A count that has to be estimated

Counting what a rerandomisation has left is done by enumeration. Sixteen units split evenly is 12,870 assignments; check each against the tolerance, count the survivors, and the answer is exact. The count falls as constraints are added — 3,874, then 1,006, then 314, then none — and the field stops there, at a rule with no randomisation left.

Two hundred units cannot be walked. C(200, 100) is about 9·10⁵⁸, which is not a number any procedure gets through, so the count has to come from an acceptance rate: draw assignments at random, count how many are admissible, and multiply the share by the total.

The obvious worry about that is where the estimate breaks. The obvious answer is where the count is small, which is the regime the sixteen-unit measurement was about, and it would make the whole construction useless exactly where it is wanted.

That answer is wrong, and the way it is wrong changes the recommendation the previous field ends on.

A rate with no n in it #

Standardise each basis function’s imbalance by the spread a coin would give it. For an orthonormal basis those standardised imbalances are asymptotically independent standard normals, so a tolerance box of half-width a on k functions admits

p(k, a) → ( 2Φ(a) − 1 )^k

— a function of the number of constraints and the tolerance, and of nothing else. There is no n.

Three functions at a tolerance of one coin-spread: the closed form is 0.31818. The exact walk gives 0.22533 at sixteen units and 0.23016 at twenty. Sampling gives 0.29665 at forty, 0.30875 at a hundred, 0.31555 at two hundred and 0.31885 at four hundred.

So from a hundred units on, the rate sits on a limit that does not move. And the two trials small enough to walk are 29% and 28% below it — the small-sample measurement understates the rate rather than bracketing it, so extrapolating downwards from sixteen units is extrapolating in the wrong direction as well as from the wrong quantity.

Both exact figures are averaged over eight draws of the units, which matters more than it sounds: at sixteen units one design is a long way from another, and a rate read off a single draw of sixteen numbers is a fact about those sixteen numbers.

What an estimated count costs, in draws #

The precision of an estimated share has a closed form, and it says the cost of the estimate in the only unit that matters.

Drawing M assignments and observing a share p̂, the relative standard error is $\sqrt{(1-p)/(Mp)}$ — which for small p is $1/\sqrt{Mp}$, and depends on M and p only through their product.

So a count to ten per cent relative precision costs a hundred admissible draws, whatever p is, and a count to one per cent costs ten thousand. The number of draws that takes is 100/p and 10,000/p.

At a share of one in two hundred and sixty-eight, ten per cent precision is about 27,000 draws and one per cent is 2.7 million.

That reframes where the estimate breaks. It is not where the count is small — the count is astronomical at two hundred units however tight the rule is. It is where p is small, and the two are only the same question at a fixed total.

It also names what can and cannot be checked. The count’s relative precision is knowable and its absolute value never is: nothing will ever verify a figure of $2 \times 10^{56}$, so the estimate is useful exactly as a ratio and never as a number.

The count and the rate are different quantities #

Here is the reversal.

The admissible count is the rate times C(n, n/2), and the second factor is astronomical. At sixteen units the count is about 2^11.50, which is the couple of thousand the enumeration finds. At a hundred units it is 2^94.65. At two hundred, 2^194.19. At four hundred, 2^393.70.

The count grows by three hundred and eighty-two bits — a hundred and fifteen decimal orders of magnitude — across the same range in which the rate does not move at all. A rerandomisation on a real trial does not run out of assignments. It cannot: the total is growing far faster than any fixed acceptance rate can shrink it.

So the exhaustion the sixteen-unit field measures is real and it is a fact about sixteen units. Reading it as a property of the rule — as this many constraints leaves a rule with no randomisation — is an extrapolation from a quantity that grows with the trial to one that does not, and this field refuses it rather than qualifying it.

Why the rate does not move, in one paragraph #

The mechanism is worth stating because it is what makes the closed form believable before anything is measured.

For a random equal split of n units, the imbalance in a standardised covariate has a standard deviation of exactly 2/√(n − 1) — a closed form with no distributional assumption in it, and the anchor the whole balancing field is stated against. Divide by it, and the standardised imbalance has variance exactly one at every n. It is only the shape of its distribution that depends on n, and that shape converges: a sum of n terms with random signs is asymptotically normal.

