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
Best Answer
Let the referee's number be $Y$, and your choice $X$. Then you win if nobody's choice is in the interval from $X$ to $2Y-X$. If $0 \le 2Y-X \le 1$, that interval has length $2|Y-X|$. But if $2Y-X < 0$, its intersection with $[0,1]$ has length $X$, and if $2Y-X > 1$ it has length $1-X$. The probability that none of the other players make a choice in an interval of length $L$ contained in $[0,1]$ is $(1-L)^n$. Thus the conditional probability that you win, given $X=x$ and $Y=y$, is $$ \cases{(1-2|y-x|)^n & if $x/2 \le y \le (1+x)/2$\cr (1-x)^n & if $y \le x/2$\cr x^n & if $y \ge (1+x)/2$\cr}$$ Integrating over $y$, I find that the probability that you win if you choose $x$ is $$ f(x) = {\frac { \left( x\,n+2\,x-1 \right) \left( 1-x \right) ^{n}+ \left( - n-2 \right) {x}^{n+1}+2+ \left( n+1 \right) {x}^{n}}{2\,n+2}} $$ An optimal strategy would be to choose $x$ that maximizes this.
One critical point is $x=1/2$, but $f''(1/2) = (n-4) n 2^{1-n}$, so if $n > 4$ this is a local minimum. For example, if $n=5$ we have $$f(x) = {\frac { \left( 7\,x-1 \right) \left( 1-x \right) ^{5}}{12}}-{\frac {7\,{x}^{6}}{12}}+{\frac{1}{6}}+{\frac {{x}^{5}}{2}}$$ whose maximum on $[0,1]$ is at the real roots of $7\,{x}^{4}-14\,{x}^{3}+18\,{x}^{2}-11\,x+2$, approximately $0.2995972362$ and $0.7004027638$.
On the other hand, for $n \le 4$ the optimal solution is $x = 1/2$.