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?
Best Answer
It's true that $$ P(S_1 S_2 S_3) = \frac{C(13,3)}{C(52,3)}$$ and also that $$ P(S_1^c S_2 S_3) \neq \frac{C(39,1) \cdot C(13,2)}{C(52,3)}.$$
In each case, the denominator on the right-hand side counts the number of possible three-card hands, ignoring the order in which the cards are drawn. The numerator $C(13,3)$ counts the number of hands with three spades; although it again disregards the order in which the cards were drawn, the event that you draw a spade followed by two others is exactly the event that you draw three spades, so indeed the order of drawing does not matter for this particular probability.
But suppose we wanted the probability that you drew a hand with exactly two spades in it. Examples of such hands are $(2\spadesuit, 9\spadesuit, 3\diamondsuit)$, $(5\spadesuit, 9\heartsuit, \mathrm J\spadesuit)$, and $(9\heartsuit, 4\spadesuit, \mathrm 8\spadesuit)$ -- shown here in the order the cards were drawn, but for this particular probability all these hands are considered "two spades and one non-spade."
To compute the probability of any hand with exactly two spades, combinations are fine: to get a random three card hand we can simply line up the $52$ cards and pick three at random, with $C(52,3)$ equally-likely possible outcomes. The hands with exactly two spades will occur when two of the three choices occurred among the spades (there are $C(13,2)$ possible ways to do this) and one choice occurred among the other three suits (there are $C(39,1)$ possible ways to do that). The total number of hands with exactly two spades, disregarding the order of drawing them, is $C(39,1) \cdot C(13,2)$, and the probability of drawing such a hand is $$ \frac{C(39,1) \cdot C(13,2)}{C(52,3)}. $$
The probability that all three cards are spades, given that at least two of the cards are spades, is in fact $$\frac{C(39,1) \cdot C(13,2)}{C(52,3)}.$$
The only thing wrong with that answer is that the question specifically said the second and third cards must be spades. In the usual language of card probabilities, that means the second and third cards in the particular sequence in which the cards were drawn. Cards drawn in sequences such as $(2\spadesuit, 9\spadesuit, 3\diamondsuit)$ and $(5\spadesuit, 9\heartsuit, \mathrm J\spadesuit)$ are not cases in which the second and third cards are spades but the first is not; they are not part of the event $S_1^c S_2 S_3.$
There are several ways to fix this. One way is to forget about using combinations and just use permutations for everything, since there are cases where the order of drawing matters.
Another way to fix this is to use the probability $$\frac{C(39,1) \cdot C(13,2)}{C(52,3)},$$ but also account for the fact that independent of which set of three cards was chosen for the hand, there is only a $\frac 13$ probability that the single non-spade in that hand happened to have been drawn first. Therefore $$P(S_1^c S_2 S_3) = \frac13 \cdot \frac{C(39,1) \cdot C(13,2)}{C(52,3)}.$$
Another fix is to partially consider the order of drawing cards when drawing a non-spade followed by two spades. That is, in the denominator we consider all the choices that could have been made for the first card ($C(52,1)$), and all the choices for the other two cards from the remaining cards after the first is chosen ($C(51,2)$). That is, the number of ways to choose one "first" card and then choose two others is $C(52,1)\cdot C(51,2),$ so $$P(S_1^c S_2 S_3) = \frac{C(39,1) \cdot C(13,2)}{C(52,1)\cdot C(51,2)}.$$
Personally, I find all of the above (including the idea of using permutations) to be more complicated than necessary. We have a simple sequence of independent events: draw a non-spade, then draw a spade from the remaining cards, then draw one of the remaining spades from the remaining cards. Taking these events one at a time, the probability comes to $$P(S_1^c S_2 S_3) = \frac{39}{52} \cdot \frac{13}{51} \cdot \frac{12}{50}.$$
One can easily confirm that all these ways of computing $P(S_1^c S_2 S_3)$ give the same answer.