I understand this proof and it is quite slick and throws up a counterintuitive result.
However, before looking at the solution, I tried my own ham-fisted approach of finding P(AoS immediately drawn after first ace), and wasn't sure where my mistake is.
Finding number of orderings s.t. AoS follows first ace.
$$\text{Let } A_1 = \text{first ace, which is not ace of spades}$$
$$\text{Let } A_s = \text{ace of spades} $$
Scenario 1: A_1 and A_s drawn first:
$$\text{A_1, A_s, ... = 50!*3 orderings (3 choices for A_1)}$$
Next scenarios:
$$\text{[X],} A_1, A_s, ... \text{ where [X] is a card other than an ace = 48*49!*3} \text{ orderings.}$$
$$\text{[X], [X]}, A_1, A_s ... = \frac{48!}{(48-2)!} * 48! * 3 \text{ orderings.}$$
...
Final scenario:
$$\text{[X], [X] ... [X],} A_1, A_s ... = \frac{48!}{(48-48)!}*2!*3$$
Sum these terms up. A general term can be written as:
$$\frac{48!}{(48-r)!} * (50 - r)! *3 = \frac{48!}{(48-r)!} * (48 - r)! * 49 * 50 *3 = 50! * 3 $$
Summing 49 of these terms would lead to an answer of 147*50!, which is not what the author had for the number of orderings that verify this event (51!).
What's gone wrong with my reasoning?