So the acceptance region {|z_j| ≤ a} is a fixed region in a space where the distribution has a fixed variance and a converging shape. Nothing in that setup gives the rate anywhere to go as n grows, and the only reason the small trials are low is that the shape has not converged. The rate is flat because the standardisation already did the work of removing n.

Which also says why the rate depends on k and a in the way it does. Under orthonormalisation the components are asymptotically uncorrelated, so the acceptance probability of the box factors, and factoring is what makes it geometric in the number of constraints.

Two routes, three sizes apart #

The measurement has this site’s usual shape and it is worth spelling out, because the three routes are independent in a way that is unusual.

A walk. At sixteen and twenty units every equal split is enumerated and checked. Nothing is estimated; the count is the count.

A sample. At forty and above, twenty thousand assignments are drawn and the share admitted is counted, with a relative standard error of about 1.04% attached.

A formula. (2Φ(a) − 1)^k, which never touches an assignment at all.

The three agree in the range where two of them exist, and the disagreement at sixteen units is explained by the third — the normal approximation, which is what the formula assumes and what sixteen units do not supply. That is a stronger position than two routes agreeing, because the place they disagree has a stated reason and the reason predicts the sign.

The sign is the part worth carrying. The small trials are below, not scattered around, at both sizes and at every tolerance measured. A finite sum of signed terms has lighter tails than a normal in the region that matters here, so a fixed box catches less of it than the limit says — and the correction is towards more assignments, never fewer.

Where the rate does fall #

None of that says constraints are free. The rate falls, and it falls geometrically.

At two hundred units with a tolerance of one, one constraint admits 0.68380, two admit 0.45410, three 0.25940, four 0.17125, five 0.11225, six 0.06570. Each step multiplies by roughly two thirds, which is what a product of k factors does.

And the count over the same range goes 2^195.3, 2^194.7, 2^193.9, 2^193.3, 2^192.7, 2^191.9. It loses about three quarters of a bit per constraint out of a hundred and ninety-six.

What a constraint costs is not assignments. It is attempts. Collecting a thousand admissible draws takes about 1,465 attempts at one constraint and 9,878 at six, and that is the currency the next essay is denominated in.

What an estimated count is worth, and when it stops being worth anything #

The estimate has a standard error and it is worth writing down, because the failure it is supposed to have is a real failure somewhere.

Drawing M assignments and accepting a share p̂ gives a relative standard error of √((1 − p)/(Mp)). At twenty thousand draws and a rate of a third that is about 1.04% — a count known to a per cent, which is far more precision than any use of it requires.

The relative error blows up as p falls, and it blows up like 1/√p. At a rate of 10⁻⁴ and twenty thousand draws the expected number of acceptances is two, and the estimate is worthless. So there is a regime where the sampling route fails, and it is worth being precise about where it is: it is a regime in k and a, not in n. Since p ≈ (2Φ(a) − 1)^k, a rate of 10⁻⁴ needs about ten constraints at a tolerance of one, or about four at a tolerance of a tenth.

That is a real place to be. A rule balancing ten functions tightly is not absurd. But it is not the place the sixteen-unit measurement pointed at, and the difference matters: the small-trial exhaustion is a shortage of assignments, and this is a shortage of draws. The first cannot be bought and the second can, at a linear price.

And when no draw is accepted at all, the answer is not zero. Twenty thousand draws with no acceptance gives an upper bound of 1.5·10⁻⁴ by the rule of three, and multiplying that by C(200, 100) still leaves about 2^183 assignments. Nothing found is not nothing there, and the arithmetic that distinguishes them is one line.

The box is not what the theorem is about #

There is a distinction here that the one-covariate field’s central theorem makes it very easy to miss.

That theorem says a balancing criterion cannot tell one basis from another with the same span: what a rule removes of a shape is a squared multiple correlation on the span, and any recombination of the columns leaves it alone.

A tolerance box can tell them apart, exactly. |z_j| ≤ a for each j is a box in the coordinates it is written in, and a box is not carried to a box by a rotation. Take two orthonormal bases for the same plane, sixteen units, all 12,870 splits: the box admits 2,800 under one and 3,138 under the other, with 2,342 in common — 65.1% agreement. A third of the admissible assignments change, inside a span that has not moved.

