Say Six: The Only Opening That Wins the Race to Fifty

Two players count up to fifty. The first names any integer from 1 to 10. From then on each call must exceed the previous one by between 1 and 10, and nobody may go past 50. Whoever says 50 wins. Do you want to go first?

Yes, and there is exactly one opening that works: 6. Not 1, not 10, and not any of the other eight. That uniqueness is the part of the answer worth proving, because a strategy that happens to win is a different thing from a strategy that cannot lose.

Work from the end

The only way to see this problem is from 50 backwards. Say the position is the last number named, and think about whether the player who has to speak next is in trouble.

From any position in 40 through 49 you simply say 50 and win. So a player who is handed a position in that range is fine, and a player who has to hand one over is not. Which positions force you to hand one over? Exactly 39: from 39 the legal replies are 40 through 49 and nothing else, so whatever your opponent says, you win on the next call.

Repeat the reasoning from 39 instead of from 50 and the same thing happens eleven lower. The positions you want your opponent to face are

Why eleven and not ten

The spacing is the whole idea, and it is one subtraction:

The gap between stations is one more than the largest legal step. That means a single move can never cross a gap. Your opponent, standing on a station, must land strictly inside the gap above it, and you can always finish the gap because the amount left is between 1 and 10.

The invariant also explains why the opening is unique. The five stations are the residue class $6 \bmod 11$, and a legal first call is any integer from 1 to 10. A residue class modulo 11 meets an interval of ten consecutive integers at most once, and here it meets it exactly once, at 6. Every other opening leaves your opponent a station to take, and a good opponent takes it.

The reflex opening loses

Most people open with 10, on the reasonable-sounding grounds that the biggest step must be the strongest. Check it against the invariant: $50 - 10 = 40$, and 40 is not a multiple of 11. Your opponent replies 17 and from that moment you are the one being managed. Opening with 1 fails the same way, since $49$ is not a multiple of 11 either. Only 6 works.

This is the pattern behind most first-move puzzles of this shape. The strong move is not the biggest or the smallest, it is the one that lands on the arithmetic structure, and the structure is invisible until you look at the game from the finish line.

The general subtraction game

Replace 50 by a target $T$ and the step range by $1$ through $s$. Everything above goes through with 11 replaced by $s+1$:

The first player wins if and only if $T$ itself is not a multiple of $s+1$, and then the unique winning opening is the remainder $T \bmod (s+1)$. With $T = 50$ and $s = 10$ that remainder is 6, which is the answer above.

This also locates the honest boundary of the result. Keep the same steps and move the target to 55. Now $55 = 5 \cdot 11$, so the starting position 0 is itself a station, the first player has no winning opening at all, and the advantage flips to the second player, who simply answers every call $s$ with $11 - s$. One in every eleven targets is like that. The first player is not favoured by the rules of counting games in general, only by this particular target.

Changing who wins

Flip the goal so that whoever says 50 loses. The analysis is identical with the target moved down by one: you now want to hand your opponent 49, so the stations become

and the unique winning opening is 5 rather than 6, since $49 \equiv 5 \pmod{11}$. The first player still wins. Both versions were solved by backward induction over all 51 positions with no arithmetic assumed, and the station sets came out of the game tree rather than being fed into it.

What breaks the ladder

The invariant needs the set of legal steps to be every integer from 1 to $s$. Punch a hole in it and the periodicity can change completely. If the legal steps were $\{1, 2, 5\}$, for instance, the losing positions are no longer an arithmetic progression and have to be computed by the same backward pass, which is what the theory of subtraction games is for. The period is usually still eventually regular, but its length is not something you read off the largest step.

The other thing the invariant needs is that overshooting is forbidden. If a player were allowed to name a number above 50 and simply lose, the analysis would not change, because a player sitting on 40 through 49 always has 50 available and would never choose to overshoot. That is worth checking rather than assuming, and it is the kind of rule you should pin down out loud before you start counting.

Sources and further reading

• The method in figure 1 — Backward induction
• The family this game belongs to, and what happens when the step set has holes — Subtraction game
• The language of equation (3) — Modular arithmetic

The strategy was verified rather than argued: every position from 0 to 50 was solved by backward induction, the losing set came out as exactly the five stations, and the station strategy was then played against every legal reply at every turn. All 10,000 complete lines of the game were enumerated and the first player won all of them. The general rule in equation (3) was checked against the solver for every target up to 60 and every step limit up to 10.