The Verification Venue · pointed at a game with no decisions in it

The Board That Loiters at Sixty

You cannot play snakes and ladders badly. There are no choices, so the whole game is a machine, and a machine can be solved rather than played. This page prints its board, solves the machine in your browser, and shows you where a long game actually lives: not crawling toward the end, but turning back, over and over, to a preferred stretch of the board.

Everything below is computed the moment the page loads, from the layout printed under the board. Press Play one game to watch a single token ride the chutes, or Solve the chain already done: the exact distribution of game lengths is the gold histogram, and pressing Spin 5,000 games lays a seeded simulation over it so you can watch the two agree.

square 1 · spinner 1–6 exact-on-finish rule →

Expected spins to finish, exact

direct linear solve of (I − Q)h = 1, this layout only

Simulated mean (modelled)

no games spun yet

The layout this page solves, printed in full

computing…

This is one specific arrangement of chutes and ladders. Commercial boards differ from it and from each other, so every number on this page is a theorem about this object and about nothing else. No layout here is attributed to any manufacturer. ↓

← spins to finishgold: exact · blue: simulated →

Mean from the exact distribution

second method, must agree with the solve above

Mass beyond 160 spins

the long tail the chutes near the finish make

The rules this page implements, stated so there is no ambiguity. One token starts on square 1. Each turn a fair six-sided spinner (or die, the maths does not care) gives 1 to 6. Move forward that many squares. If the landing square is the foot of a ladder, climb it; if it is the head of a snake, slide down; chains do not cascade, a single application per turn. You must land exactly on square 100 to finish. If a spin would carry you past 100, you do not move at all and the turn is spent. That last rule matters enormously: change it to "reach or pass 100 wins" and the answer below drops by several whole spins. Every published analysis you have ever skimmed made some choice here, and they do not all make the same one.

With no decisions anywhere, the game is an absorbing Markov chain. Number the 99 non-final squares as transient states and build Q, the 99 by 99 matrix where Q[i][j] is the chance that a turn spent on square i ends with the token on square j. Each row of Q plus the chance of finishing that very turn sums to exactly 1, which is a conservation law the check panel tests. The expected number of turns from every square at once is the vector h satisfying:

h = (I − Q)¹¹ · 1, solved by Gaussian elimination with partial pivoting

No simulation is needed for that number. It is a theorem about the printed layout, and the page solves it directly, then hands you the simulation anyway so you can watch the histogram converge onto the exact curve. The distribution is not bell-shaped. Its mode sits well below its mean, and the mean is dragged upward by a long thin tail: the snakes on 93, 95 and 98 sit close enough to the finish that a game can be one bad spin from done and then lose fifty more turns. That tension between "usually quick" and "sometimes enormous" is the whole character of the game.

Where the long games live. Here is the thing no amount of playing will show you. Take the chain restricted to the 99 transient squares and ask: conditioned on not having finished yet, where is the token? As the condition "still going" persists, that conditional distribution converges to the quasi-stationary distribution, the dominant left eigenvector of Q. Press Paint the residence above and the board repaints itself with it. A long game is not a slow walk to the finish. It is a residence: the token keeps coming back to the far stretch of the board while the game refuses to end. How tightly it bunches is a computed quantity, not a flourish, and words make it easy to overstate. The busiest ten squares hold the share printed in the check panel below, weighed there against a plain yardstick: twice the share that a perfectly even spread over all 99 squares would give any ten squares. The residence clears that bar, so the weighting is real, and it is spread wider than the word "cluster" suggests. The peak square is named in the caption under the board.

The check — every number recomputed in front of you

Every quantity below was computed by this page's own script, at load time, from the printed layout. The status column is a real comparison against a stated tolerance, not decoration: a failing row would say FAIL in words as well as colour.

quantityvaluetolerancestatus

What each row is: the exact mean comes from the linear solve; the distribution mean is the same physics computed a second way (iterating the state vector until the tail vanishes) and must land on the first; the conservation row is the largest violation, over all 99 states, of "row of Q plus finish-chance equals 1"; the eigen-residual measures how well the painted distribution satisfies wQ = λw; the residence-share row weighs the busiest ten squares of the quasi-stationary distribution against twice the uniform share, a deliberately modest bar, because this layout's residence is weighted toward the far end of the board rather than packed into one neighbourhood. What would have to be true for this page to be wrong: the layout printed above would have to differ from the layout in the code, or the arithmetic would have to disagree with itself at far larger than double precision round-off. An independent verifier that re-derives all of it, including a seeded Monte Carlo anchor, lives beside this page.

run it yourself: node research/the-board-that-loiters-at-sixty/verify-the-board-that-loiters-at-sixty.mjs

The simulation is modelled, seeded, and deterministic for a given press count; the exact figures are computed; nothing on this page is measured.

What's idealised here, and what's exactly true

Exactly true, for the printed layout. The expected length from the linear solve, the entire exact distribution of game lengths, and the quasi-stationary distribution are consequences of the transition matrix alone. Given the layout and the rules as stated, they are theorems. The only numerical approximation is ordinary double-precision floating point, and the check panel prints the residuals so you can see how far from exact the round-off leaves us (far below any digit shown).

Idealised. One token, no interactions. A perfectly fair spinner with six equally likely faces. Single application of a chute or ladder per turn (landing on a square that is both a snake head and a ladder foot is impossible in this layout; the code would resolve a conflict in favour of the ladder, and the printed layout contains no such square). The exact-on-finish overshoot rule, stated above, is a choice: other rules give other answers, honestly computed elsewhere.

Representative, not universal. Every number here is about the layout printed on this page. Other boards have means a few spins either side of this one. The qualitative claims, no bell curve, a long tail from late snakes, a quasi-stationary residence weighted toward the far end of the board, survive the variation between published layouts; the specific digits do not, and are not offered as universal.