In Texas Hold'em, dealt a pocket pair, how do you find the probability that the flop contains at least 1 card of your denomination In Texas Hold'em, dealt a pocket pair, how do you find the probability that the flop contains at least 1 card of your denomination. 
Appreciate any help explaining this problem, as I am stumped trying to use the complement rule.
 A: Say you have a pair of Jacks in hand.
First, there are $\binom{50}{3}=\frac{50\cdot 49\cdot 48}{6}=19600$ possible flops (only 50 cards to choose from because you already have two in your hand).
Then there are 48 flops giving you 4 of a kind.  How many give you 3 of a kind?
Well, you can choose either of the remaining Jacks (a factor of two).  Then you can choose two other cards in the flop, out of the 48 (not 49, because that last card is a Jack) cards. So you can flop a single Jack in $2\cdot \binom{48}{2} = 2\cdot \frac{48\cdot 47}{2} = 2256$ ways.  
So the answer is $\frac{2256+ 48}{19600} = \frac{6}{50}\cdot \frac {48}{49} = 11.76\%$
A quick and easy way to get a good estimate is to say that you have a 2 out of 50 chance to hit your J on each of three cards, giving 6 out of 50 or 12%.  This over-counts one case of getting 4 Jacks.  
A: We have suppose two same cards of rank $\bf X$, now dealer deals the cards(out of $52-2=50$), he'll deal 3 cards of flop so choosing 3 cards out of remaining 50:
$$\binom{50}3=19600$$
If some cards of $\bf X$ are dealt(out of $4-2=2$ remainign of $\bf X$) and then rest out of the pack, possible ways are.There are two remaining cards of $\bf X$ as we already have the other two, so it may be possible that dealer deals 1 of $\bf X$ and rest 2 of other and so on:
$$\underbrace{\binom{2}{1}}_{\text{1 of ${\bf X}$ out of 2}}\times\overbrace{\binom{52-4}{2}}^{\text{2 remaining out of non-${\bf X}$(52-4)}}+\underbrace{\binom{2}{2}}_{\text{2 of ${\bf X}$ out of 2}}\times\overbrace{\binom{52-4}{1}}^{\text{1 remaining out of non-${\bf X}$(52-4)}}=2304$$ 
Total probability:
$$P=\frac{2304}{19600}=\frac{144}{1225}\approx11.76\%$$

Note: similar explanation holds for this too.
We have suppose two same cards of suit $\bf X$, now dealer deals the cards(out of $52-2=50$):
$$\binom{50}3=19600$$
If some cards of $\bf X$ are dealt(out of $13-2=11$) and then rest out of the pack, possible ways are:
$$\binom{11}{1}\binom{50-13}{2}+\binom{11}{2}\binom{50-13}{1}+\binom{11}{3}\binom{50-13}{0}=9526$$ 
Total probability:
$$P=\frac{9526}{19600}=\frac{4763}{9800}\approx48.6\%$$

