Combinatorial Game Theory
who wins, and by how much
Field

Where it stops

Misère play and loopy games: the two places the theory itself gives way — values that stop composing, and a recursion with no bottom.
What reversing the ending destroys. Everything that makes normal play tractable is a theorem about who moves last, and misère play contradicts every one of them. The positions are unchanged; the means of evaluating them is gone, and what replaces it is far heavier.

Misère play

Change one word — the player who cannot move wins — and the games are identical, the strategies are not, and almost every theorem of the normal-play theory stops being true. It is the cheapest possible modification and the most expensive.

 A position that comes back. Three positions whose moves lead round in a circle. Every value in this subject is defined by recursion on the options, and that recursion assumes play ends — here it need not, so the definition has nothing to stand on and the outcome may be a draw, which normal-play theory has no name for.

Loopy games

The whole theory assumes play stops. Allow a position to recur and the induction that every value rests on has nothing to stand on — and a fifth outcome appears that normal-play theory has no name for.

 The misère quotient of Nim, heaps up to 2. Each row and column is a class of positions that no sum in this universe can tell apart, and each entry is the class their sum falls into. The shaded classes are the ones a player wants to hand over. Under normal play the same positions need only the Nim values; the extra classes here are what misère play costs.

What survives misère play

Misère play destroys the value theory, and something much smaller grows back. Fix one game, look only at sums of its own positions, and the classes that behave alike form a monoid — computed here, and larger than the normal-play answer every time.

 a loop with a way out: what the backward analysis settles. A position graph in which the moves can lead back to where they started. The labels are the order in which a backward analysis settles each position, starting from the ones where a player has already run out of moves. Positions the analysis never reaches are drawn — and there is no test for that; being unreachable is what a draw is.

Start at the end and work backwards

When play can return to where it started there is no bottom for the recursion to stand on. What replaces it begins at the positions where somebody has already lost and propagates outwards — and the positions it never reaches are exactly the draws. There is no test for a draw, and there does not need to be.

 a cycle of three: what the backward analysis settles. A position graph in which the moves can lead back to where they started. The labels are the order in which a backward analysis settles each position, starting from the ones where a player has already run out of moves. Positions the analysis never reaches are drawn — and there is no test for that; being unreachable is what a draw is.

An outcome with no value behind it

Retrograde analysis labels positions in rounds, outward from the ones already lost. Whatever is still blank when nothing more can be deduced is a draw — and there is no separate test for a draw, because a draw is exactly the residue the method never reaches.

 Poker Nim from 3, 5, 7, with reserves of 4 and 4. Nim with one extra kind of move: a player may put any number of counters back onto a heap from a private reserve. It looks as though a losing player could stall for ever. They cannot, and the winner is decided by exactly the same nim-sum as ordinary Nim — checked here over every position within a stated range rather than argued.

The condition the recursion rests on

Not that the moves run out, and not that the options are few. Poker Nim's heaps can grow without bound and it ends; the game called `on` has one option and never does. What every value on this site needs is that no infinite run of moves exists — and there are three separate ways to fail it.

 on + off: what the backward analysis settles. A position graph in which the moves can lead back to where they started. The labels are the order in which a backward analysis settles each position, starting from the ones where a player has already run out of moves. Positions the analysis never reaches are drawn — and there is no test for that; being unreachable is what a draw is.

One part that never ends

The game called `on` has one move and it is back to itself. Add anything to it — a star, a point, its own mirror image — and the whole board is drawn. So `off` is exactly the negative of `on` and their sum is not zero, which is the group law failing for a reason that has nothing to do with who is winning.

 The mirror strategy, and the ending that punishes it. A position beside its negative and the sum of the two, with the outcome under both endings. Under normal play the sum is worth zero every time, because the second player answers every move with its mirror image. Under misère the same answers are available and the same player runs out last, so every one of these sums is a first-player win — there is no zero, and no subtraction.

