What is the probability of getting at least one pair in Poker? Problem:
A 5-card hand is dealt from a perfectly shuffled deck of playing cards. What is the probability of each
of the hand has at least two cards with the same rank.
Answer:
By a rank, I mean a card like a $2$ or a king. The set of all poker hands is ${52 \choose 5}$. Let $p$ be the
probability we seek. I note that there are $13$ ranks of cards and for each rank there are $4$ cards.
\begin{eqnarray*}
p &=& \frac{ 13{ 4 \choose 2 }{ 50 \choose 3  } } { { 52 \choose 5 } } \\
{ 4 \choose 2} &=& \frac{4(3)(2)}{2} = 12 \\
{50 \choose 3} &=& \frac{50(49)(48)}{3(2)} = 25(49)(16) \\
%
{52 \choose 5} &=& \frac{52(51)(50)(49)(48)}{5(4)(3)(2)} = \frac{52(51)(10)(49)(48)}{4(3)(2)} \\
 &=& \frac{52(51)(10)(49)(12)}{3(2)}  = 52(51)(10)(49)(2) \\
p &=& \frac{ 13(12)(25)(49)(16) } { 52(51)(10)(49)(2)  } = \frac{ 13(12)(25)(49)(8) } { 52(51)(10)(49)  } \\ &=& \frac{ 13(6)(25)(8) } { 52(51)(5)  } = \frac{ 13(3)(25)(8) } { 26(51)(5)  } \\ &=& \frac{ 13(3)(5)(8) } { 26(51)  } \\ &=& \frac{ 1560 } {1586 } \\
\end{eqnarray*}
I am fairly sure that my answer is wrong but I do not understand where I went wrong. I am hoping somebody can tell me where I went wrong.
Thanks,
Bob

As Eric Fisher pointed out in his answer, the probability of obtaining at least one pair can be found by subtracting the probability of obtaining no pairs from $1$.  If no pairs are obtained, we must select five different ranks.  There are $\binom{13}{5}$ ways to select the five ranks.  For each rank, we must select one of four suits.  Hence, the number of hands that do not contain a pair is 
$$\binom{13}{5}4^5$$
so the probability of not obtaining a pair is 
$$\frac{\dbinom{13}{5}4^5}{\dbinom{52}{5}}$$
Hence, the probability of obtaining at least one pair is 
$$1 - \frac{\dbinom{13}{5}4^5}{\dbinom{52}{5}}$$
