 15 and 30 turns show where players get stuck. Open in the protected viewer
6. The expected number of turns — and a mistake
In the full exemplar (about 1.5 pages). Proves E(T) = Σ P(T > n), finds an off-by-one error through the simulation and corrects it. Open in the protected viewer
7. Checking with a simulation
In the full exemplar (about 1 page). 20,000 simulated games and 30 real ones, compared with the model. Open in the protected viewer
8. Which snake matters most? And what about several players?
In the full exemplar (about 2 pages). Removes each snake in turn, extends to two players, and measures the effect of the exact-finish rule. Open in the protected viewer
9. Designing a shorter board
In the full exemplar (about 1 page). Tests all 256 subsets of snakes to design a board with about 25 turns on average. Open in the protected viewer
10. Conclusion and reflection
On my grandmother's board, with our rules, a one-player game lasts on average about 36.7 turns, with a median of 32 and a long tail: roughly one game in ten lasts more than 63 turns. The snake from 87 to 24 is responsible for the largest share of the length, and our exact-finish rule adds about 6 turns. So our abandoned game was unlucky, but games of that length are a normal part of this board.
The Markov chain worked well because the game really has no memory and no decisions; I would be much less confident applying it to a game like Monopoly, where players choose what to buy. The main assumptions — a fair die and independent rolls — are reasonable for a real die, although in real games children sometimes re-roll a “bad” throw, which the model cannot capture.
The most valuable part of the project for me was finding my own mistake through the simulation. If I extended the work, I would use the fundamental matrix of an absorbing Markov chain, which I read about but which is not in my course, to find the expected number of turns from every square, and I would like to play the redesigned board from Section 9 with my family before my grandmother repaints anything.
A Answers both parts of the aim clearly and briefly.
D Evaluates when the method is and is not appropriate (Monopoly), and returns to the lesson learned.
Appendices
In the full exemplar (about 2 pages). Appendices 1–3. Open in the protected viewer
What would push it higher?
- Use one name for the transition matrix throughout and define every event precisely (T ≤ n, T > n).
- Justify that the game ends with probability 1 and bound the error from stopping the sum at n = 600.
- Use the fundamental matrix (I − Q)⁻¹ to find expected turns from every square, explaining it as new mathematics.