Every chance but a certainty

The number a coin gives a position is an average of two numbers of the same kind, with nought and one at the bottom of the recursion. So every one of them is a fraction over a power of two, and the possible answers form a grid.

Which points of that grid actually arrive is a different question, and it has a count for an answer.

Neither end #

The strongest thing in the picture is what is missing. No position is a certainty.

That is not a shortage of positions. It is a fact about the recursion: the number is an average of two others, so it can only be one if both of them are, and it can only be nought if both of them are. Tracing that back, a chance of one would need every position below it to be worth one as well, all the way down to a position where the coin cannot name Right — and there is no such position, because the coin names whoever it likes.

Put as play: a lead of three free moves is lost one time in sixteen, a lead of ten one time in two thousand and forty-eight, and a lead of a million still has a chance attached to it. However decided a position is, a long enough run of flips exhausts it, and a fair coin gives every run some chance.

So the image is the open unit interval, in the exact sense that the two endpoints are never reached and everything between them that the grid allows is.

Every interior point #

The other half of the picture is the completeness, and it is worth separating because it did not have to hold.

On day two the grid has eighths in it and all seven interior eighths arrive: an eighth, a quarter, three eighths, a half, five eighths, three quarters, seven eighths. Nothing is skipped.

On day three the grid has sixteenths and all fifteen interior sixteenths arrive. The denominator doubles with the birthday, the endpoints remain out of reach, and the interior fills.

That gives the image a shape that is easy to state and easy to get wrong. It is not more values spreading further; the range is the same range. It is the same range measured more finely, with the grid halving each day and the ends never reached. A reader who expected the chances to spread towards nought and one as the games got deeper would be reading the birthday as an amount of advantage, and it is not: a deeper game is a game with more structure in the middle.

There is a small piece of arithmetic worth doing here because it says how little room the picture has. The day-two pool holds twenty-two values and the grid holds nine points, two of which are unreachable. So seven slots have to absorb twenty-two values, and the surprise is not that they are all occupied but that the occupancy is so uneven — one value at each end and eight in the middle. On day three, 1,474 values go into fifteen slots and the same shape appears, sharper.

That unevenness is the reason the completeness is worth stating at all. If the values were spread evenly there would be no question of a slot being missed. They are spread very unevenly indeed, and the thin ends are exactly where a missing point would be expected to turn up — so the claim is being checked where it is weakest rather than where it is safe.

Why the denominator is what it is #

The denominators are worth one paragraph, because the bound and the attainment are different claims and only one of them is arithmetic.

Each step of the recursion averages two numbers, which at worst halves the precision once, and the base cases are whole numbers. So a value born on day $d$ has a chance whose denominator divides $2^{d+1}$ — one more doubling than the birthday, because the average at the root counts as well. That is a bound and it is proved by the recursion.

What is not proved by the recursion is that the bound is met, and on both pools it is met exactly: the finest fraction on day two is an eighth and on day three a sixteenth, with nothing finer and nothing left out. The grid is not merely an upper bound on where the answers can be; it is where they are.

What shares a chance #

Twenty-two values over seven chances means the number is coarse, and the shape of the coarseness is the interesting part.

The groups run 1, 2, 4, 8, 4, 2, 1. Symmetric about a half, largest in the middle, and doubling on the way up.

The symmetry is not a coincidence and is the one thing here with a proof. Negating a position exchanges the two players and sends the coin’s chance to one minus itself — checked over all 1,496 values of both pools, with no exception — so the group at three eighths and the group at five eighths are the same group read backwards, and the middle group is closed under negation.

The doubling is a different matter. It is exactly what a reader would predict from a pool of twenty-two, and it is what the larger pool refutes.

Day three gives 1, 2, 4, 20, 73, 204, 251, 364 and the mirror image. The first three steps are the same and the fourth is twenty where a doubling would have given eight. The pattern that looked like a law on twenty-two values does not survive its first real test.

The extremes are what stay small. Exactly one value on each pool has the highest chance and exactly one the lowest, and they are the deepest numbers available — a lead of two on day two and a lead of three on day three. Nothing else can be that decided, because being that decided requires spending the entire birthday on free moves.

What the ends are made of #

The thin ends are worth one paragraph because they are the only part of the image with a description rather than a count.

The single value at seven eighths on day two is a lead of two free moves. The single value at fifteen sixteenths on day three is a lead of three. In both cases it is the largest integer the day can build, and nothing else on the pool comes near it.

The reason is that reaching an extreme chance takes a run of unanswered moves, and a run of unanswered moves is exactly what an integer is. A position with anything else in it — a star, a switch, an option for the other player — has a branch where the coin helps the opponent, and one such branch is enough to pull the average away from the edge. So the ends of the image are populated by the positions neither player wants to move in, and by nothing else.

That also predicts what day four does at the ends: one value at thirty-one thirty-seconds, a lead of four, and its mirror. The prediction is cheap and it is the only part of the next pool this essay can say anything about.

Inside one group #

A group is a set of positions the coin cannot tell apart, and the question worth asking of it is whether anything else can.

The eight values at a half are ranked by the game order on twelve of their fifty-six ordered pairs. Up is strictly greater than zero, zero than down, and up than down — the ordering no number can express — and all three sit at a half.

That is the exact sense in which the chance throws information away. It is not that the positions are alike; they are strictly ordered, and the order is the relation the whole value theory is built on. It is that the coin’s number is a function that fails to be injective on a set the order separates.

The smaller groups are flatter. At three eighths, two of twelve ordered pairs are ranked; at a quarter, none of two; at an eighth there is nothing to rank. So the collapse is concentrated at the middle of the range, which is where the positions whose content is a parity all end up.