Misère play has no negatives

Put a position beside its own mirror image and answer every move with the mirror move. Under normal play the answerer wins and the sum is worth zero. Under misère the answerer still has every reply and loses because of it — so there is no zero, no subtraction, and no comparison, which is why the misère theory had to be rebuilt rather than adjusted.

 The genus of Kayles ·77, heap by heap. One row per heap: the genus symbol, the misère outcome it implies, and whether the symbol is one a Nim heap has. A game all of whose positions are tame is played in a misère sum exactly as Nim is; a single wild heap ends that, and the normal-play Grundy value gives no warning of which heaps those will be.

Tame and wild

The genus is a Grundy value with a tail — the misère values of the position with 0, 1, 2, … heaps of ∗2 added — and a game is tame when its symbols are the ones Nim heaps have. Computed here for seven games over heaps 1 to 14: Kayles goes wild at heap 5, Dawson's chess at heap 9, the octal game ·6 at heap 7, and heaps 3 and 11 of Dawson's chess are both worth ∗2 under normal play with only one of them tame.

 What the two outcome classes of the parts settle. For each pair of outcome classes, the set of outcomes the sums actually took. A cell with one letter is a pair of classes that decided the answer; a shaded cell with several is a pair that did not. Both conventions have ambiguous cells — the difference is that normal play repairs them with values and misère play has nothing to repair them with.

Two misère outcomes are not enough

Knowing who wins each part does not say who wins the sum. Over 676 sums built from a pool of twenty-six positions, nine of the sixteen pairs of outcome classes settle the answer under normal play and not one of the sixteen settles it under misère — and the nine that work are theorems about a value being zero, which is exactly the thing misère play does not have.

 Top Entails, one heap at a time. Each heap with the outcome of playing it alone, the Grundy value an ordinary solver would give it, and the moves that win from it. Taking the top coin of a heap forces the opponent to answer in that heap, which is a kind of move no other game on this site has.

A move that must be answered

Every argument on this site about sums assumes the parts are independent: a move in one leaves the others alone, and the reply may go anywhere. Top Entails denies it — take the top coin of a heap and the opponent must answer in that heap. The nim-sum then misreads 9 of 36 two-heap positions, and two heaps of two coins are a first-player win, which no impartial game the theory covers can be.

 a loop with a way out under three rules for never ending. One graph, one labelling, and three ways of reading the residue the labelling never reaches. A draw is not a computed outcome here — it is what is left over — so declaring infinite play a win for one side is a legal alternative that costs no extra search and changes who wins.

When never ending is a win

Retrograde analysis labels a position a win when somebody can force the opponent to be stuck, and leaves everything else blank. Calling the blanks draws is a rule from outside the game — and two other rules are available. The labelling does not change under any of them; only the residue does, and on a three-cycle that residue is every position on the board.

 Kayles ·77: what each heap may be replaced by. Each heap with its genus, the Nim position carrying that genus, and the Nim heap a reader would substitute from the normal-play value alone. The two columns agree except where the genus belongs to no single heap — and there the second one is wrong, in sums, by exactly the amount the census counts.

What a tame heap may be replaced by

Calling a heap tame is only worth anything because a tame heap can be swapped for a Nim position with the same genus in any misère sum. The swap is not always a single heap: Kayles' heap of eight is worth ∗ under normal play and carries the genus of 2 + 3, and substituting ∗ instead gets three of the twenty-eight Kayles pairs wrong.

 What the outcome of a loopy sum can be. One row and one column per loopy outcome class, and each cell lists every outcome a sum of two such positions was found to have. Most cells hold several. The cell where both parts are drawn holds one.

The one outcome that adds

Finite outcomes do not add: two first-player wins can sum to anything. Loopy play has seven outcome classes instead of four and adds even less — of the 28 cells in the table, eleven hold several answers. Two do not, and they are the two worth having: a second-player win added to anything leaves the outcome alone, and a draw added to a draw is a draw. A draw added to anything else is not.

 The same position, two conventions, two winners. Three-player Nim with the last counter winning. The two columns differ only in what a player does when they cannot win themselves, which is a question the rules do not answer — and the answer decides who wins.

