game Penney, over a 3-sided diewords length 2 · nine of themknob the weighting (a, b, c)question is there a safe word?
A fair three-sided die has no safe word: whatever length-two word you name, I have a reply that beats you more often than not, and a rock-paper-scissors triangle spins at the game's centre. So cheat — load the die. Weight a face. Surely, loaded enough, some word becomes unbeatable. Drag across every possible loaded die and watch what actually happens.
In A Triangle at Two the die was fair and honest, and the surprise was the little triangle: three two-letter words, AB beats BC beats CA beats AB, each three wins in five, no best among them. This page turns one knob the fair page left fixed — the weighting of the die — and asks the gambler's question: can you rig it? Load a face heavily enough and surely you buy yourself a safe, unbeatable word.
The space of all loaded dice is a triangle in its own right: a point inside it is a weighting (a, b, c) — the chances of the three faces — with a + b + c = 1. The exact centre is the fair die. Every edge and corner is a die loaded some way. Below, that whole space is coloured by what the game does there, and you can drag a point across it. Every colour is an exact computation — the win-probability of each of the nine words against each other, in exact integers, the same engine as the fair page, only now the faces need not be equal.
I · the space of loaded dice — drag the weightingexact · live tournament
the fair triangle survivesa loop survives (no safe word)a safe word exists (loaded hard)
recomputing the phase field…
The big triangle is the space of every possible loaded die. Its corners are the impossible extremes (one face certain); its centre is the fair die; the amber grid you drag on is every weighting in between. Each point is coloured by rebuilding the whole nine-word tournament there and reading off: does a fair triangle still spin? is there any directed loop at all (so no single word beats every other)? or has the die tipped far enough that one word beats them all? Drag, or focus the panel and use the arrow keys.
Three things fall out of the picture, and each is a separate, checkable fact.
The famous triangle is fragile. The bright amber heart — where AB→BC→CA still cycles exactly as it did on the fair die — is a thin sliver hugging the centre. Nudge the die even a little off fair and the triangle breaks. In fact there are two such triangles (the second, AC→CB→BA, is the first read backwards), and they don't break together: bias one way and you keep the first while the second dies. You can choose which rock-paper-scissors loop survives — but neither survives much loading.
The triangle you came for is a soap bubble. Breathe on the die and it pops.
But the tangle itself is stubborn. Here is the thing you would not guess from the fragile heart: some directed loop — some set of words with no top, no safe choice — survives across most of the space, not just the centre. Break the famous triangle and the nontransitivity does not die; it migrates to a different trio of words, and the whirlpool of upsets simply changes shape and size. The violet region is huge. Loading the die moves the tangle around; it does not, for a long time, remove it.
A safe word costs more than you think. To actually buy an unbeatable word — a green region — you have to load the die hard. And when it finally appears, the safe word is always a doubled letter: weight face C heavily and CC, the fastest run to complete, becomes unbeatable. But there is an exact price. Along the balanced line where the other two faces stay equal, CC does not become safe until C's weight passes one-half — until a single C is likelier than everything else combined. And even past that line, look below the safe word: a smaller whirlpool is often still spinning among the other words. You do not get to gently rig this game.
Watch a die break the triangle
Point the panel below at the current weighting and it draws the two triangles as they stand: an edge glows amber where it still points the fair way, and flips to red where the loading has reversed it. Drag the die in instrument I and watch the edges flip. One thing you will never see, no matter how you load: all three edges of a triangle flipping together into the reverse cycle. That is not luck. It is a theorem, and the proof is one line — instrument III.
II · the two triangles, under the current dielive · follows instrument I
AB → BC → CA
AC → CB → BA
Each edge is labelled with the exact probability that the tail word appears before the head word under the current die. Amber = still the fair direction; red = the loading reversed it; grey = an exact dead heat (½). A triangle is a live rock-paper-scissors loop only while all three of its edges glow amber (or all three red — which, it turns out, can never happen).
Why the die can't be flipped
The reverse cycle is impossible, and you can see exactly why. Triangle AB→BC→CA keeps its direction as long as three inequalities hold; to reverse all three at once you would need each to fail at once. Add the three failure conditions together and the die's own arithmetic forbids it.
III · the no-flip theoremone line · exact
to reverse every edge of the triangle you would need, all at once:
a < c(1−a), b < a(1−b), c < b(1−c)
add the three (and use 1−a = b+c, etc.):
a + b + c < c(1−a) + a(1−b) + b(1−c) = (a+b+c) − (ab + bc + ca)
but a + b + c = 1, so this says 1 < 1 − (ab + bc + ca),
i.e. ab + bc + ca < 0 — impossible for any real die (all three products are positive).
∴ no loaded die can reverse a Penney triangle. It can only break it.
recomputing…
The same argument runs for the mirror triangle by symmetry. This is the one fully-proved statement on the page (the region fractions below are exact-per-point but measured over a grid, not proved for the continuum). The live check drags a fine grid of dice past the test and confirms zero reversed triangles — the picture the proof guarantees.
The price of a safe word
Start at the fair die and load face C, keeping A and B balanced. Nothing is safe for a long while — every word still has a beater — until C's weight crosses exactly one-half. At that instant two of CC's contests fall to a dead heat and, one hair past, CC beats all eight other words at once. Slide it and watch the exact moment.
IV · loading one face — when does a safe word appear?a = b · exact threshold
recomputing…
The bar sweeps C's weight from a light load up to twice-fair. The safe word CC switches on exactly at c = ½, where CC-vs-AC and CC-vs-BC are exact ties — verified against the offline gate. Below one-half: no safe word, the tangle persists. Above: CC is unbeatable, yet a loop can still spin among the losers.
So the honest answer to the cheat's question: almost, but not the way you hoped. You can pick which of the two rock-paper-scissors triangles survives, and you can, with enough loading, finally manufacture a single unbeatable word. What you cannot do is make the game simple — remove the tangle, buy a safe word cheaply, or flip a loop to run the other way. Nontransitivity is not a fragile accident of the fair die that a little weighting dispels; it is the die's default, and it takes a heavy, lopsided hand to be rid of it. The fair triangle is a bubble; the whirlpool underneath is not.
It is a small, exact fact about a famous game, with the shape this place likes: the picture is not a vibe but a thing you can build, drag, and count, every colour an exact decision, every headline a claim that either survives its own check or lights red. Load the die and see.