TITLE: A Game in Probability
QUESTION [7 upvotes]: You and n other players ( $n \ge 1$ ) play a game. Each player chooses a real number between 0 and 1. A referee also chooses a number between 0 and 1. The player who chooses the closest number to the referee's number wins. What should be your choice. Will it depend on $n$ ? 
Assume that the n players (other than yourself) and the referee choose the numbers uniformly between 0 and 1. 
Can someone give a sketch of the solution? I've been trying this but unable to produce an answer. 
Edit:You may ignore two or more players choosing the same number

REPLY [2 votes]: Building on the very nice solution by antkam, we can calculate the asymptotics for the optimal choice and the corresponding winning probability.
So let $x$, $y$, $z$ be as defined in antkam’s answer. For large $n$, we can approximate the other players’ guesses as a Poisson process with density $n$. (The total number of guesses is fixed, but the resulting correlation between guesses appearing at different locations can be neglected as $n\to\infty$, since a guess at one location only reduces the density remaining for another location by a fraction $\frac1n$.) To simplify the calculations, I’ll do them for density $1$ and then rescale.
So we choose $0$ on the circle and then cut at $x$. The basic winning probability is $\frac1{n+1}\sim\frac1n$, and we need to calculate two corrections to this, dependent on $x$.
The case where we gain due to the cut is the one where $0\lt y\lt x\lt z$ with $y\gt z-y$. The probability for this to occur, conditional on $y$ being the first of the $n+1$ values, is
\begin{eqnarray}
\int_{\frac x2}^xf_Y(y)\left(F_{Z-Y}(y)-F_{Z-Y}(x-y)\right)\mathrm dy
&=&
\int_{\frac x2}^x\mathrm e^{-y}\left(\mathrm e^{-(x-y)}-\mathrm e^{-y}\right)\mathrm dy
\\
&=&
\int_{\frac x2}^x\left(\mathrm e^{-x}-\mathrm e^{-2y}\right)\mathrm dy
\\
&=&\frac x2\mathrm e^{-x}+\frac12\left(\mathrm e^{-2x}-\mathrm e^{-x}\right)\;.
\end{eqnarray}
The case where we lose due to the cut is the one where $0\lt x\lt y\lt z$ with $y\lt z-y$. The probability for this to occur, conditional on $y$ being the first of the $n+1$ values, is
\begin{eqnarray}
\int_x^\infty\mathrm f_Y(y)\left(F_{Z-Y}(\infty)-F_{Z-Y}(y)\right)\mathrm dy
&=&
\int_x^\infty\mathrm e^{-y}\mathrm e^{-y}\mathrm dy
\\
&=&
\int_x^\infty\mathrm e^{-2y}\mathrm dy
\\
&=&
\frac12\mathrm e^{-2x}\;.
\end{eqnarray}
Subtracting the second result from the first yields
$$
\frac x2\mathrm e^{-x}-\frac12\mathrm e^{-x}\;,
$$
and setting the derivative to $0$ yields the optimum at $x=2$. This needs to be rescaled by the density $n$, so the optimum asymptotically occurs at $\frac2n$, in agreement with antkam’s findings. The difference in the winning probability also has to be multiplied by $\frac1{n+1}\sim\frac1n$, the probability for $y$ to be the first of the $n+1$ values. So the optimal winning probability is asymptotically
$$
\left(1+\frac{\mathrm e^{-2}}2\right)\frac1n\approx\frac{1.068}n\;.
$$