Question from Sheldon Ross's First Course in Probability, Chapter 2:
If it is assumed that all $\binom{52}{5}$ poker hands are equally likely, what is the probability of being dealt two pairs? (This occurs when the cards have denominations $a, a, b, b, c,$ where $a$, $b$, and $c$ are all distinct.)
1st approach:
Out of the 13 types in each deck possible, I select any one card (A,1,2,3...,K,Q,J) by $\binom{13}{1}$ ways, and take 2 of these selected out of 4 possible decks (Clubs, Spades, Diamonds, Heart) using $\binom{4}{2}$ ways. This gets me the first pair.
Similarly, select another card from the remaining 12 types using $\binom{12}{1}$ ways, and make another pair by $\binom{4}{2}$ ways.
Since the 2 types taken above cannot be chosen again, I exclude all the 2 types (8 cards in total), and select the one remaining card in $\binom{44}{1}$ ways.
This gives me the numerator as : $\binom{13}{1} * \binom{4}{2} * \binom{12}{1} * \binom{4}{2} * \binom{44}{1}$ ways which will be divided by $\binom{52}{5}$.
2nd approach:
Out of the 13 types in each deck possible, I select 2 card types (A,1,2,3...,K,Q,J) by $\binom{13}{2}$ ways, and for each type I select 2 cards out of 4 possible decks (Clubs, Spades, Diamonds, Heart) using $\binom{4}{2} * \binom{4}{2}$ ways. This gives me both pairs.
Again, I select the remaining one card in $\binom{44}{1}$ way, by following the previous approach.
The final numerator returns: $\binom{13}{2} * \binom{4}{2} * \binom{4}{2} * \binom{44}{1}$.
Why is both the solution different, and what is the right approach? I want to know the underlying concept underneath it - if I am unconsciously solving the question sequentially or what not.