Three players and no answer

Every theorem here is about two players, and the reason is not convenience. With two players the game is zero-sum, so 'play well' needs no further explanation. Add a third and the winner of a Nim position becomes a fact about the convention: two reasonable ones disagree on 56 of the 71 positions swept. The one question no convention touches — can a player force a win against the other two together — is answered 'nobody' in 65 of the 71.

 A pass that may not end the game is not a component at all. The same grouping with the pass forbidden as the final move. Each group now holds several values, and a group with several values is a proof that the parts do not determine the whole.

A pass is not a move

Put a single pass token on a Nim board and one clause decides everything. If it may be taken at any time — including as the move that ends the game — the value of the whole is the nim-sum with a one added, in all 120 positions swept: the pass is a heap of one. Forbid it as the final move and the value stops being a function of the nim-sum at all, and 3 and 1 + 2 come apart.

 What the auction can and cannot see. Values under both conventions. The Richman value is the share of the money the second player needs; a half means the position itself decides nothing and whoever has more money wins. Every infinitesimal on the list, and zero with them, comes out at a half.

Nobody has to move

Every convention here rests on one sentence nobody examines — the players move alternately. Replace it with an auction and a position stops having an outcome class and starts having a number: the share of the money the second player needs. The 22 values born by day two collapse to seven of those numbers, eight of them landing on exactly a half; the new number respects the game order on all 179 comparable pairs, and is not determined by the parts under addition on 14 of 49.

 Hydras, and how long each takes to kill. Six small hydras with the ordinal the termination proof assigns to each and the exact number of chops it takes to finish it. Two of them are not finished here: the fight is guaranteed to end and the machine runs out of memory long before it does, which is the gap between a termination proof and a bound.

It ends, and nothing says when

The recursion this site runs needs every line of play to reach a position with no moves, and the condition is usually met by an obvious decreasing quantity. The hydra meets it with no such quantity anywhere: the tree grows at nearly every step and the fight ends regardless, because the only thing that decreases is an ordinal. A four-node hydra dies in twenty chops; one level deeper and 279 chops reach forty thousand nodes with no end in sight.

 How much company equality needs. Each row restricts the quantifier in the definition of equality to the games named, and counts how many of the 22 values born by day two survive as distinct. The bar is the same number drawn; the jump from nine numbers to four games is the whole argument.

Equal in this company

Equality quantifies over every game there is, and the quantifier can be made smaller. Restricted to a company of nine numbers, the twenty-two values born by day two collapse to seventeen; restricted to four games — nought, one, minus one and star — they stay twenty-two, and no three of the four will do. The company that decides equality is tiny, and it has to contain a star.

 Two ways to be certain and ignorant at once. Ten positions from two games that both terminate for reasons no bound comes out of. Sylver Coinage's proof counts something that goes down and can be counted; the hydra's counts an ordinal, which cannot, and the last column shows what that difference is worth.

Two ways to end with no bound

Sylver Coinage and the hydra are both guaranteed to finish and neither will say when. The difference is that one of them carries its own bound: every move in Sylver removes at least one gap, the gaps can be counted in a moment, and over ten openings the longest play uses every single one. The hydra has no decreasing quantity a solver can hold — three hydras of five nodes each take seven chops, twenty-one, and a number past two hundred and seventy-nine that this machine never reaches.

 The genus of a sum. Every pair of heaps up to 9 counters, from nine impartial games, filed by the genus symbols of its two parts. The claim under test is that the file determines the answer; it does, and neither half of the symbol determines it alone.

The genus of a sum

A genus symbol is meant to be carried one per heap, so that a solver never has to look at the heap again. That is a claim that the pair of symbols determines the sum's, and across nine games and 405 pairs it holds without exception — while the bases alone determine it in only 38 of 50 cases and the superscripts alone in 70 of 74. Both halves of the symbol are load-bearing, and two wild heaps can add to a tame sum.

