Double occupancy of the Hubbard dimer ($P_{\mathrm{double}}=\frac{1}{2}(1-U/\sqrt{U^2+16t^2})$)

The ground-state probability of both electrons occupying one site is $P_{\mathrm{double}}=\frac{1}{2}[1-U/\sqrt{U^2+16t^2}]$. It equals the derivative of the relative ground energy with respect to $U$. It tends to zero for $|t|/U\to 0$ and to $1/2$ for $|t|/U\to \infty$.