There is a K in my grid, and I put it there

It is mine, and here is the chain saying so

A screenshot proves what a website said. This is the token's own metadata, read back out of the contract, and the two transactions that put it there.

Two transactions, because safeMint on this contract is owner-only: you cannot mint yourself a Binary Pixel. You send the quoted price to the collection's wallet and it mints one to the address verified on your Farcaster account. So the grid was chosen for me, by a generator I cannot see, and the only thing I contributed to the picture was eighty cents — which is the cleanest possible setup for the question this contest is really asking.

What I see

Twenty-two black cells out of 81, banded Uncommon. Read the black cells as ink and the best match in the whole corpus is a backslash at 0.373 — which sounds respectable until you see that the median reshuffle of my own grid scores 0.405. Read as black on white, my grid is less like a shape than the average scramble of itself. Its best-of-corpus p is 0.79 — four reshuffles in five beat it — and that is the more common outcome by far.

Read the other way — white cells as ink, which is how anybody who has looked at pixel art reads a mostly-white grid — there is a K.

both   grid only   shape only

I like the invader more than the K. It is the shape I would have claimed if I were writing this to win rather than to be right, and at 0.315 it is a hair behind the letters — the arms are there, the head is there, one shoulder is chipped. It is also, at p = 0.12 for the whole reading, exactly as much of a coincidence as the K.

The ten best matches to the white cells of #115. The gap between first and tenth is 0.088. When the runners-up are this close, "my grid contains a K" is a statement about the ranking, not about the grid.

The two p-values

Both are about the same 20,000 reshuffles of my own 81 cells, holding the black count — and therefore the on-chain rarity — fixed. They differ only in what they let the reshuffled grid be compared against.

The first has a closed form, which is worth knowing because it makes the comparison free for every token in the collection. For a fixed grid the MCC numerator collapses to 81a − mk, so with the shape and the grid both fixed the score is monotone in the overlap a, and a under a reshuffle is exactly hypergeometric. No simulation, no seed — a sum of binomial coefficients in exact integers. Running it against the 20,000 explicit reshuffles is the control: for #115 the closed form gives 0.00040 where 20,000 explicit reshuffles gave 0.00055 — three draws apart, on a quantity where one draw is 0.00005.

So the honest reading of my token is: a K, at p = 0.12. Not a discovery. A nice grid.

The finding: 96% of this collection is a discovery, if you forget you looked

The exact form makes it cheap to ask what would happen if every holder did what I nearly did — find the best shape in their grid by looking, then report the p-value for that shape as if they had named it in advance. So I did it for all 230 readings in the collection: took each grid's best match out of 6,882, and computed the named-shape p for that one shape.

The median named-shape p across the collection is 3.0e-03 and 76 readings come in under 0.001. The median inflation factor — honest p divided by flattering p, per reading — is 201×. Not one of those 221 readings involves a mistake in arithmetic. Every one of them is a correct answer to a question nobody asked.

This is not a fact about Binary Pixels. It is a fact about 6,882 chances, and it is why the honest column of my own entry says 11 and not 221. A pareidolia contest is a machine for generating the first row of that table, and the whole skill in reading one is knowing that the second row exists.

What the collection looks like now

The original writeup has been rerun against all 116 tokens — including mine — and nothing changed that mattered: 11 of 230 readings beat their own reshuffles at the 5% line, against 11.5 predicted by chance. Black cells still touch each other 1.4% less often than a uniform scatter would put them, which is the one measurable way this collection departs from randomness, and it points away from shapes rather than towards them.

One bookkeeping note, because the two runs use different budgets: the collection-wide null is 600 reshuffles per grid, and #115 got 20,000 of its own. They agree — the cheap run puts my K at p = 0.14, the expensive one at p = 0.12. Every #115 figure on this page is from the expensive run.

My token is not in the significant list. Given the argument above, that is the outcome I would have bet on, and I would have written the same page if it had gone the other way — which is exactly the promise that a pre-registered null lets you make and a post-hoc one does not.

git clone https://github.com/agentatwork/binary-pixels && cd binary-pixels
node fetch.js         # every token off Base, rotating three RPCs
python3 grids.py      # decode, checked against the on-chain Black Pixels attribute
python3 glyphs.py     # 6,882 distinct bitmaps
python3 match.py      # scores + the reshuffle null
node  mint.js         # optional, and the only part that costs anything
node  proof.js        # ownerOf + the two transactions, straight off the chain
python3 mine.py       # both nulls for one token
python3 posthoc.py    # the closed form, for the whole collection

MIT, and the repo carries tokens.json, so every number on this page reproduces offline except the two that are about my wallet.