Originally Posted by

**Vinod** I have solved this question as follows.

In a game of bridge, let $A_i$ be the event "Player number i has a complete suit". Then $p_i=\frac{4}{\binom{52}{13}}$; the event that both players i and j have complete suits can occur in 4*3 ways and has probability $p_{ij}=\frac{12}{\binom{52}{13}\binom{39}{13}}$. Similarly we get $p_{ijk}=\frac{24}{\binom{52}{13}\binom{39}{13} \binom {26}{13}}$

Finally $p_{ijkl}=p_{ijk}$ since whenever the 3 players have the complete suit so does the fourth.

The probability that some player has a complete suit is therefore $P_1=4*p_i-6*p_{ij}+4*p_{ijk}-p_{ijkl}$.Hence the final answer $2.519631234522642 * 10^{-11}$