Expected number of people to not get shot? Suppose $n$ gangsters are randomly positioned in a square room such that the positions of any three gangsters do not form an isosceles triangle.
At midnight, each gangster shoots the person that is nearest to him. (A person can get shot more than once but each person can only shoot one person)
How many people are expected to survive? (I.e. what is the expected value of the number of people who do not get shot?)
E.g. For one person, the expected value is 1. For two people, it is zero since they both get shot. For three, the value is 1 since they form the vertices of a scalene triangle. I'm just interested in what happens as $n \rightarrow \infty$. 
Thanks for your help!
 A: ${\bf Update~Oct~19}$: Added some analysis on the $D\to\infty$ limit of $\frac{E[n]}{n}$ for the original game and to the slightly modified case where the geometry is that of a torus.

Martin have done a great job in exploring the problem and as mentioned in his answer there seem to be some disagreement in the litterature about the exact value of $\lim\limits_{n\to\infty}\frac{E[n]}{n}$ where $\frac{E[n]}{n}$ is the fraction of survivors in a game with $n$ players. I will here try to nail down this limit to $4-5$ decimal places for the first few values of $D$ - the dimension the game is played in.

Numerical algorithm
The bootleneck in the numerical calculation is finding the closest neighbor of any given player. A brute force search scales as $O(n^2D)$ which on a single CPU becomes way to slow for $n\gtrsim 10^2-10^3$. To improve on this I added a uniform grid with $M^D$ gridcells covering $[0,1]^D$ to the simulation. $M$ is choosen such that $n_{\rm per~cell}  = n/M^D \sim $ a few. Players are added to the cells and when we search we normally only need to go through the neighboring $3^D$ cells to find the closest neighbor to any given player. This brings the number of operations down to $O(nD3^D)$ allowing us to go to much larger $n$ than with brute-force search. The code I used to do this, written in c++, can be found here. The code is only suitable to explore relatively low values of $D \lesssim 5$ as the grid needed to speed up the calculation becomes too memory expensive for large $D$ (plus the algorithm scales exponentially with $D$).

Results
In the figure below one can see the cummulative average of $\frac{E[n]}{n}$ as function of the number of samples for $\{D=1,~n=10^3\}$ (left) and $\{D=2,~n=10^4\}$ (right).


In the figure below I show the evolution of $\frac{E[n]}{n}$ as a function of $n$ for different values of $D$. For each $n$ I performed $N$ samples (varying from $10^2-10^8$ depending on $D$ and $n$) until the desired accuracy was reached. The error bars shows $3\hat{\sigma}$ ($99.7\%$ confidence) of the standard error $\hat{\sigma} = \frac{\sigma}{\sqrt{N}}$ where $\sigma^2 = \frac{1}{N}\sum_{i=1}^N(f_i-\overline{f})^2$ and $f_i$ is the fraction of survivors in one single run and $\overline{f} = \frac{1}{N}\sum_{i=1}^Nf_i$ is the cummulative mean. To be able to show all in one plot I have subtracted $f$ (the value given in the table below).
$~~~~~~~~$
This gives me the following result:
\begin{array}{c|c}
D & f=\lim\limits_{n\to\infty}\frac{E[n]}{n} \\ \hline
1 & 0.25000 \pm 10^{-5} \\ \hline
2 & 0.28418 \pm 10^{-5}\\ \hline
3 & 0.30369 \pm 10^{-5}\\ \hline
4 & 0.3170  \pm 10^{-4}\\
\end{array}
The quoted error is the $99.7$% confidence statistical error plus the estimated error in the evolution with $n$ (only relevant for $D=4$ as we have convergence to within the statistical error for lower $D$). Note that for $D=1$ we have the analytical result $f=\frac{1}{4} = 0.25$ which serves as a test of our numerical analysis.
The large $D$ limit
I also did a some simulations for low values of $n\lesssim 10^3$ looking at the evolution of $\frac{E[n]}{n}$ with $D$ - the dimension the game is played in. For these calculations I just used a brute forced neighbor-finding algorithm. 
I considered both a closed box and a box with periodic boundary conditions (i.e. $x_i=0$ is the same point as $x_i=1$). The latter situation is equivalent to doing the game on a torus. The results are seen below
$~~~~~~~~~~~~~~$
For low values of $D$ the results between the two geometries are pretty similar, but we start to see some big differences for large values of $D$ (and $n$). The reason the boundary effects becomes more and more important for large $D$ can be understood by considering a sphere at the center of our box (with radius $1/2$). As we increase $D$ we find that the volume of the sphere to the total volume of the box goes to zero so (loosely speaking) most of the volume of the box is in the corners. A player in a corner is less likely to get shot than a person close to the center thus a common sitation for large $D$ is that we have many players in different corners shooting players close to the center giving us a high survival percentage.
For the torus geometry a person in the corner is just as likely to get shot as anybody else and if $A$ is $B$'s closest neighbor than $B$ is $A$'s closest neighbor with probabillity $1/2$ as $D\to\infty$ (see this related question) which implies that $$\lim\limits_{D\to\infty}\frac{E[n]}{n} \to \left(1-\frac{1}{n}\right)^{n-1}$$ which converges to $\frac{1}{e}$ for $n\to\infty$.
A: Here's a simulation in Octave. "Model" is the linear model $\frac{2}{7}n$ which martin showed above. Maybe it would be interesting to also investigate the spread and how it changes with $n$?

Even if the mean value of surviving gangstas is pretty close to $\frac{2}{7}n$ as martin discovered in his answer, we can also see that there is a quite large spread of gangsta fatalities. A spread which seems to increase with increasing number of gangstas.
Octave code:

N_lst = 1:45;
s_lst = 1:25;
data_table_ = zeros(numel(N_lst),numel(s_lst));
for i_N = 1:numel(N_lst);

for i_s = s_lst;

N_ = N_lst(i_N);
d_x = rand(N_,1)';
d_y = rand(N_,1)';
D = abs((vec(d_x)'-vec(d_x)) + 1i*(vec(d_y)'-vec(d_y))) + eye(N_)*2;
[v,i] = min(D,[],1);
data_table_(i_N,i_s) = numel(unique([i(1,:)]));

endfor

endfor

