Let $X_1, X_2, Y_1, Y_2$ are {\it i.i.d.} r.v.s obeying the same probability destribution. Player I [II] looks privately $(X_1, X_2)= (x_1, x_2), [(Y_1, Y_2)=(y_1, y_2)]$. They choose a single common number $\theta$, and player II opens the nearest number to $\theta$ among $y_1$ and $y_2$ and covers the other number. If II's opened number is $>(<)\theta$, then I gets as his payoff, the opened (covered) number. The problem is to determine $\theta$ under which the expected payoff $M_1(\theta)$, I can get, is maximized. The maximization problem for II is quite similar as for I. Because of symmetry between the two players, our problem essentially reduces to computation of $M_1(\theta)$ and finding the $\theta$ which maximizes $M_1(\theta)$. This game is solved for (1) uniform distribution on [0, 1], (2) exponential distribution on $[0, \infty)$, (3) normal distribution on $(-\infty, \infty)$, and (4) some other distributions around them.