The problem:
A player is randomly dealt 13 cards from a standard 52-card deck. What is the probability the 13th card dealt is the first king dealt?
I understand the solution which is based on the Total Probability Theorem which is the following:
We need to count the number of ways to get 12 cards so that there is exactly one king in the deck. There are $48 \choose 12$ ways of doing this. So the probability of getting this is $\frac{48 \choose 12}{52 \choose 13}$. Next we can choose one out of 4 kings with probability $\frac{1}{13}$ which makes the total probability $4 \cdot \frac{1}{13} \cdot \frac{48 \choose 12}{52 \choose 13}$.
What I don't understand is why we can't just use counting to calculate probability here? We know that sample space is uniformly distributed, so we just need to count the total number of ways to get 13 cards out of 52 which is $52 \choose 13$ and the number of decks where the king is the last one which should be equal to $4 \cdot \frac{48 \choose 12}{52 \choose 13}$. Then divide the numbers.
Would appreciate any help to get an intuition.
Best Answer
Think about it this way.
The total number of orders is $13! \binom{52}{13}$ - no issues there.
To construct a "successful" ordering, you need to first pick a set of $13$ cards that has exactly one king, then make sure that the king is actually the $13$th card. There are $4$ ways to choose the king (pick its suit), and then $\binom{52 - 4}{12} = \binom{48}{12}$ ways to pick the other cards. The number of orders is thus $12! (4) \binom{48}{12}$ - here we are multiplying by $12!$ because we know the king has to go last. We only have freedom to permute the other (non-king) cards.
The probability is thus $P = \frac{12! (4) \binom{48}{12}}{13! \binom{52}{13}} = \frac{4 \binom{48}{12}}{13 \binom{52}{13}}$, as desired.
The problem with your approach is that you do not consider that the king has to be the last of the $13$ cards in the order.
Your answer would be right had the question been, "What is the probability that exactly one king is drawn within the first $13$ cards?"