Poker probabilities 3 of a kind, 2 pair, 1 pair I'm finding it hard to understand why the answers are what they are. 
correct # ways to get 3 of a kind: $13C1·4C3·12C2·4C1·4C1$
my thinking: $13C1 * 4C3 * 12C1 * 4C1 * 11C1 * 4C1$
the logic is that the value of the card that repeats 3 times can be out of 13 possible choices, and you choose 3 suits. The 4th card can be out of 12 choices, an you choose out of suit. The 5th card can be out of 11 choices, and you chooseone suit. 
If indeed this method leads to ovecounting, why is the solution not 
$13C3 * 4C3 * 4C1 * 4C1$
The logic above is that out of the 5 cards we choose, we only want 3 unique numbers. The first unique number repeats 3 times, so we have 4C3 options. The second unique number repeats once, so we have 4 options. 
Im getting the same wrong answers with the question to # of ways to get 2 pairs. 
my thinking: $13C1 * 4C2 * 12 * 4C2 * 11 * 4$. 
My thinking for # ways to get 1 pair: $13 * 4C2 * 12 * 4 * 11 *4 * 10 *4$
Again, I see the correct answer, and I wonder why its not $13C4 * 4C2 * 4C1 * 4C1$
Thanks in advance! 
 A: To obtain three of a kind, you must draw three cards from one kind and one card each from two of the remaining suits.  
Using that reasoning, you realized that there are $\binom{13}{1}$ ways of selecting the kind, $\binom{4}{3}$ ways of choosing three of the four suits for that kind, $\binom{12}{1}$ ways of selecting one of the other kinds for the fourth card, $\binom{4}{1}$ ways of selecting its suit, $\binom{11}{1}$ ways of selecting one of the remaining kinds for the fifth card, and $\binom{4}{1}$ ways of selecting its suit.  That led you to the apparent answer 
$$\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{1}\binom{11}{1}\binom{4}{1}$$
However, you have overcounted.  Consider a particular hand.
$$J\clubsuit, \color{red}{J\heartsuit}, \color{red}{J\diamondsuit}, 5\spadesuit, \color{red}{2\diamondsuit}$$ 
Once you have selected the three Jacks, you could select a $5\clubsuit$ then a $\color{red}{2\diamondsuit}$ or you could select a $\color{red}{2\diamondsuit}$ then a $5\clubsuit$.  You will obtain the same hand.  Thus, you must divide your answer by $2$. Equivalently, after you choose three of a kind in $\binom{13}{1}\binom{4}{3}$ ways, you choose two of the remaining $12$ kinds, then select one of the four suits for each, yielding 
$$\frac{\binom{13}{1}\binom{4}{3}\binom{12}{1}\binom{4}{1}\binom{11}{1}\binom{4}{1}}{2} = \binom{13}{1}\binom{4}{1}\binom{12}{2}\binom{4}{1}\binom{4}{1}$$
ways of selecting three of a kind.
Your answer $\binom{13}{1}\binom{4}{3}\binom{4}{1}\binom{4}{1}$ fails to take into account that there are $48$ cards from which the fourth card can be selected (those that are not of the kind from which three suits have been selected) and $44$ cards from which the fifth card can be selected (those not of the two kinds already selected).  Since the order in which the fourth and fifth cards are selected does not change the hand, the number of hands containing three of a kind is also
$$\frac{\binom{13}{1}\binom{4}{3}\binom{48}{1}\binom{44}{1}}{2}$$ 
As you can check, the three answers are equivalent.
You can calculate the correct number of one pair and two pair hands using similar reasoning.  
