Is the machine fair on every day of the year?

Testing a partition of the calendar

The distance. Two sets of draws are compared with the whole-draw maximum mean discrepancy: for a kernel $k$ on draws and sets $A$, $B$,

$\mathrm{MMD}^2=\frac{1}{|A|^2}\sum_{i,j\in A}k_{ij}-\frac{2}{|A||B|}\sum_{i\in A,\,j\in B}k_{ij}+\frac{1}{|B|^2}\sum_{i,j\in B}k_{ij}.$

It compares whole draws rather than a chosen statistic, so it does not need anybody to have guessed in advance what a calendar-conditioned fault would look like — which is the point, since a fault nobody anticipated is exactly the kind that survives.

One matrix, many partitions. The naive route rebuilds an MMD for every partition of every shuffle, and that is a full pass over the kernel matrix each time. With $S$ the matrix total and $r_i$ its row sums, the three blocks all follow from a sum over $A$ alone:

$\textstyle\sum_{A,A}=\sum_{i,j\in A}k_{ij},\quad \sum_{A,B}=\sum_{i\in A}r_i-\sum_{A,A},\quad \sum_{B,B}=S-2\sum_{i\in A}r_i+\sum_{A,A}.$

For seven weekdays that is a forty-ninth of the work, which is what makes two hundred shuffles of six families cheap enough to run while a page renders.

Calibrating the maximum, not a partition. A family's statistic is the largest of its partitions' distances, and quoting "the biggest of seven" against a null built for one would be cheating. So the labels are shuffled across draws with the partition sizes held fixed, the maximum is recomputed inside each shuffle, and

$p=\frac{\#\{\text{shuffled max}\ \ge\ \text{observed max}\}+1}{\text{shuffles}+1}.$

The $+1$ on both sides is what stops a finite simulation ever reporting an impossible zero.

Correcting once. Each family's p is multiplied by the number of families tested, and read against a plain 5% — not against 5% divided by that number as well. Applying Bonferroni at both ends demands $p<\alpha/m^2$, which on six families means a raw p below one in seven hundred before anything is said. That is not caution; it is a test that cannot fire.

The correction is also achievable by construction: a permutation p cannot fall below $1/(\text{shuffles}+1)$, so with six families and 199 shuffles the smallest possible adjusted p is $6/200=0.03$, comfortably under the level.

What it cannot do. Tipton's tampering touched roughly two per cent of the draws he had access to. A test of this kind does not find two per cent, and the guard tests in this repository pin exactly that: the same rig at 100% of one weekday is flagged, and at 2% it is not.

The man who only cheated on three days a year

Eddie Tipton wrote the random number generator for the Multi-State Lottery Association. He was its information security director. And between 2005 and 2011 he used it to win at least $2.2 million.

What makes his case worth a page rather than a paragraph is how he did it. He did not make the machine crooked. He installed a rootkit that lay dormant almost all the time and woke on three days of the year — 27 May, 22 November, 29 December — after 8pm, and only in a non-leap year. On those nights the generator abandoned its usual behaviour and drew from a small set of predictable combinations. On every other night of every other year, it was honest.

That is a design aimed squarely at statistics. Pool six years of draws and test them for fairness and you will find nothing, because a handful of tampered nights sit inside thousands of clean ones and are averaged into invisibility. This site's own Observatory would not have caught him. Neither would its Bias Watch, its frequency tests, or any of the instruments that ask "are these numbers fair" without asking "fair when".

So this page asks the second question. Six ways of splitting the calendar — the day of the week, the month, the part of the month, public holidays, which draw of the day it was, and how recently the machine was changed — and for each, whether the draws on one side differ from the draws on the other by more than shuffling the labels can produce.

Now the part that matters. This test would not have caught Tipton either.

He tampered with something like two per cent of the draws available to him. The guard tests behind this page make that concrete: rig every single Saturday and the court finds it instantly; rig two per cent of Saturdays and it finds nothing at all. That is not a flaw to be apologised for, it is the sensitivity of the instrument, and it is stated because a page that reported "nothing found" without saying what it could not have found would be lying by omission.

What this can find is a systematic difference — a calendar-conditioned fault that nobody built to hide. Nobody has one, on any game here. And when a family does get flagged, remember what that means: this page corrects for its own six tests, but the site holds hundreds of games, and at a one-in-twenty level about one game in twenty will show something. A single flag is a thing to look at twice, and never on its own a finding.

Tipton was caught because a man walked into a Des Moines petrol station wearing a hooded sweatshirt to buy the winning ticket, and the security footage was still there a year later. Not by mathematics. That is worth remembering about all of this.