Even rows always reward the move

Milnor’s 1953 theory of scoring games — the ancestor of every mean and temperature in this subject — comes with a hypothesis attached. A game must have a non-negative incentive to move: whoever is to move does at least as well as they would if the opponent moved first. Without it, a sum of games can be worth something far from the sum of their means, and counting at the end found coin rows where exactly that happens. A hypothesis has to hold all the way down then showed that checking the condition on the row in front of a player is not enough — it has to hold at every position the play can reach — and the restriction that buys the most turned both readings into classes and scored them.

One cell of that last table deserves a second look. On rows of four coins from $\{-2, 1, 3\}$, the class “incentive condition at the top” held eighty-one rows — every row of four there is. A restriction that restricts nothing is either an accident of one family or a theorem about it. It is a theorem, and the theorem says what the incentive condition means for coin rows.

The census #

Four coin sets — $\{-2, 1, 3\}$, $\{-3, 1, 2\}$, $\{-1, 1, 2\}$ and $\{-5, 1, 4\}$, each with one negative coin and two positive ones — and every row of one to seven coins from each. For each row the score is computed twice, once with Left to move and once with Right, and the row has an incentive to move when Left’s score moving first is at least Left’s score moving second.

The even columns are full. Every row of two coins, every row of four and every row of six, under every coin set, has an incentive to move: 9 of 9, 81 of 81, 729 of 729, four times over. The odd columns are nothing like that. Rows of one coin have the incentive exactly when the coin is positive, two of three. Rows of three have it on 11 to 16 of 27 depending on the set, rows of five on 70 to 119 of 243, rows of seven on between a fifth and two fifths.

The difference is too clean to be sampling. An even row never lacks the incentive; an odd row lacks it often. The reason is a strategy the first player has on every even row and on no odd one.

The parity strategy #

Number the coins from left to right. On a row with an even number of coins, the two end coins sit in positions of opposite parity — one odd-numbered, one even-numbered. Whichever parity the first player prefers, one of the two ends has it.

Suppose the first player takes a coin of the odd positions. What is left is a row whose two ends are both even-numbered positions of the original, so the opponent has no choice: whatever end they take, they take an even-positioned coin. That exposes a new end of odd parity, and the first player takes it. By induction the first player takes every odd-positioned coin and the opponent every even-positioned one. The same works with the parities exchanged. So the first player can guarantee the whole of whichever parity class is worth more.

The two parity classes together hold the whole row, so the larger holds at least half the total — and that is true whatever the signs of the coins, because the larger of two numbers is at least their average. So the first player’s total is at least half, the opponent’s at most half, and the first player finishes level or ahead. The strategy was drawn in counting at the end as the reason the first player never loses an even row. What it was not used for is the next step.

From never losing to the incentive condition #

The coin-row game treats the two players identically: a coin counts for whoever takes it. So the row scored with Right to move first is the row scored with Left to move first, with the roles exchanged — Right’s advantage moving first is exactly Left’s advantage moving first, and the two scores are negatives of each other. Write $A$ for the first player’s margin under best play. Then Left’s score moving first is $A$ and Left’s score moving second is $-A$.

The incentive condition asks that Left’s score moving first be at least Left’s score moving second: $A \ge -A$, which is $A \ge 0$. The incentive condition on a coin row is precisely the statement that the first player does not lose. And on an even row the parity strategy proves the first player does not lose.

So on even rows Milnor’s hypothesis is not a hypothesis. It holds for every even row, for every coin values — negative coins included, since the averaging argument never used a sign — and the class of even rows satisfying it is the class of all even rows. The cell in the earlier table that held eighty-one of eighty-one was reporting a theorem.

Why odd rows escape the argument #

The strategy fails on an odd row at its first step, and seeing exactly how explains why odd rows fail the incentive condition so often.

On a row with an odd number of coins, the two ends are in positions of the same parity — both odd-numbered, say, in a row of five: positions one and five. The odd positions hold three coins and the even positions two. The first player can take an odd-positioned coin, but can never take an even one on the first move, and after that first move the ends are one odd and one even, which hands the second player the parity choice. The second player now has an even row in front of them, and the parity strategy belongs to them.

So on an odd row the roles invert after one move. The first player takes one coin of their choosing from the ends, and the opponent then plays the parity strategy on the even row that is left, guaranteeing at least half of it. The first player’s margin is then at most the coin they took, less whatever the opponent’s half of the remainder exceeds theirs by — and when a large coin sits in the middle at the right parity, that excess is large. $1\;3\;1$ is the extreme small case: take a $1$, and the opponent is handed $3\;1$, takes the $3$, and the first player finishes one behind.

That is why the odd columns of the census are so uneven across coin sets. Whether a given odd row punishes its opener depends on how the coins are distributed between the parity classes of the even row left after the first move, which depends on the particular values. The theorem has nothing to say about it, and the counts — a fifth of rows of seven for $\{-5, 1, 4\}$, two fifths for $\{-1, 1, 2\}$ — are the measure of how often the opponent’s inherited parity choice is worth more than the opener’s first coin.

What is left for the hereditary condition #

The stronger condition — the incentive at every position the play can reach — is the one the mean-value bound actually needs, because Milnor’s argument is an induction over the play. For a single row, the positions the play can reach are its intervals: every run of consecutive coins the ends can have been eaten down to.

Half of those intervals now need no checking. Every even interval is an even row and has the incentive by the theorem. So the hereditary condition on a coin row is a condition on its odd intervals alone, and among them are the single coins.

A one-coin interval has the incentive exactly when its coin is non-negative: whoever moves must take it. So a hereditary row has no negative coin, and that part of the condition is a sign condition that can be read off the row. The longer odd intervals are where the content is. A row of three positive coins can still punish the player who moves first.

