The auction never gets to the money

Nobody has to move computes a number for every position and calls it the share of the money the second player needs. The number is real, the recursion is short, and the reading was never tested against a game.

The obvious next step is the discrete one — real auctions are held with a finite number of chips rather than a continuum, so the critical ratio should become a threshold on an integer. Played out that way, it does not become a threshold on anything.

What playing it out means #

The recursion averages two branches and stops. Playing it out asks the question the branches were standing in for.

At a position, Left holds some chips and Right holds the rest. Each names a whole number no larger than their holding; the larger bid takes the auction and pays the bid across; the winner of the auction must move, and a winner with no move has lost. Left wins from a split when some bid of Left’s answers every bid of Right’s, which is the ordinary alternating quantifier arriving inside a single turn rather than between turns.

That is the same rule the earlier essay states, with the money made countable and nothing else changed. Equal bids need a decision and the decision is made both ways, which is what turns a convention into a measurement: everywhere the two runs agree the tie-break is decoration, and everywhere they differ it is the whole answer.

A row that should have a step in it #

The position with one free move for Left is the clearest case, because the number attached to it is unambiguous: $R(1) = \tfrac34$, so Right needs more than three quarters of the pool. With eight chips that is a threshold between two and three, and the row of verdicts should step there.

It does not step. It is flat at every split — Left wins everywhere under one tie-break and nowhere under the other — and the same is true of the cold position with nothing in it and of the hottest thing born on day two.

The empty position says it plainly. Neither player has a move, so winning the auction is losing; both bid nothing; and what happens next is whatever the rule for equal bids says. A player holding every chip in the pool is in exactly the position of a player holding none.

Declining every auction wins #

The reason is one line long and needs no arithmetic.

A player who never wins an auction never has to move. The other player therefore makes every move in the game. Every move in a short game shortens it — that is the condition the whole subject rests on — so the mover runs out, wins an auction with nothing to play, and has lost.

And declining every auction is free. Bid nothing: the only way to win an auction with a bid of nothing is for the opponent to bid nothing too and the tie to fall the wrong way. So whichever player the tie-break is not fixed in favour of can refuse every auction there is, and refusing wins.

The money never enters the argument. Neither does the position, beyond its ending. The whole of it is that a game ends and that somebody has to be the one still moving when it does.

The threshold the number predicts, and where it would be #

It is worth writing out what a working critical share would look like, because the sweep is only readable against it.

With a pool of eight and $R(1) = \tfrac34$, Right needs more than six chips and Left therefore wins from two upward. That is a prediction with a location: the row of eight-chip verdicts should read as a run of Right wins, a boundary between the split two to six and the split one to seven, and a run of Left wins above it. A row with the boundary in the wrong place would mean the recursion is computing a threshold and computing it slightly wrong, which would be a finding about arithmetic.

A row with no boundary at all is a different kind of answer. It says the quantity has no location, and the four pool sizes are what rule out the boundary having fallen off the end: at two chips the prediction puts it between nought and one, at sixteen between three and four, and the row is flat at every size.

The two-move position sharpens it from the other end. Its critical share is seven eighths, so Right is said to need more than five and a quarter of the six chips — every one of them — and Left is said to win from a single chip upward. That is a boundary one step from the bottom of the row rather than one step from the top, and it is as easy a prediction to meet as the pool allows. The row is flat.

What the recursion was really computing #

None of this makes the recursion wrong. It makes the reading wrong, and the reading is wrong in a way that has a name.

Richman’s theorem is about games in which the ending declares a winner — a token on a graph with two marked vertices, one of which is a win for each player, and play stopping when the token reaches one. Under that arrangement, winning an auction is never automatically a loss: a player who runs out of moves has not necessarily lost anything, because the game ends where the graph says it ends. The critical share is then a real threshold and the theorem is a real theorem.

