What is the chance that someone is dealt a suit in the game of Bridge? In a game of Bridge, what is the probability that some player has a complete suit?
(There are four players in a Bridge card game.  A deck of Bridge cards consists of 52 cards arranged in four suits of thirteen each. Playing Bridge begins by distributing the cards randomly to four players, namely North, South, East, West, so that each receives 13 cards.)
I am referring to various books on statistics to solve this problem.
 A: Using the conditional probabilities, $P$ with a subscript $i$ $i=1, 2, \ldots, 13$ to denote the probability of getting the "right" card, we can use inductive reasoning to calculate the probability. We can assume WLOG that the player is dealt all 13 cards in sequence irrespective of what may be dealt to the other players. We care not what the suit of the first card is that is dealt, so the $P_1 = 1$, but the next card that is dealt must match that suit, which of the 51 remaining there are only 12 to draw, so $P_2 = 12/51$. Continuing in this fashion, the 13th card has probability $P_{13} = 1 / 40$ to be the correct card. Mathematically, this combinatoric reduces to (13! 39! / 52!) which is impossibly small. 
A: One player =  $\frac {\binom{4}{1}}  {\binom{52}{13}} = 6.29907808979643E-012$  
One player not having it is 1 - 6.29907808979643E-012 so that to the 4th is none
Any player $1 - (1 - 6.29907808979643E-012)^4 = 2.5196289499263E-011$
One or more is one minus the chance of no player   
This is a little odd because if 3 are suited then the 4th has to be suited  
