could somebody kindly help me understand the intuition why $P(A_i) = \frac{1}{n} = \frac{(n-1)!}{n!} $. according to the text:
"To find the probability of the union, we’ll use inclusion-exclusion. First,
$$P(A_i) = \frac{1}{n} = \frac{(n-1)!}{n!} $$
for all $i$. One way to see this is with the naive definition of probability, using the
full sample space: there are $n!$ possible orderings of the deck, all equally likely, and
$(n−1)!$ of these are favorable to $A_i$ (fix the card numbered $i$ to be in the $i$th position in the deck, and then the remaining $n − 1$ cards can be in any order). Another way to see this is by symmetry: the card numbered $i$ is equally likely to be in any of the $n$ positions in the deck, so it has probability $1/n$ of being in the $i$th spot."
i can understand the intuition behind $1/n$, however, i can't get $\frac{(n-1)!}{n!}$ and why is $(n-1)!$ of these are favorable to $A_i$
fyi, this is the first part solution of the de Montmort’s matching problem,
"Consider a well-shuffled deck of $n$ cards, labeled 1 through n. You flip over the cards one by one, saying the numbers 1 through $n$ as you do so. You win the game if, at some point, the number you say aloud is the same as the number on the card being flipped over (for example, if the 7th card in the deck has the label 7). What is the probability of winning?"
kindly advise. Regards
@Siong Thye Goh & @ N. F. Taussig. thank you for the clarifications. Here's an update from what i understand from your examples, kindly verify whether it illustrates (n-1)! in this context.
For simplicity, i'm using $n$ = 3, with i,j,k and having i as the $i$ card. Therefore, there are a total of 6 permutations for n!. where the 6 permutations are:
('i', 'j', 'k')
('i', 'k', 'j')
('j', 'i', 'k')
('k', 'i', 'j')
('j', 'k', 'i')
('k', 'j', 'i')
therefore, with (n-1)!; whether $i$ is at the first, second or third positions, there are 2 permutations from any of the 3 positions.
first position:
('i', 'j', 'k')
('i', 'k', 'j')
second position:
('j', 'i', 'k')
('k', 'i', 'j')
third position:
('j', 'k', 'i')
('k', 'j', 'i')
by fixing $i$ card to one of the three positions, gives me $\frac {(n-1)!}{n!} = \frac {2!}{3!} = \frac {2}{6} = \frac {1}{3} = \frac {1}{n}$
Best Answer
We want to find the probability, $\Pr(A_i)$, that the $i$th card is in the $i$th position.
There are $n!$ ways to arrange the $n$ cards in the deck. For the favorable case, there is one way to place the $i$th card in the $i$th position. There are $(n - 1)!$ ways to arrange the remaining $n - 1$ cards in the remaining $n - 1$ positions. Hence, $$\Pr(A_i) = \frac{1 \cdot (n - 1)!}{n!} = \frac{(n - 1)!}{n!} = \frac{(n - 1)!}{n(n - 1)!} = \frac{1}{n}$$