Counting to $P$ game - strategy? A while back I asked this question Counting to 21 game - strategy? on whether there was a strategy to stay in the 'counting to 21 game' described in that question with $N$ players.
This question will ask a more general version of that. The game is as follows:


*

*$N$ players.

*Each can say at least one number and at most $m$ (following the sequence $1,2,3,...$) in turn (i.e. player 1, then player 2,..., then player N, then back to player 1).

*The player with no alternative but to say the number $P$ is eliminated from the game.

*Every player is guaranteed to get to say at least one number (i.e. $P\gt m(N-1)$).


Given that I can chose if I go in the $l$th position. For what combinations (if any exist) of $N$, $P$, $m$ and $l$ can I always not be eliminated, regardless of how the other players play? 
 A: I believe that there is no possibility of systematic winning strategy for a single player in such a multiplayer version of the game. In a scenario with $N=2 \,$ players, the winning strategy for player A is to end his sequences on the numbers given by  $P - 1 -k (m+1) \, \, \,$, since this allows to conclude his last sequence on $P-1\,$. For example, in the case $P=21 \, $ (as in the previous question) and $m=5\,$, since player A wins if he ends his last sequence on $21-1=20 \, \,$ , stopping his second to last sequence on $21-1-(5+1)=14 \, \, \,$       allows him to be sure about the possibility of ending his successive sequence on $20$, whatever the choice of player B. Similarly, stopping his third to last sequence on $21-1-2 \cdot (5+1)=8 \, \, $ allows him to be sure to end his successive sequence on $14$ whatever the choice of player $B $, and so on. 
However, in the scenario with $\geq 3$   players, this strategy cannot be successful for a single player (let us call him again player A). As pointed out in the answer to the previous question for the original version with $m=3$  , even in this new modality of the game we can imagine that all remaining players could cooperate against him, with the advantage of having a considerably larger range of options.      Let us consider for example the case with three players. Players B and C could cooperate so that the last of them (i.e., that preceding player A) ends his last sequence on one of the numbers included in the range between  $P - 1 - (2m+1) \, \, $   and $P - 1 - (m+2) \, \, $  . In this way, whatever the choice of player A, they can force him to lose: since the distance $d $ to their target $P-1$   ranges from $m+2$      to   $2m+1$       , any successive choice of player A characterized by a sequence of $k $ terms (where $1 \leq k \leq m \, \, $  )   can be followed by two sequences of players B and C which in total include $d-k $    terms (where $2 \leq d-k \leq 2m \, \,$)     , and the target is achieved.  Note that the range between  $P - 1 - (2m+1) \,$        and $P - 1 - (m+2) \,$        includes $m $ terms, so that there is no way for player A, if the other players collaborate against him, to avoid that the player who precedes him ends within this range during the game. 
Generalizing, for the case with $N$ players where $N-1$    of them cooperate against player A, they can choose their sequences so that, at a given point of the game, the last of them ends his sequence on one of the numbers included in the range between  $P - 1 - [(N-1)m+1] \,\, \, $       and $P - 1 - (m+N-1) \, \,$      . In this way, whatever the successive choice of player A, they can force him to lose. The distance $d $ to their target $P-1$   ranges from $m+N-1 \,\,$     to $(N-1)m+1 \,\, $      . Therefore, any successive choice of player A characterized by a sequence of $k $ elements (where $1 \leq k \leq m \, \,$       )   can be followed by $N-1$       sequences of the other players including in total $d-k $    elements, where $N-1 \leq d-k \leq (N-1)m  \,\, \, \, $         , and the target is again achieved.  
