Problem 5

Alice and Bazza are playing the inekoalaty game, a two-player game whose rules depend on a positive real number

\lambda

which is known to both players. On the

n

th turn of the game (starting with

n=1

) the following happens: if

n

is odd, Alice chooses a nonnegative real number

x_n

such that

x_1+x_2+\cdots+x_n\le\lambda n

; if

n

is even, Bazza chooses a nonnegative real number

x_n

such that

x_1^2+x_2^2+\cdots+x_n^2\le n

. If a player cannot choose a suitable number

x_n

, the game ends and the other player wins. If the game goes on forever, neither player wins. All chosen numbers are known to both players. Determine all values of

\lambda

for which Alice has a winning strategy and all those for which Bazza has a winning strategy.

Step 1 of 4: Alice plays 0 forever, then unleashes one huge move

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\lambda>\tfrac1{\sqrt2}\ \Longrightarrow\ \text{Alice wins}

Detailed analysis

Suppose Alice plays $x_{2i-1}=0$ on every odd turn. Whatever Bazza does, his own constraint forces $x_2^2+\cdots+x_{2k}^2\le2k$ , so by Cauchy–Schwarz $x_2+\cdots+x_{2k}\le\sqrt{k\cdot2k}=k\sqrt2$ ; since the odd terms are $0$ , the running total after turn $2k$ is $x_1+\cdots+x_{2k}\le k\sqrt2$ . If $\lambda>1/\sqrt2$ , choose $k$ large enough that $\lambda(2k+1)-k\sqrt2>\sqrt{2k+2}$ (possible since the left side grows linearly in $k$ while the right side grows like $\sqrt k$ ). Then Alice may legally play $x_{2k+1}$ equal to any value up to $\lambda(2k+1)-(x_1+\cdots+x_{2k})$ , which is at least $\lambda(2k+1)-k\sqrt2>\sqrt{2k+2}$ ; playing such a value makes $x_1^2+\cdots+x_{2k+1}^2>2k+2$ already, so Bazza has no legal $x_{2k+2}\ge0$ on the next turn and loses.