 Where running out of moves is permanent. Eleven rulesets, each walked position by position from three small boards, with every position at which a player has no move examined for whether any continuation gives them one back. Nothing here is evaluated: dead-ending is a property of the rules, and two boards worth the same value can differ on it. 9 of the 11 are dead-ending and 2 are not.

Nobody comes back

There is a class of games in which running out of moves is permanent, and it is the setting almost every modern misère result is stated in. Nine of this site's eleven rulesets belong to it across 5,334 positions; the two that do not are Toads and Frogs and Amazons, and Toads and Frogs loses the property to a single clause — delete the hop and it joins the list.

 What a wider pool rescues. The misère outcome table built four times over, on pools of 10, 22, 100, 113 positions. A cell holds the set of outcomes that sums of its row class and column class actually took. Fifteen of the sixteen cells are short of all four outcomes on the smallest pool and none is on the largest, so every near-miss in the original table was a statement about the pool rather than about misère play.

What a wider pool rescues

The misère outcome table has sixteen cells, and over a pool of ten positions fifteen of them hold fewer than four outcomes — which looks like structure and might be a shortage of positions. Thirteen values further on there is nothing left: every pair of outcome classes takes every outcome, so the near-misses were the pool, and the prediction the rung below made was right.

 What a component has to carry. Four impartial games, one of which is Nim. In the other three a component cannot say what its own legal moves are without knowing something about the past or about the rest of the board, so the Sprague–Grundy recipe does not apply — and the table says by how much. Every outcome was obtained by solving the sum outright rather than by any formula.

What a component has to carry

Three impartial games on this site break the sum, and they break it for the same reason: a component cannot say what its own legal moves are. Measured with one instrument — one number per part, exclusive-ored — the failure rate runs from a quarter to nearly half, against a control where the same recipe is a theorem and is never wrong.

 One number per heap, and one number per state. Sums of Fibonacci Nim components solved in full, against two predictions. Giving each component the number its heap size suggests gets a quarter of the pairs wrong; giving it the Grundy value of its state — the pair of heap size and cap — gets every pair and every triple right.

What restores the theorem

Fibonacci Nim breaks the recipe every impartial game is supposed to obey: one number per heap, exclusive-ored, gets a quarter of two-heap sums wrong. Index the recursion on the pair of heap size and cap instead and the recipe is exact on every pair and every triple — and the number a heap of nine carries turns out to be five rather than one.

 How two genus symbols make a third. The composition rule for genus symbols, stated with its cases and checked on every pair of heaps of nine games. The base exclusive-ors, the sum is fickle only when every component is, and the symbol follows from those two.

The rule the symbols follow

Two genus symbols make a third by three lines and no lookup table: the base exclusive-ors, the sum is fickle only when every component is, and the symbol follows. Checked on 252 pairs across nine games it is right on 238 — and the fourteen failures are exactly the fourteen pairs with a wild heap in them, which is the boundary the genus is defined up to arriving as a measurement.

 What a finite closed company is made of. The finite closed companies found by the search, counted by the properties they share. Every one of them consists of games equal to their own negatives and has a size that is a power of two, and not all of them are made of nimbers.

The company that is closed

Restricted equality licenses substitution only inside a company closed under addition, and none of the five companies this site computes in is closed — day two keeps a quarter of its own sums. Searching for companies that are closed finds seven, at one, two, four and eight members, and every member of every one of them is its own negative.

 The second closure picks out the nimbers. The seven finite companies closed under addition, tested for closure under forming options. The four that are groups of nimbers keep every option; the three containing plus-or-minus one lose theirs.

The closure that picks the nimbers

Closure under addition lets a sum be rewritten and turned out to admit companies that are not nimbers at all. Closure under forming options lets a subposition be rewritten, and it pulls the other way: every company this site computes in has it and none has the first, and among the seven finite addition-closed companies, keeping every option is exactly being a group of nimbers — four of seven, both directions, no exception. Demand both at once and nineteen of twenty-two day-two values generate nothing finite.

