First to Call 50

Problem statement

Informal analysis

Let the calls during the game be denoted \[ C_1, C_2, \dots, C_w, \] where $C_1$ is the first call and $C_w = 50$ is the winning call. The rules imply \[ 1 \le C_1 \le 10, \qquad 1 \le C_i - C_{i-1} \le 10 \quad \text{for } i \ge 2. \]

We analyze the game by backward induction: identify positions (numbers) that are winning positions for the player about to move, meaning that from such a number the player to move can force a win with perfect play.

Backward induction (key observation)

To call $50$ on your turn, the previous call must lie in the range $[40,49]$. Therefore, if you can force the opponent to hand you any number in $[40,49]$, you will win by calling $50$.

To force the opponent into $[40,49]$, you would like to hand them the number $39$, because from $39$ the opponent’s allowed responses are $[40,49]$ and then you can call $50$.

Repeating this backward: \[ 39,\; 28,\; 17,\; 6 \] are the critical numbers: if you call one of these on your turn, the opponent is forced into an interval that allows you to move to the next critical number (adding $11$ each time), eventually reaching $39$ and then $50$.

Thus the numbers \[ 6,\, 17,\, 28,\, 39,\, 50 \] form the thread of winning positions spaced $11$ apart.

Formal justification

Claim

If a player on their turn calls a number congruent to $6 \pmod{11}$ (i.e., of the form $6 + 11k$), then with perfect play that player can force a win; conversely, if the player to move holds a number not congruent to $6 \pmod{11}$, the opponent can (by proper play) move to a number congruent to $6 \pmod{11}$.

Proof

Suppose on your turn you call $n = 6 + 11k$ for some integer $k \ge 0$ (so $n \le 50$ as relevant). The opponent must respond with a number $m$ satisfying \[ n+1 \le m \le n+10. \] All such $m$ satisfy \[ m \equiv n + r \pmod{11} \quad \text{for some } r \in \{1,\dots,10\}, \] so $m \not\equiv 6 \pmod{11}$. In particular, no response of the opponent can be congruent to $6 \pmod{11}$. Therefore, you can always respond by calling \[ n' = n + 11 = 6 + 11(k+1), \] which is legal because $n' - m \le n' - (n+1) = 10$ and $n' - m \ge n' - (n+10) = 1$. Thus you move to the next number in the sequence $6, 17, 28, 39, 50$.

This induction continues until you call $50$ (when $n'$ reaches $50$). Hence control of numbers congruent to $6 \pmod{11}$ lets you force the win.

Conversely, if it is your opponent’s turn and they hold a number not congruent to $6 \pmod{11}$, you can always choose a response (within $1$–$10$ added) that lands on the next $6 \pmod{11}$ number, so the other player can be forced back into the losing class.

Therefore the positions congruent to $6 \pmod{11}$ are precisely the winning positions for the player about to move.

Concrete strategy

Winning sequence (example)

The following table shows the moves if both players follow the described structure (opponent’s calls are shown as allowed ranges):

Why the strategy works

Two simple facts suffice:

1. On any move you can increase the current number by $1$–$10$.
2. The residue classes modulo $11$ partition the integers into blocks of size $11$. By forcing the running total into the residue class $6 \pmod{11}$ on your turns, you deny the opponent the ability to do so on theirs; therefore you can maintain control and eventually reach $50$.

Conclusion

Remark. The same backward-induction method generalizes: if on each move a player may add between $1$ and $M$, and the target is $T$, then winning positions are those congruent to $r \pmod{M+1}$ where $r$ is the smallest positive integer such that $T \equiv r \pmod{M+1}$ and that is reachable from a legal initial move.