A poker hand is defined as drawing 5 cards at random without replacement from a deck of 52 playing cards. Find the probability of the following poker hand: Four of a kind (4 cards of equals face value and 1 card of a different value)?
at first I tought maybe 13!/9! (13p4)* 52!/51! (52p1) but that was wrong? any tips solutions?
Best Answer
There are $\binom{52}{5}$ ways to select a hand, and there are $\binom{13}{1}\cdot \binom{48}{1}$ ways to choose a hand with four of a kind. So: $P(A) = \dfrac{\binom{13}{1}\cdot \binom{48}{1}}{\binom{52}{5}}$, whereas $A$ = getting a four of a kind hand.