The Mahalanobis rule z′R⁻¹z ≤ c does not have that problem. On the same splits and the same two bases it admits 2,766 under each, and the two sets are the same set, assignment by assignment.

So the span-invariance the field rests on belongs to the criterion, not to the acceptance rule, and an experimenter who chose a basis on the grounds that the choice does not matter and then wrote a per-function tolerance has made the choice matter again. It is one of the cleanest examples this site has of a correct theorem applied one step past where it holds.

The closed form is a limit, and the basis decides how fast it arrives #

The last measurement is the one that stops this essay from being a recommendation.

The closed rate is a normal-theory limit. How fast a standardised imbalance becomes normal is a fact about the tail of the function being balanced, not about the size of the trial — and the difference is dramatic.

A rule reading the covariate itself is a weighted sum of Gaussian weights, which is Gaussian at every trial size: it admits 0.68250 against a closed 0.68269. The square gives 0.65750, the cube 0.68040, a median split 0.67450.

A cut at two is a different animal. About 2.3% of units lie above it, so at two hundred units the imbalance is a sum over four or five of them and takes a handful of values. Under a box it admits 0.78625 — far too much. Under an ellipsoid whose radius came from the same limit it admits nothing at all in twenty thousand draws, where the closed form promised 5.0%.

That last one is the sharpest failure in this field. The honest report is not a rate of zero; it is an upper bound of 1.5·10⁻⁴ from the rule of three, and the difference between the count is zero and none were found is the whole of it. A median split under the same ellipsoid goes the other way, at 10.6% against a nominal 5%, for the same reason with the opposite sign.

A rate computed from a limit is a rate about a limit, and a basis function that only a few units distinguish has not reached it at any trial size anybody runs.

What this changes about the recommendation #

The one-covariate field ends with a rule and this one ends with a different one, and the difference is worth stating as a change rather than as an addition.

Before: constraints exhaust the randomisation, so a balancing rule may read only a few functions, and the product dictionary’s twenty-four terms are out of the question.

After: on any trial large enough to be worth randomising, constraints do not exhaust anything. Six constraints at two hundred units leave about 2^192 admissible assignments, and the interactions can be balanced if somebody wants them balanced.

What replaces the old constraint is a cost in attempts, which is a computational budget rather than an inferential wall, and a cost in the granularity of the reference distribution, which is inferential and is the next essay. Neither is a reason to stop at three functions.

There is one place the old rule survives and it should be said plainly: a genuinely small trial. Sixteen units, six constraints, and the admissible count really is under one. Cluster trials with a dozen clusters, a first-in-human design, anything where the randomisation unit is scarce — the exhaustion is real there, it was measured by walking rather than by extrapolation, and nothing in this essay touches it. The correction is to the extrapolation, not to the measurement.

What is claimed here, and what is not #

This essay takes the admissible count of a rerandomisation at a real trial size, and the claims are the rate’s independence of n, the count’s growth past every scale, the geometric fall in the number of constraints, the box’s failure to be basis-invariant where the ellipsoid succeeds, and the closed form’s failure for a function only a few units distinguish.

What stays out and is named as a decision: what the sampling costs and what it does to a p-value, which is the next essay; any question of how tightly to set the tolerance, which is the one-covariate field’s dial and is unchanged by anything here; and the distribution of the covariates, which is stated as standard normal throughout and is an assumption rather than a measurement.

The boundary against the field that counts by walking is the trial size. The tolerance, the standardisation in units of a coin’s own spread, and the meaning of the admissible count are established there, and where the enumeration stops being possible at all is where this field begins.

The checks, and the refusals that make them mean something #

Three claims are gated in this field’s library. The rate is required to sit on the closed form from a hundred units up and to be strictly below it at every size small enough to walk. The count is required to grow at least eightfold at every step in the trial size, and six constraints on four hundred units are required to leave more than 2^300 assignments. And the ellipsoid is required to accept the identical set under two bases for the same span while the box agrees on under four fifths of it, which is the distinction between a criterion and an acceptance rule.

Two refusals. A sixteen-unit exhaustion extrapolated to a real trial: the check computes both — under one and about 2^190 — and refuses the extrapolation, because it is the conclusion this field was built to test and it is wrong. And a closed rate quoted for a basis a normal approximation has not reached: the check finds the nominal 5% and the observed zero, and requires the rule of three rather than the formula.