For integers $n$ such that $\:3< n\:$,$\:$ what is known about the following 2-player game:


Player\_1 and Player\_2 take turn choosing points on $\mathbb{R}^2$ that were not previously chosen, with Player\_1 going first. $\:$ If there is ever a segment with $n$ points chosen by Player\_1 and zero points chosen by Player\_2, then Player\_1 wins. $\:$ If infinitely many turns go by without that happening, then Player\_2 wins.
  
  
  

(That includes, if Player\_1 has a win, how long the win should take.)



$n$-in-a-row game on $\mathbb{R}^2$

> 
> 
> ```
> 
> The points on the lines and the lines through the points...
> The lines should be many with numerous joints,
> the points should be few and should sit at right places.
> What else? We can use high-dimensional spaces,
> if we do not mind doubling enemy's strength.
> It would not be bad if we bound the length
> of games, tell who wins, and what way he should play.
> The questions are many, but short is the day.
> One cannot resolve all the problems at once 
> by being just smart. Better leave them to Chance!
> The law of large numbers is our best friend.
> Who is unpredictable beats every trend.
> Of course, he can lose, but the God in the sky
> does know enough to condition, not try,
> thus weaving each random irrational state
> into the unerring decisions of Fate.
> 
> ```
> 
> 


The first player wins playing more or less randomly after he prepares the playing field in the right way.


As it has already been said, we can play this game in $\mathbb R^d$. Since, when projecting a high dimensional configuration to the plane, we can avoid extra triple intersections but not extra double intersections of lines, we should allow the opponent to make two moves for each our move to get a stronger version of the game.


Fix $n$ (the number of points on the line we want). Choose small $\delta>0$ and large integer $d>0$ in this order. Consider the cubic lattice in $\mathbb R^d$ with $n$ points along each side ($n^d$ points total). Make $\delta n^d$ moves choosing a random lattice point at each move. If this point is already occupied, just put a point somewhere far away and forget of it. However, mark the occupied point as "intended". Let the opponent move twice after each of your moves. At the end of the game, just look at what you got. Pay attention only to the fully filled (with either actual, or intended points) lattice lines in coordinate directions. You should see the following.


1) The probability to fill each particular line is about $\delta^{n}$, so you should expect $L=n^{d-1}d\delta^{n}$ filled lines. Moreover, the probability to fill less than half of this amount is extremely small because filling two disjoint lines are negatively correlated events and intersecting pairs are few, so the expectation of the square is pretty close to the square of the expectation if $n,\delta$ are fixed and $d\to\infty$.


2) The probability that a given grid point has $k$ filled lines passing through it is at most $\delta\cdot \frac{[d\delta^{n-1}]^k}{k!}$, which is exponentially small in $d$ if $k>2ed\delta^{n-1}$ in the same regime when $n,\delta$ are fixed and $d\to\infty$.


3) Each opponent's point that doesn't lie on the grid can interfere with at most one line.


4) Each opponent's point that lies on the grid can interfere with one of the filled lines only if it was intended during one of your moves. However, since during the whole game only $3\delta n^d$ points have been used, the chance to hit an already occupied point has never been higher than $3\delta$, so it is highly unlikely that the opponent could score much more than $6\delta^2 n^d$ such points.


Now it is time to count the blocked lines. First, $e^{-c(n,\delta)d}dn^d=e^{-c(n,\delta)d}\delta^{-n}nL$ come from "high efficiency" points (let the opponent grab them all, it is still nothing compared to $L$ if $d$ is large enough!). Second, the points put off the grid can block only $2\delta n^d=d^{-1}\delta^{-n+1}n L$ lines, which isn't much either. Last, the "efficient" but not "highly efficient" points on the grid block at most $6\delta^2 n^d\cdot 2ed\delta^{n-1}=2e\delta n L$ filled lines. Here we do not gain from $d$, but if we start with choosing $\delta$ so that $12e\delta n\ll 1$, we are still fine. Thus, normally at most $L/2$ lines will be blocked and, with high probability, the outcome is that we have won the game by this moment.


Since the game is now of fixed length, one of the players must have a deterministic winning strategy. But the second player loses against the random strategy with positive probability no matter what he does, so it isn't he. The deterministic winning strategy for the first player can be defined explicitly in terms of conditional probabilities to win the random game from the current position (just move to the point that gives you the best chance to win the random game in the maximin sense (max over locations, min over all possible opponent's strategies) but the computation of those conditional probabilities is well beyond the human abilities. 