 How often one position beats another. Misère comparison inside each ruleset's own universe. A quarter to a half of ordered pairs compare, and the ruleset that is not dead-ending is in the middle of the range.

What the class does not buy

Dead-ending is the hypothesis several modern misère results are stated under, and the rung below sorted this site's games into it without running the comparison those results are about. Running it: a quarter to a half of ordered pairs compare inside a ruleset's own universe, which is a great deal — and the ruleset that is not dead-ending sits in the middle of that range. Ten comparisons are lost when a universe is enlarged, and every one is lost to a dead-ending company.

 A function on the wild side too. Every pair of heaps filed by the pair of genus symbols it is made from. No file holds two different sums, including the sixteen with a wild symbol in them.

A function with no formula

The rung below's composition rule is exact on tame pairs and wrong on all fourteen wild ones, which looked like an exact boundary. Two heaps further it is wrong on 34 of 35 and right on one — Kayles' five and nine — so the boundary was a boundary of the pool. What survives is stronger and stranger: the pair of symbols still determines the sum on the wild side, and no rule of that shape describes it.

 A product against a sum. The mean cost ratio on two components and on three. The saving from substituting grows with the board rather than staying a fixed factor.

A product against a sum

A company closed under both addition and options licenses a solver to rewrite any subposition, and the rung below found that exactly the nimber groups have both closures. Priced on Cram boards, that licence is the difference between walking a product of position sets and walking their sum — four to twenty-five times on two components, twenty-four to a hundred and sixty-one on three — and it is available to impartial games because their class representative is a heap rather than a form.

 Not closed, and not nearly. Where the table's answers live. None is a symbol a wild heap carries; some are symbols tame heaps carry; the rest are symbols nothing in the sweep carries.

The wild side does not close

The rung below asked for the wild composition table and for two things about it: whether the wild genus symbols form a small closed set, and whether that set is a misère quotient in disguise. Building the table needed a wider sweep — nine counters a heap gives a diagonal rather than a table — and both answers are no. Not one of the twelve entries is a symbol any wild heap carries, and two wild heaps added together are tame two thirds of the time.

 Four solvers on one sum. The states each solver has to distinguish on a three by four board plus a three by five, with one more substitution allowed at each step. A million and a half becomes fourteen.

Half a licence is nearly all of it

The rung below priced the substitution licence a restricted universe gives a solver and asked what half of one is worth — the licence to rewrite components but not subpositions. It is worth nearly the whole saving. Rewriting components collapses a million and a half states to three thousand six hundred; rewriting subpositions collapses those to eight hundred and eighty-four, and splitting the pieces takes it to fourteen.

 A gap that widens without bound. Both savings as the number of components grows, enumerated where possible and given by the closed forms beyond.

One half multiplies, the other adds

The rung below priced the two halves of a substitution licence on sums of two Cram boards and predicted that the first half's saving would grow with the number of components while the second's would not. It is right, and both halves have closed forms: the component licence saves s^(k−1)/k and the subposition licence k·s over a shape count that never moves.

 One-sided, all three. The three option tests with their disagreements split by direction. None ever refuses a comparison that holds.

Wrong in one direction only

The rung below asked for the simplified comparison test the dead-ending hypothesis is supposed to license, and predicted it would agree with the quantifier on the dead-ending rulesets and not on Toads and Frogs. Written three ways and scored on 492 pairs, it agrees best on the ruleset that is not dead-ending — and never once refuses a comparison that holds, which makes it a sound filter and not a test.

 Who gains, and how much. How much each test improves when dead-endedness is turned on, with the class-specific test beside the others.

The clause that turns the class off

Three rungs failed to find the dead-ending class doing measurable work, and each time the population was blamed. Toads and Frogs with and without the jump is the matched pair the anchor wanted — the same board with the class switched on and off — and on it the test the class licenses gains less from the class than a control that has never heard of it.

 Two readings of one sequence. The three licences with their savings and their tables, which order them oppositely.