$1\;3\;1$ is the smallest example: three positive coins, and the player who opens takes a $1$, exposing the $3$ to the opponent, then takes the last $1$ — two against three. Moving first costs one. A row that contains $1\;3\;1$ as an interval, like $2\;1\;3\;1$, fails the hereditary condition even though it has an incentive to move itself, as every row of four does. The hidden odd interval is exactly the thing a hypothesis has to hold all the way down found breaking Milnor’s bound: a row fine at the top with a punishing interval underneath.

The bound, pair by pair #

The measurement that made the hereditary condition matter was a count of pairs of rows breaking Milnor’s bound, sorted by where each row satisfies the incentive condition.

That figure is on rows of three coins from $\{-2, 1, 3\}$, and the theorem accounts for every bucket in it. A row of three has three kinds of interval: the row itself, two intervals of two coins, and three single coins. The intervals of two are even and satisfy the condition automatically. So on a row of three, “fine at the root and broken underneath” can only mean fine at the root and containing a negative coin — a single-coin interval where the mover must take a cost. The figure’s example is exactly that: $-2\;{-2}\;{+1}$ rewards the opener as a whole and contains a lone $-2$.

And “fine everywhere” is then the positive rows of three that do not punish their opener, which is the eight rows built from $1$ and $3$ less the one that does, $1\;3\;1$ — seven rows, as the figure counts. The bound held on all twenty-eight pairs of those seven, and broke on twenty-four of the ninety-two pairs involving the eight rows with a negative coin underneath. None of that required the survey once the theorem is known; the survey is what made it visible that there was something to explain.

The hereditary class, taken apart #

With the theorem in hand the hereditary class can be described in two parts, and the census separates them.

The first part is the sign condition: no negative coin. The second is that no odd interval of length three or more punishes its opener. For the coin set $\{-2, 1, 3\}$ the second part bites immediately — the positive rows of three are eight and the hereditary ones seven, the missing one being $1\;3\;1$ — and by seven coins only 54 of the 128 positive rows are hereditary. For $\{-1, 1, 2\}$ the positive coins are $1$ and $2$, too close together for any short odd interval to punish its opener, and the two counts agree through four coins, parting at five where 31 of the 32 positive rows are hereditary, and at six where 60 of 64 are.

That is a much more specific description than “the incentive condition holds everywhere”, and it says what the condition is really about on coin rows: the relative sizes of coins that sit an odd distance apart. A large coin with smaller ones on either side, at odd spacing, is what creates a punishing odd interval, and the hereditary class is the rows where no such configuration appears.

What the earlier table was comparing #

The four classes of the restriction that buys the most can now be read again. On rows of four:

Every row and the incentive at the top are the same class, by the theorem. The table’s second row was a copy of its first.

The incentive everywhere is the rows with no negative coin and no punishing odd interval of three — twelve rows, all built from $1$ and $3$, with $1\;3\;1$ nowhere inside.

The cancelling rows are the fifteen whose coins pair off, which cancelling is not pairing found to be a coincidence of that length.

So the table compared one genuine restriction with a sign condition, a copy of the baseline, and a class that coincides with a simple pattern on that length alone. None of that undoes its measurements. It does change what they mean: the finding that the cancelling class was larger than the hereditary one was a comparison between a class of fifteen and a class that, on rows of four from that coin set, could not have held more than sixteen.

The surprising connection #

The incentive condition was introduced for a mean-value theory — a statement about temperatures and the value of sums, in a tradition that runs from Milnor through Hanner to the thermographs of normal-play theory. On coin rows it turns out to be a statement about who wins: the incentive condition holds exactly when the first player does not lose, and the proof that it always holds on even rows is the oldest trick in the game, the one a child finds for taking coins from a row — take all the odd ones or all the even ones, whichever is worth more.

That makes the mean-value theory’s hypothesis, for this family, a theorem about parity. It is the kind of connection the scoring games keep producing: the theory built for sums and temperatures turns out, on the simplest game, to reduce to the question of which player can force a particular set of coins.

The convention named #

Take-the-ends, with a coin counting for whoever takes it and the score being Left’s total minus Right’s. The incentive condition is Milnor’s: Left’s score with Left to move is at least Left’s score with Right to move. The parity strategy needs only that the row have an even number of coins; it uses no property of the coin values, which is why the theorem holds for negative coins too. The hereditary condition asks the incentive of every interval, which for a single row is every position the play can reach. For a sum of rows the positions are products of intervals, and nothing here says the theorem extends to sums.

What the census cannot show #

The census cannot show anything about sums. A sum of even rows is a position with an even number of coins, but the parity strategy does not apply to it — the first player cannot keep both rows’ parities under control at once, and whether a sum of even rows always rewards the move is a separate question. Milnor’s bound is a statement about sums, so this theorem settles the hypothesis for rows and leaves it open for the positions the bound is actually about.

Nor does the theorem say anything about odd rows beyond the fact that they can fail. The census counts them; it does not predict which odd rows reward the move, and the obvious description — the parity class containing both ends is worth more than the other — is not tested here. It is not even obviously right: the first player on an odd row does not get that parity class, only its first coin, and what happens after is decided by the opponent’s choice on the even row that remains.

Still open: sums of even rows #

The next measurement is direct. Take two even rows, put them side by side, and ask whether the first player can always avoid losing — which, by the same symmetry as for one row, is whether the sum has an incentive to move. If every sum of even rows does, the incentive condition is automatic on a much larger family and the hypothesis of Milnor’s theorem is satisfied by construction for every position built from even rows. If some sum does not, then two rows that each reward the move can combine into a position that punishes it — a scoring-game counterpart of the way outcomes do not add in normal play — and the parity argument has found its limit at the first sum.