Conditional probability of a Royal flush What is the probability of drawing a 10, jack, queen, king, and ace of all the same suit from 5 random cards given that one card drawn is an ace of spades?
Solution
Let $R$ denote the event that the hand is a royal flush and let $S_A$ be the event that one card is ace of spades.
Clearly, $$P(R) = \frac{4}{{52}\choose{5}}.$$
Now we intuition we see that 
$$P(R|S_A) = {{51}\choose{4}}^{-1},$$
because there are only 51 cards left after $S_A$.
Now let us compute this conditional probability using the usual definition. The textbook (Stirzaker, Elementary Prob., 2nd ed, p. 53)  correctly says:
$$P(R|S_A) = \frac{P(R\cap S_A)}{P(S_A)}=\frac{{{52}\choose{5}}^{-1}}{{{51}\choose{4}}\big/{{52}\choose{5}}}.$$
Now, while I see that $P(R\cap S_A) = {{52}\choose{5}}^{-1}$ I would say that $P(S_A)  = 1/52 \neq {{51}\choose{4}}\big/{{52}\choose{5}},$ because there is only one ace of spades in a standard deck of 52 cards. But this does not give the correct conditional probability of $R$ given $S_A$. What am I missing?
 A: Your $1/52$ is effectively the probability that the first card is ace of spades.
Consider $P(S_A)$ when drawing the cards one by one where the order matters. Since we are considering the probability and not the number of permutations, this will be equivalent to the "draw $5$ cards at once" where the order doesn't matter.
The probability that the first card is ace of spades is $\frac1{52}$. Then the following cards don't matter so $\frac1{52} \cdot 1 \cdot 1 \cdot 1 \cdot 1 = \frac1{52}$.
The probability that the second card is ace of spades is $\frac{51}{52} \cdot \frac{1}{51}$ and then the following cards don't matter so $\frac{51}{52} \cdot \frac1{51} \cdot 1 \cdot 1 \cdot 1 = \frac1{52}$ .
Similarly, the probability that the third card is ace of spades is  $\frac{51}{52} \cdot \frac{50}{51} \cdot \frac1{50} \cdot 1 \cdot 1 = \frac1{52}$.
The probability that aces of spades appears on the 4th and 5th draw are also $\frac1{52}$.
The five situations listed above are disjoint, thus $\frac1{52} + \frac1{52} +\frac1{52} +\frac1{52} +\frac1{52} = \frac5{52}$, which is also ${{51}\choose{4}}\big/{{52}\choose{5}}$.
When one wants to take the approach of "draw $5$ cards at once" where the order doesn't matter, the denominator must be ${{52}\choose{5}}$ and the numerator shall be the number of combinations of whatever event is requested. One cannot directly obtain $\frac1{52}$ under this approach.