The licence that weighs nothing

The third substitution licence is constant in the number of components, exactly as predicted, and it saves under two times where the first saves seventy-six million. Priced by its table instead of by its saving it is the only one of the three whose cost does not run away — which reverses the order three rungs of this anchor have put them in.

 One licence, five prices. The third licence measured by saving, by table size, by expansions avoided, and by work under two implementations.

The price of asking what the parts are

The third licence lets a solver look up a region rather than a position, and the rung below priced it by the entries it stores. Priced by the work it costs, it saves between a third and two thirds of the expansions and pays for them with a flood fill at every node — six times the total. A square would have to be ten times cheaper than a table probe before it broke even.

 Neither quotient identifies anything. The number of misère-equivalence classes on each side of the matched pair, against the number of distinct positions.

A quotient that identifies nothing

The dead-ending class is famous for quotients rather than comparisons, so the matched pair was asked the question its own subject is about. Neither quotient identifies a single pair of positions, and both are separated by exactly five addends — because a quotient is small when its universe is poor, which is a choice of company and not a property of a class.

 The genus of Kayles ·77, heap by heap. One row per heap: the genus symbol, the misère outcome it implies, and whether the symbol is one a Nim heap has. A game all of whose positions are tame is played in a misère sum exactly as Nim is; a single wild heap ends that, and the normal-play Grundy value gives no warning of which heaps those will be.

Closing the wild side

The twenty-two wild genus symbols are not closed under addition, and the rung below offered two answers: a monoid nobody had guessed, or no algebra at any size. Neither. Five of the six games with wild heaps close at three or four heaps, with closures of two to five symbols, and the sixth is still growing.

 on: which positions play can return to. A position graph with the moves of both players drawn, and beside it the shortest sequence of moves that gets back to each position. A position play can return to is a position whose value is defined in terms of itself, so the recursion every value in this subject is built from has no base case there. A position with no way back is one the ordinary recursion terminates on.

A stopper and how to find one

The class a value theory for loopy games would need is the ones with no infinite alternating run, and the qualifier does the work: seventy-nine of the two hundred and fifty-six two-node loopy games qualify and seventy-two of them have a cycle. Every one has a decided outcome, and under eight tests the seventy-nine collapse to six.

 Two ways to be certain and ignorant at once. Ten positions from two games that both terminate for reasons no bound comes out of. Sylver Coinage's proof counts something that goes down and can be counted; the hydra's counts an ordinal, which cannot, and the last column shows what that difference is worth.

Which games end at which level

Between a game that ends within a computable bound and one that ends with no bound at all there are levels, each corresponding to a strength of induction. This site's games sit at three of them, and which level a game is at is decided by exhibiting its termination measure and checking that every move lowers it.

 Two clauses, and what each is about. Four rulesets against the two clauses of the condition. A ruleset passes both or the one-number-per-component recipe fails on it, and the two clauses fail for different reasons: locality is about the state proposed, isolation is about the rule.

Two clauses and a third question

A component can carry its own rule when two things hold: its moves are a function of what it carries, and a move in it leaves every other component alone. Two rulesets built to fail one clause each are both caught on a named witness. The four real games sort exactly — every one the recipe gets right fails no clause, every one it gets wrong fails one — and the two clauses still miss something, because Fibonacci Nim and a held pass fail the same clause and only one of them can be repaired.

 The money played out, and it never mattered. The bidding rule played move by move with a countable pool of chips, at every way of splitting it. The verdict is constant across the splits and opposite under the two ways of resolving equal bids, so what settles these positions is the tie-break rather than the money.

The auction never gets to the money

The critical fraction is computed and never played. Played out with a countable pool of chips — twelve positions, four pool sizes, every split of the chips, every bid answered — the verdict does not move with the money on a single one of the forty-eight sweeps, and the rule for equal bids settles all forty-eight. The reason is one line long: declining every auction wins, and bidding nothing declines.

 The same number, from a rule that needs no tie-break. The number computed twice: once as the critical share of a pot under the auction, and once as the probability that Left wins when a fair coin decides who moves at each turn. They agree on every position, and only the second derivation survives being played out.

