I'm trying to understand why my method doesn't work to answer the question: What is the probability of having 3 cards of the same suit and $2$ cards of the same suit (but a different suit than the first three) in a $5$ card hand from a standard $52$ card deck?
The method that seems to work uses combinations. You do: $$ {4 \choose 1}\cdot{13 \choose 3}\cdot{3 \choose 1}\cdot{13 \choose 2} = 267,696 $$ To find the number of successful possibilities and then $$ {52 \choose 5} = 2,598,960 $$ To find the number of total possibilities. So the answer is: $$ \frac{267,696}{2,598,960} $$
I'm wondering why figuring out the individual card chances and then multiplying them doesn't work. My method goes like this: $$ \frac{12}{51}\cdot\frac{11}{50}\cdot\frac{39}{49}\cdot\frac{12}{48} $$
After you've picked your first card you have a $\frac{12}{51}$ chance of getting a matching second card, then you've a $\frac{11}{50}$ chance of getting your third card to match.
For the other two cards you've a $\frac{39}{49}$ chance, because you still have three suits left. You need the next one to match, so now you have a $\frac{12}{48}$ chance.
Shouldn't multiplying all of those individual probabilities yield the same answer as the first method? I realize that there are different ways to order the individual probabilities, but when they're all multiplying, shouldn't it all work out to be the same?
I've noticed that multiplying the answer to my method by $10$ makes it equal to the first method's answer. Is there a reason that works?