Required probability $=P($ When two persons have same birthday) $+P$ (When three persons have same birthday) $+P$ (When four persons have same birthday)
$={ }^{4} C_{2} \times \frac{1}{365} \times\left(\frac{364}{365}\right)^{2}+{ }^{4} C_{3} \frac{1}{(365)^{2}} \times \frac{364}{365}$
$+{ }^{4} C_{4}\left(\frac{1}{365}\right)^{3}$
$=\frac{6(364)^{2}+4 \times 364+1}{(365)^{3}}$
$=\frac{794976+1456+1}{(365)^{3}}$
$=\frac{796433}{48627125}$
$=0.016$