In Poker, I receive $2$ cards and they are a pair of $8$s. What are the chances of me getting, just a pair, 2 pair, and three of a kind(but nothing higher) as the other three cards are placed down?
I went about this problem for just the pair: $\binom{12}{3}\binom{4}{1}\binom{4}{1}\binom{4}{1}$ that we need to just choose three cards and pick random suits for them. We don't need to pick a suit for our $8$s because we already have them?
The second part, since we already have a pair, to get two pair: $\binom{4}{2}\binom{12}{1}\binom{11}{1}\binom{4}{1}$ we choose two suits for the two cards of the same rank, and then just choose a random card that is not the same rank as the first or second pair.
For the last part, since we already have a pair: $\binom{13}{1}\binom{4}{1}\binom{12}{2}\binom{4}{1}\binom{4}{1}$ we choose a card that is the same rank as our pair, choose its suit, and then pick two random cards and pick their suits.
Therefore, the number of ways is: $\binom{12}{3}\binom{4}{1}\binom{4}{1}\binom{4}{1} + \binom{4}{2}\binom{12}{1}\binom{11}{1}\binom{4}{1} + \binom{13}{1}\binom{4}{1}\binom{12}{2}\binom{4}{1}\binom{4}{1}$.
I am not very good at counting and keep practicing, so that is why I am asking is my logic right for this? For already having a pair is neat and giving me a new perspective on how counting problems work.
What you did was an attempt to count the number of possibilities for each outcome.
The first is correct: pick three different ranks that are not $8$, and pick the suit for each.
The second is also correct: pick a rank that is not $8$ for the second pair, pick the suits for each card in the second pair, and a third card that has a different rank, and its suit.
For the third: Pick another $8$ (there are two left to choose from), and then two different ranks, any suit: $2 \times {12 \choose 2} \times 4 \times 4$.
Finally, to get the probability of each, you divide by the total number of ways that you can choose three of the remaining $50$ cards without any restrictions. This is simply ${50 \choose 3}$.
A simpler example of why you divide is rolling an even number on a six-sided die. There are three even numbers between $1$ and $6$, and there are a total of $6$ possible outcomes. So the probability is $3/6 = 1/2$.