Why the middle is heavy #

The shape of the group sizes has an explanation and it is the same one that explains the collapse of the infinitesimals two essays below.

A chance near the middle is what a position gets when its two branches are close together. A chance near an end is what a position gets when both branches are near that end, which requires the whole position to be one-sided all the way down. Most positions are not one-sided all the way down: they have an option for each player, the options point in different directions, and the average lands near the middle.

So the heaviness of the middle is not a fact about the coin. It is a fact about what a game is — two players with options — and the coin is reading it off. Alternating play reads the same fact differently and reports it as the first-player class being the largest: 1,039 of day three’s 1,474 values are first-player wins, which is seventy per cent of the pool, and the middle five chances hold a similar share.

The two readings of that fact are not the same statement and they point the same way. Most positions are contested, under whatever convention is asked.

What the grid is not #

One reading of the picture is available and wrong, and it is worth blocking because the shape invites it.

The chances are dyadic rationals and so are the numbers the value theory produces — halves, quarters, eighths, the same family. It is tempting to read the coin’s number as a value of the same kind, on the grounds that both live on the same grid.

They are not the same kind of object and the grids coincide for unrelated reasons. A game value is dyadic because the simplicity rule picks the simplest number in a gap, and the gaps are produced by comparisons between games. A chance is dyadic because a fair coin halves things. The first is about an order and the second about a measure, and a position’s value and its chance are not functions of each other in either direction: a half as a value is a position with a genuine half-move in it, and a half as a chance is a position the flip decides.

The clearest evidence is the middle group itself. Eight day-two values share the chance one half and their values are nought, star, up, down, up-star, down-star, star-two and the hottest switch of the day. Not one of those is the number one half, and the number one half is over at five eighths.

The classes inside a group #

One column of the grouping is worth reading on its own, because it is where this essay meets the one below it.

The group at a half contains all four outcome classes at once — a Left win, a Right win, a first-player win and the second-player win, given the same number. The groups either side of the middle contain two classes each; the outermost contain one.

On day three the mixing goes further. The group at seven sixteenths holds Left wins, Right wins and first-player wins together, which is where the seven positions Left wins whoever moves and loses more often than not live. A group holding a Left win and a Right win at a chance below a half is a group in which the class and the number are pointing different ways, and it exists only on the larger pool.

What a reader can check by hand #

The two endpoint claims are the ones with an argument, and both can be run without a table.

For the top: a chance of one would need both branches of the average to be one. The branch where Right moves is worth one only if every option of Right’s is worth one, or if Right has no option at all — and if Right has no option at all the branch is one by the base case, which is the losing end of the recursion for Right. Follow that down and a position worth one needs Right to have no option anywhere below it, which is a position with no Right options at any depth. There is such a thing: a lead of $n$. And a lead of $n$ is not worth one, because the other branch — the coin naming Left — spends a move and arrives at a lead of $n-1$, which is smaller. The recursion cannot get to one because it always has that second branch pulling it down.

For the bottom: the same argument mirrored, and the negation identity makes it free.

Those two paragraphs are the whole of the endpoint result, and they explain why it is a fact about the recursion rather than about the pool. Everything else on this page is a count.

How a grid is derived rather than predicted #

Random turns for the chances and alternating play for the classes, with normal play throughout: whoever is named and cannot move has lost.

Two conventions of the census.

Values are counted, not forms. Two expressions naming one value have one canonical form and therefore one chance, so the twenty-two and the 1,474 are counts of distinct values rather than of the many more ways of writing them.

The grid is derived from the attained denominators rather than predicted from the birthday. Predicting it from the birthday alone is off by one doubling — the average at the root is a step too — and a grid one step too coarse would report both pools as reaching points outside their own grid.

What a census of constructed values misses #

Two pools. Everything here is a statement about day two and day three, and the interesting claims — the interior filling, the ends staying empty, the group sizes — are counts on those. Only the endpoint claim has an argument behind it that does not depend on the pool.

Nothing here is about real games. These are the values the construction produces, and the values a ruleset produces are a different list. A census over positions of an actual game would weight the middle of the range far more heavily, because real positions are mostly close.

And a group is not an equivalence. Two positions sharing a chance are not interchangeable in company: there is no sum theory here, so a group says what the coin cannot see and nothing about what happens when the positions are added to something.

The one place the two pools disagree about shape #

Everything above holds on both pools except the group sizes, and it is worth being precise about the disagreement because it is the only one.

The endpoints are missing on both. The interior is complete on both. The symmetry about a half holds on both, exactly, with a proof. The largest group is the middle one on both, and the extremes hold exactly one value each on both.

The sizes are 1, 2, 4, 8 up the left half on day two and 1, 2, 4, 20 on day three. Four numbers, three of them agreeing and the fourth off by a factor of two and a half.

That is what a sweep of one pool is worth in general, and this pool is a favourable case: twenty-two values is small enough to hold in view and large enough to look like a pattern. Three terms of a sequence agreeing with the powers of two is not evidence of much, and it took the next pool to say so. The safe claims here are the ones with an argument attached and the count is not one of them.

Still open: where a group stops growing #

The group sizes are the number with the most room left in it, and the question is what they are converging to.

At day two the middle group holds eight values of twenty-two, a share of thirty-six per cent. At day three it holds 364 of 1,474, a share of twenty-five. The share is falling and the grid is doubling, so the two effects pull against each other and neither is obviously winning.

What would settle it is the shape of the distribution rather than another count. A binomial spreading over a doubling grid would have the middle share falling like the reciprocal of a square root; a distribution with a fixed number of heavy points would have it settling. Day three’s 1, 2, 4, 20, 73, 204, 251, 364 is not enough to tell those apart, and the numbers that would tell them apart are on day four, where the enumeration used here does not reach.