This is Example 5j, from Sheldon Ross's First Course in Probability 8th ed, page 38. I don't understand why the following is true.
Solution.To determine the probability that the card following the first ace is the aceof spades, we need to calculate how many of the (52)! possible orderings of the cardshave the ace of spades immediately following the first ace. To begin, note that each ordering of the 52 cards can be obtained by first ordering the 51 cards different from the ace of spades and then inserting the ace of spades into that ordering.
I don't see how this (sentence in italics) can be true. For example if we have $S=\{ 1, 2, 3\} $ the number of orderings that can be obtained are $3!=6$. Following the solution's reasoning we could calculate the orderings for $S$ by ordering the cards different form $3$ and then inserting in into that ordering, that is $ 2!$ . What am I missing? perhaps the sentence in italics does not mean what I think it does?
Also, the solution given is a probability of $ \frac{1}{52} $ for both, I understand why but I have a different solution that also seems valid:
My solution
Ordering in which the card following the first ace is the ace of spades; We have 3 other aces so we put $A_i A_s $, with $i = c, d, h $, together as one unit and count the number of permutations $ = 51! $. As we have three of these such pairs $$ P(N_a) = \dfrac{3\cdot 51!}{52!} $$
Ordering in which the card following the first ace is the two of clubs By a similar argument we put $A_i A_s $, with $i = c, d, h, s$, so $$ P(N_c) = \dfrac{4\cdot 51!}{52!} $$
Can someone tell me what is the error in this reasoning?