Consider the following scenario:
You and other n-1 players play a game together, where each come up with a number in the range 0~100. At
the end of the game, the player with the number nearest to the mean of all player's number will win the game.
Knowing nothing but the distribution of other players' distribution is Unif[0,10]. What should you come up with so that you can
win the game with max probability?
Give it a try:
Try some "lucky-numbers" reported by previous players:
Are they, or any of them are truly lucky?
Mechanism behind
Denote each player as $U_i \sim \text{Unif}[0,100]$ and any two are i.i.d., we have:
$$\begin{align*}
&\text{argmin}_X \left|X-\frac{\mathbb{E}\left[\sum_{i=1}^{n-1} U_i \right]+X}{n}\right|\\
&= \text{argmin}_X \left(X-\frac{\sum_{i=1}^{n-1} \mathbb{E}\left[U_i \right]+X}{n}\right)^2\\
&= \text{argmin}_X \left(X-\frac{(n-1) \mathbb{E}\left[U_i \right]+X}{n}\right)^2 \\
&= \text{argmin}_X \left(\frac{n-1}{n}X-\frac{n-1}{n} \mathbb{E}\left[U_i \right]\right)^2 \\
&= \text{argmin}_X \underbrace{\left(\frac{n-1}{n} (X- \mathbb{E}\left[U_i \right])\right)^2}_{\mathcal{J}(X)} \\
\end{align*}$$
$$\begin{align*}
\Rightarrow \left . \frac{\partial \mathcal{J}(X)}{\partial X} \right|_{X=\hat{X}}&=0\\
\frac{\partial \left(\frac{n-1}{n}\right)^2 2 \left( X-\mathbb{E}[U_i]\right)}{\partial X}&=0,\\
\hat{X}&=\mathbb{E}\left[U_i \right],\\
\end{align*}$$
where in our case, $\mathbb{E}\left[U_i \right]$ is just $50$