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_{Disclaimer: This is an open-problem, I don't have a complete solution for this puzzle yet.}

We are playing a $2$-player game: you as a challenger and me as a judge. Initially, there is an empty infinite $2$-dimensional plane.

Alternately starting from you, put a single point. Your points will be denoted as $O$-points and mine will be $X$-points. The point we put can be anywhere — it may have Real coordinate — as long as it is not overlapping with any previous points.

You will win if and only if at some point of time, you can draw a line segment passing $N$ number of $O$-points without passing any $X$-points. Thus, you will lose if and only if I can make sure you can't reach your goal at any point of time.

Questions:

1. If $N = 4$, can you win this game?


2. If $N = 5$, can you win this game?


3. Can you win this game for any $N$? If yes, then what is the strategy? If no, then what is the smallest number of $N$ which is impossible for you to win?

This is an example of the game for $N = 3$:

I think you can

win even if you need to connect a kajillion dots.

Here's my reasoning:

Well, any two points can be connected by a line, right?

Then you can see that

When you put a point down, you can draw a line from that point to any other point you already put down. important

The challenge is that your opponent can block the connections with a point of their own.

But as you remember about

the birthday paradox

Because of the important knowledge above, you see that

you can connect lines to $n-1$ dots when you put the $nth$ dot down.

Making the equation

When you put the $nth$ dot, you have 1 + 2 + 3 + ... + (n-1) connections.

Let's do 4.
By putting

101 dots down, you have 5050 connections!

Note: In this step don't put any dots in 3 in a row.

That's a lot of connections you can make with lines.

Since your opponent has only put

101 blocking dots down

you can

make a 3 in a row with 2 of the remaining connections you have by looking at 2 non-parallel connections. 2 non-parallel lines will connect, right?

Your opponent can't block both lines, so you got a 4 in a row!

The strategy for $x$ dots in a line is

put many many dots to make a kajillion connections. Maybe even Graham's number.

Then

make 3 in a row with $2x-3$ non-parallel connections.

Then

make 4 in a row with $2x-4$ non-parallel connections.

Eventually you

make 2879 in a row with $2x-2879$ non-parallel connections.

And then you make

$x$ in a row with $2x-x$ connections. (The hint there is that $2x-x=x$.)

So you can get infinity in a row! YAY!