Normal play does not supply an ending of that kind. Its ending is whoever must move and cannot has lost, and that phrase is about whose turn it is. Bidding is precisely the rule that removes whose turn it is, so the ending has to be re-specified, and the earlier essay re-specifies it in the only way that seems natural: a player who wins an auction and has no move has lost.

That specification is what makes declining profitable. It converts the end of the game from a fact about the position into a penalty for winning an auction, and a penalty for winning an auction is something a player can decline to incur, for nothing, for ever.

So the hypothesis the encoding breaks is the one about endings, and it breaks it silently — nothing about the recursion looks different, every number it produces is the number it always produced, and the object those numbers are about has quietly stopped existing.

What survives of the critical share #

It is worth separating what this leaves standing, because most of it stands.

The recursion is a well-defined function on games, and every count made of it holds. Twenty-two values born by day two do produce seven numbers; eight of them do land on exactly a half; the order is respected on all 179 comparable pairs; addition fails on fourteen of forty-nine pairs. None of those is a claim about play and none of them moves.

The collapse of the infinitesimals is real and is about the recursion. Zero, star, up, down and the rest all come out at a half because their options all come out at a half, which is arithmetic and stays true.

What goes is the sentence attaching the number to a quantity of money. The critical share, the threshold, the reading in which a player with more than so much wins — those describe a game that has been played nowhere and, under this convention, cannot be.

That is a narrow loss and a sharp one. A number with no interpretation is not much, and the earlier essay hands out several interpretations: it is not a value, it is not a strategy, it is a probability. The last of those is the one that survives, and it survives for a reason the play-out above makes exact — a coin has no tie-break to fix and therefore nothing to refuse.

The position where both players want to move #

The hot position is the interesting test, because it is the one case where the account above might have been expected to fail.

At a position with a whole unit at stake, moving is worth something to both players. Left moving reaches a position worth a point; Right moving reaches one worth a point the other way. Neither is refusing out of laziness; each has a reason to pay.

They still refuse, and the reason is that the reward for moving is finite and the penalty for being the last one moving is total. A player who buys every move buys the whole game and then loses it, however good each individual move was. So a player who can refuse does, whatever it costs in the short run — which is the bidding version of a fact the alternating theory states as zugzwang, and it is very much stronger here: under alternating play a player in zugzwang must move anyway, and under bidding they never have to.

Being obliged to move is what makes an advantage an advantage, and bidding removes the obligation without replacing it.

What would put the money back #

Two repairs are available and they are repairs of different things.

Declare the winner at the end. Add a rule saying who has won when neither player can move — a score, a territory count, a marked position — and the game stops being a normal-play game and starts being the kind of game Richman’s theorem is about. The critical share then means what it says. This is not a small change: it is the difference between a scoring game and a normal-play one, which is most of a subject.

Make the auction’s winner able to pass. If the winner of an auction may move rather than must, then refusing costs nothing and is available to both, and the game never ends at all. That is worse rather than better, and it is the same failure a pass is not a move records from the other direction.

Neither repair keeps normal play. Which is the honest conclusion, and it is a conclusion about a convention rather than about a theorem: bidding and normal play do not compose, and the number the recursion computes is about something else. What that something else is has a good answer, and it is the subject of the essay that follows.

Why the collapse was not visible from the recursion #

The recursion never mentions a bid. It has two branches, a maximum, a minimum and an average, and every quantity in it is a share rather than an action — so there is nothing in it that could have looked wrong.

What it silently assumes is that the two branches are both available to be chosen between. The average is justified by nobody knowing which of the two will happen, and the whole content of the play-out above is that one of them is never going to happen: a player who can refuse will refuse, so the branch where they win the auction is not on the table.

An average over branches one player can veto is not an average. It is the branch that player did not veto, and the recursion has no way to express that, because the veto is a fact about the ending rather than about either branch.

That is a general hazard worth carrying out of this, and it is the same shape as the sum theory failing where a capture keeps the turn: a recursion that is correct about its own two steps, applied to a game whose rules make one of those steps unreachable. In both cases nothing in the arithmetic is wrong and nothing in the arithmetic applies.