A coin needs no tie-break

The same recursion has a second derivation: a fair coin decides who moves at each turn, a player whose turn it is with no move has lost, and both play to win. Written from those rules it comes out identical on every position — and it needs no rule for equal bids, because there are no bids. The number is a probability, it belongs to Left rather than Right, and the empty position is the one where the coin decides everything.

 A position Left always wins, and not always. Values grouped by the outcome class alternating play assigns them, with the range of probabilities the coin gives Left inside each class. A class that alternating play calls a win for Left every time holds no position the coin makes certain.

Left always wins, and loses more often than not

Alternating play answers with one of four classes and the coin answers with a chance, and the two do not have to agree. Over the twenty-two values born by day two they never disagree and the margin is exactly nothing — the lowest chance on a position Left wins whoever moves is a half. Over the 1,474 born by day three, seven of them sit at seven sixteenths, and seven mirror them on the other side.

 Every chance the coin gives, by day 2. The probabilities the coin produces over all the values born by a given day, drawn on the unit interval. They fall on a grid of dyadic fractions, every interior point of it is reached, and the two ends never are — so no position is ever a certainty under random turns.

Every chance but a certainty

The coin's number lands on a grid of dyadic fractions, and which points of that grid arrive is a count rather than a guess. Over the 1,474 values born by day three it reaches every one of the fifteen interior sixteenths and neither end — no position is ever certain. The groups sharing a chance run 1, 2, 4, 8 on the small pool, which looks like doubling, and 1, 2, 4, 20 on the large one, which is not.

 What the flip is worth, and what is at stake. Left's chances if Left moves and if Right moves, with the gap between them beside the position's temperature. The two are answers to the same question computed by different routes, and they do not order the positions the same way.

The coldest position has the biggest swing

How much a flip is worth is the gap between the coin's two branches, and it is a rival to the temperature — both answer how much is at stake. They disagree at once: the empty position has the lowest temperature there is and a swing of one, twice the hottest thing born on day two. On the small pool the two quantities look like a perfect three-way correspondence, and 255 of day three's values break it.

 The coin's move is often a blunder. Positions where Left has a choice and at least one option wins under alternating play, with how often the option maximising Left's chance under random turns is an option that loses the alternating game outright.

The best chance is the wrong move

Maximising a probability and denying an opponent a reply are different objectives, and on 189 of the 904 day-three positions where Left has a choice and a winning move, the option the coin prefers is one that loses the alternating game outright. The smallest case is two options and one line of arithmetic: five eighths beats a half, and a half is the move that wins.

 Two heaps and a held pass. Every pair of heaps up to 16 with one pass available that may not be the last move. Filled cells are the pairs the player to move loses: the empty board and the pairs one and two, three and four, five and six, and so on. Outlined cells are the equal pairs Nim calls lost, all of which are wins once the pass is there.

Three heaps and a pass

Nim with a single pass that may not end the game is easy on one heap and on two: a heap swaps each odd size with the even one above it, and two heaps lose exactly at (2k − 1, 2k). On three heaps the losses are known only as a list. Fix the smallest heap and each slice of the list settles into a pattern after an irregular start — period 4, 8, 10, then 160 at a smallest heap of ten, and nothing visible from eleven.

 What a held pass can tell apart. Nim heaps, Kayles rows and heaps of Dawson's chess of sizes one to 8, grouped by whether any company of up to two of them gives a different outcome with a held pass on the board. The groups outnumber both the Grundy values and the pairs of Grundy value and held-pass value.

What a component would have to carry

For a held pass to be decided by a summary of each component, the summary must separate every pair of components some company tells apart. The Grundy value does not — Nim 1 and Kayles 8 are equal games that a held pass separates beside a single Nim heap of two. Nor does the Grundy value with the component's own held-pass value: Kayles 3 and Kayles 6 agree on both and are split by a company of two Nim heaps. Over twenty-four components, fifteen classes against fourteen pairs, and the gap widens as the pool grows.

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