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
 A: The denominator is correct.  So the numerator must be wrong.  
The probability of not getting a pair is:
$$
\dfrac{48}{51} \dfrac{44}{50}\dfrac{40}{49}\dfrac{36}{48}\approx 0.507
$$
It's complement is $0.493$.
A: One error is that you said $\dbinom 4 2 = 12,$ whereas in fact $\dbinom 4 2  =6.$
A sublter error is that as an attempt to count the number of ways to get at least one pair, $13\dbinom 3 2\dbinom{50} 3$ is an overcount.
In the first place, if you choose the remaining cards out of $50$ that are not included in the pair, you may get two of the same rank from which the pair was taken, so you don't have a pair, but rather three or four of that same rank.
Secondly, your other three cards may contain a pair of a different rank (hence you said "at least one", but that hand then gets counted twice.
For example, suppose you choose one rank out of $13$ and it's $6.$ Then you choose two of the four $6$s. Among the hands containing those two $6$s is another with two $7$s.  But then when you list all the hands that contain a pair of $7$s, that same hand gets counted again.
The answer by Eric Fisher shows that this can be done by finding the probability of getting no pairs and then subtracting that from $1.$
A: 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 
$$1 - \frac{\dbinom{13}{5}4^5}{\dbinom{52}{5}}$$