The auction, stated exactly #

Every sweep here is under normal play with the bidding rule of the earlier essay: the higher bid takes the auction, pays the bid to the other player, and must move; a player who wins an auction with no move has lost.

Three conventions of the sweep itself.

Bids are whole numbers of chips and a player may bid nothing. Allowing a bid of nothing is what makes refusal available, and it is not an artefact of discreteness — in the continuous setting the infimum of the available bids is nothing as well, and a player refusing bids arbitrarily little rather than nothing at all, which changes nothing about who ends up moving.

Equal bids are resolved by a fixed rule and the sweep runs both. A rule that alternated, or a coin, would be a different game — and the coin is the one worth having, which is why it gets its own essay rather than a footnote here.

The pool is fixed and the payments conserve it. Every chip paid goes to the other player, so nothing leaves the table. A rule paying into a bank would shrink the pool and is a different convention with a different answer; nothing here measures it.

What forty-eight sweeps do not establish #

Twelve positions and four pool sizes. The argument above does not depend on the sweep, and the sweep is what makes it an observation rather than a claim — but a proof of the general statement is not on this page, and the positions swept are all values born by day two.

Nothing here says the number is uninteresting. It is a well-defined function on games, it respects the game order on every comparable pair, and it collapses the infinitesimals in a way worth understanding. What is established is only that the money reading of it does not survive being played, under this convention.

And the discrete and continuous settings are not being distinguished. The sweep is discrete because a computer can play a discrete auction and cannot play a continuous one. The refusal argument is not about integrality and would run identically with real-valued bids, so nothing here is a claim that a finite pool is what breaks it.

The one thing the tie-break is, stated plainly #

A rule for equal bids is normally a technicality: two players happen to name the same number, somebody has to move, and a coin or a marker settles it. Here it is the entire game.

Under the rule that Left takes equal bids, Right refuses every auction and Right wins every position swept. Under the rule that Right takes them, Left refuses and Left wins every position swept. Forty-eight sweeps, two rules, and the two answers are each other’s negatives at every split of every pool.

That is worth stating as what it is: the convention has a parameter, and the parameter is the whole answer. A game with that property is a game about the parameter. The alternating convention has a parameter of the same kind — who moves first — and the difference is that alternating play’s parameter decides only the positions whose outcome class is first-player or second-player, which on the day-two pool is ten of twenty-two rather than all of them.

A rule that decides everything is not a tie-break. It is the ruleset.

How a reader can check it without a computer #

The whole argument fits in a position anybody can hold in their head, and it is worth running once by hand because the sweep is otherwise a table to be trusted.

Take the single free move for Left, with eight chips split seven to one in Right’s favour — far past the three quarters the critical share says Right needs. Left bids nothing. Right can bid nothing, in which case the tie-break decides; or Right can bid something, in which case Right takes the auction, has no move, and has lost on the spot. So Right bids nothing and the tie-break decides.

Now give Left every chip instead. Left bids nothing. Right, with nothing to bid, bids nothing. The tie-break decides. The two splits are as far apart as the pool allows and the play is identical.

The only quantity that changed anything in either line was the rule about equal bids, and it changed everything. That is the flat row of the second figure, arrived at without the table — and it is also the reason the sweep bothers to run four pool sizes rather than one, since a single flat row is a row whose step might be just off the end.

Still open: the ending that would fix it #

The repair worth measuring is the first one: attach a winner to the end of the game and see what the auction does then.

The natural attachment in this collection is a score. Dots and Boxes is a scoring game with a normal-play skeleton under it, so the same position exists in both worlds, and a bidding version of the scoring game has a declared winner at every ending while the skeleton does not. Whether the critical share means anything there — whether the row of verdicts finally steps — is a measurement nobody here has made, and it is the one that would say whether bidding is a convention about games or a convention about a particular kind of ending.