I have a question regarding a standard deck of 52 cards.
Say that you draw 5 cards. You are given that 4 of the cards are spades and then need to work out the probability that all 5 cards are spades. Naturally the probability would be the number of combinations of 5 spades over the sum of the number of combinations of 4 spades and the number of combinations of 5 spades.
I understand the above, but what I don't quite understand - but know is wrong - is the following. If 4 cards are spades then there should be 9 spades left in the deck. So the odds of drawing a fifth spade would be 9/48. 9/48 is a vastly different probability from what the above gives you. It would be very helpful if anyone had a fairly intuitive explanation as to why this second option doesn't work.
Edit: I meant a suite not a specific card.
This is drawing without replacement
Your $\frac 9{48}$ attempt, equal to $\frac 9{9+39}=0.1875$, is answering a different question, namely the probability that the fifth card drawn is a spade given that the first $4$ drawn were all spades.
But in the original question the (at least) $4$ spades can be any of the $5$, and if the other card is not a spade then it can be in any of the $5$ positions, which makes the answer $\frac 9{9+5\times 39} = \frac{9}{204} \approx 0.044$. There are several other ways of doing the calculation which produce the same result, including:
as combinations: $$\frac{{5\choose 5}{13 \choose 5}{39 \choose 0}}{{5\choose 5}{13 \choose 5}{39 \choose 0}+{5\choose 4}{13 \choose 4}{39 \choose 1}}$$
as probabilities: $$\tfrac{\frac{13}{52}\frac{12}{51}\frac{11}{50}\frac{10}{49}\frac{9}{48}}{\frac{13}{52}\frac{12}{51}\frac{11}{50}\frac{10}{49}\frac{9}{48}+\frac{13}{52}\frac{12}{51}\frac{11}{50}\frac{10}{49}\frac{39}{48} +\frac{13}{52}\frac{12}{51}\frac{11}{50}\frac{39}{49}\frac{10}{48} +\frac{13}{52}\frac{12}{51}\frac{39}{50}\frac{11}{49}\frac{10}{48} +\frac{13}{52}\frac{39}{51}\frac{12}{50}\frac{11}{49}\frac{10}{48} +\frac{39}{52}\frac{13}{51}\frac{12}{50}\frac{11}{49}\frac{10}{48}}$$
and you can see the extra factor of $5$ appearing in both