  Given integer $n$ and $k$ such that $0 < k \leq n$ and $n$ piles of stones,
two players alternate turns. By one move it is allowed to choose any $k$ piles
and remove exactly one stone from each. The player who has to move but cannot
is the loser. in the normal version of the game and (s)he is the winner in the
mis\`ere version. Cases $k=1$ and $k = n$ are trivial. For $k=2$ the game was
solved for $n \leq 6$. For $n \leq 4$ the Sprague-Grundy function was
efficiently computed (for both versions). For $n = 5,6$ a polynomial algorithm
computing P-positions was obtained for the normal version. \newline Then, for
the case $k = n-1$, a very simple explicit rule that determines the Smith
remoteness function was found for the normal version of the game: the player
who has to move keeps a pile with the minimum even number of stones; if all
piles have odd number of stones then (s)he keeps a maximum one, while the $n-1$
remaining piles are reduced by one stone each in accordance with the rules of
the game. \newline Computations show that the same rule works efficiently for
the mis\`ere version too. The exceptions are sparse and are listed in Section
2. Denote a position by $x = (x_1, \dots, x_n)$. Due to symmetry, we can assume
wlog that $x_1 \leq \ldots \leq x_n$. Our computations partition all exceptions
into the following three families: $x_1$ is even, $x_1 = 1$, and $x_1 \geq 3$
is odd. In all three cases we suggest explicit formulas that cover all found
exceptions, but this is not proven.
