The $n$ player 1-dimensional unit game
The $n\ge 2$ players take turns placing their own point on the unit segment $[0,1]$. (Hence, 1-dimensional.)
The closest distance they can place their point to someone elses placed point is some small $\varepsilon\gt 0$.
A player owns the part of the segment closest to their own point. The goal is to own the largest part of the segment. That is, say $a,b$ are closest points to your own point $x$. Then:
If your point ends up between two points $x\in(a,b)$ then your score is $s_x=\frac12(b-a)$.
If your point ends up in $x\in[0,a)$, then your score is $s_x=\frac12 (x+a)$.
Symmetrically, if your point $x\in(b,1]$, then your score is $s_x=1-\frac12 (x+b)$.
Which player(s) have the winning strategy? (Have highest chance of highest score, then maximize that score.)
WLOG (Symmetry) first player plays in $x_1\in[0,\frac12]$.
I've observed so far that the optimal play (and winniners) should be:
- $(n=2)-$ Is a win for the $1$st player. Player plays $x_1=\frac12$.
- $(n=3)-$ Is a win for either $1$st or $2$nd player (coin flip). Players play $x_1=\frac14+\frac12\varepsilon$ and $x_2=1-x_1$, then $x_3\in(x_1,x_2)$ uniformly randomly (we assume if a player is presented with equally good choices, he picks uniformly randomly), which results in $0.5$ chance (a coin flip) for the one of the $x_1,x_2$ to win (have highest score).
More details are below. Do correct me if I missed anything about $(n=3)$ there.
Can we find strategies and winners for $n\ge 4$ ?
$2$-player case is a win for $1$st player
For $n=2$, it is not hard to see that the first player has a winning strategy by playing $x_1=\frac12$.
The most the second player can score then is by playing $x_2=\frac12-\varepsilon$.
This nets them $s_2=\frac12 (\frac12-\varepsilon+\frac12)=\frac12-\frac12\varepsilon$ score, while the first player now has $s_1=\frac12+\frac12\varepsilon$.
$3$-player case is a win for either $1$st or $2$nd player with $0.5$ probability
For $n=3$, this strategy does not work for the first player anymore.
The third player would play $x_3=\frac12+\varepsilon$. (Mirror the second player.)
The second and third player now share the win by both having $\frac12-\frac12\varepsilon$ score.
The first player is now in $x_1\in(x_2,x_3)$ and their score is reduced to a minimal $\varepsilon$.
So WLOG if in $n=3$ the winning strategy exists for the first player, they will play in $x_1\in[0,\frac12)$.
Now the question is, what would be the optimal response of the second player to this?
The first player plays $x_1=\frac12-\delta$. Assuming $\delta$ is small enough, the second player can play $x_2=\frac12+\delta+\varepsilon$ to win the game. Notice that this play makes $[0,x_1)$ more favorable for the third player, than the $(x_2,1]$, meaning the third player would leech off the first player.
That is, by small $\delta$ we are assuming that the score $s_3$ of $x_3$ if they would play in $x_3\in(x_1,x_2)$ is smaller than if they would play optimally in $[0,x_1)$. In other words we are assuming $\delta \lt \frac14-\frac12\varepsilon$ here:
$$s_3((x_1,x_2))\lt s_3([0,x_1)) \iff \delta+\frac12\varepsilon\lt \frac12-\delta-\frac12\varepsilon \iff \delta \lt \frac14 -\frac12 \varepsilon$$
This forces $x_3$ to play in $[0,x_1)$, namely $x_3=\frac12-\delta-\varepsilon\implies s_3=\frac12-\delta-\frac12\varepsilon$.
The score of the second player will be $s_2=\frac12-\frac12\varepsilon$, and the first player will have $s_1=\delta+\varepsilon$.
We have $s_1\lt s_3 \lt s_2$. This now gives a winning strategy for second player if $x_1\in(\frac14+\frac12\varepsilon,\frac12]$.
So WLOG if in $n=3$ the winning strategy exists for the first player, they will play in $x_1\in[0,\frac14+\frac12\varepsilon]$.
If $x_1=\frac14+\frac12\varepsilon$ and $x_2=(1-x_1)+\varepsilon$ still, then $x_3$ would play uniformly randomly in $x_1\in(x_1,x_2)$ for $s_3=\frac14+\varepsilon$. This will result in $x_1$ winning with slightly larger probability than $0.5$ (depending on the size of $\varepsilon$), and other times $x_2$ will win. Can $x_2$ now change its strategy to improve its odds?
- If it increases the $\varepsilon$ increment, is has worse odds. (Depending on the size of the $\varepsilon$.)
- If it removes it, both players $x_1,x_2$ have exactly $0.5$ chance of winning. ($x_2$ mirrors $x_1$)
- If it decreases the $\varepsilon$ increment, is now always loses because $x_3\in(x_2,1]$ is forced.
That is, for $x_1=\frac14+\frac12\varepsilon$ it seems the optimal strategy for second player is $x_2=1-x_1$, to mirror it. This forces the $x_3$ to play uniformly randomly in $(x_1,x_2)$, giving both $x_1,x_2$ equal chances of winning (probability $0.5$).
The question now remains, can first player do better than $0.5$ if he picks $x\in[0,\frac14+\frac12\varepsilon)$ ?
It seems that if $x_1$ plays in this interval (is closer to $0$), this allows the $x_2$ to decrease the mentioned increment (come closer to the $\frac12$) without making $(x_2,1]$ more valuable than the $(x_1,x_2)$ for the $x_3$ to play in. Hence $x_2$ has higher probability than $0.5$ of winning now.
From all of these cases for $x_1$, it follows that the optimal play would be $x_1=\frac14+\frac12\varepsilon$, then $x_2=1-x_1$ and finally $x_3\in(x_1,x_2)$ does a coin flip to decide if it wants to steal more from $x_1$ or $x_2$, giving the other one the win.
$n\in\mathbb N$ players case
Is it possible to solve this for general $n$ ?
I'm wondering if there is a way to find conclusions about the game in general, other than tackling individual $n$ cases.