I am trying to solve the following probability problem:
Given a regular deck of playing cards containing 52 cards, we draw 3 cards without replacement. Given that the second and third cards are spades, what is the probability that the first card is also a spade?
I am aware that the same question has been asked here
I understand the solution that is being presented in the referenced link.
However I attempted to solve the problem using combination rather than permutation:
Let $ S_i $ denote the event that the $ i^{th} $ card is a spade and $ S_{i}^{c} $ otherwise.
By Bayes' theorem:
$$ \begin{aligned} P(S_{1} | S_{2}S_{3}) &= \frac{P(S_{1}S_{2}S_{3})}{P(S_{2}P_{3})} \\ \\ &= \frac{P(S_{1}S_{2}S_{3})}{P(S_{1}S_{2}P_{3}) + P(S_{1}^{c}S_{2}P_{3})} \\ \\ &= \large\frac{\frac{C(13,3)}{C(52,3)}}{\frac{C(13,3)}{C(52,3)} + \frac{C(39,1) \cdot C(13,2)}{C(52,3)}} \\ \\ &= \frac{11}{128} \end{aligned} $$
Correct answer is $ \frac{11}{50} $
After some checking, I realized that the term $ \frac{C(39,1) \cdot C(13,2)}{C(52,3)} $ is causing my method to produce the wrong answer. But I can't seem to understand why this term is incorrect. Could someone please explain to me?
I'm not sure that that term is the one causing you trouble! It is the other term. The key is that order of cards does matter. In a typical poker hand order doesn't matter.
This gives us that you should be tripling your values for each collection of three spades; the collection Ace of Spades, 2 of Spades, and 4 of Spades has three distinct choices for first card.
The correct calculation is \begin{equation*} \mathbb{P}(S1|S2S3)=\frac{3\cdot C(13,3)}{3\cdot C(13,3)+C(39,1)\cdot C(13,2)} \end{equation*}
To simplify, imagine that you have a twelve card deck with four suits of three cards. The desired probability should be $\tfrac{1}{10}$. Choose the first card, and then choose the remaining two together:
Four hands begin with a spade as a choice, and there is only one way to end up with all spades. Using your calculation, you would only have one way to arrive at this sequence.
Twelve hands begin with no spade, and there are $C(3,2)$ ways to draw the next two spades. \begin{equation*} \mathbb{P}(S1|S2S3)=\frac{4}{4+36}=\frac{1}{10} \end{equation*}