I have a deck on $n$ cards and choose the $i$-th card of the deck. After looking at the card I put it back and shuffle the deck. What is the probability, that the $i$-th card is the same as before.
My thoughts so far/what I (think) I know:
- I have $n!$ possible permutations of the deck.
- Selecting one card that doesn't permute reduces the number of possible permutations to $(n-1)!$
This leaves me thinking that, with selecting position $i$ with a probability of $1/n$, the probability that the card stays the same simply results in $P[\text{position i doesn't change}]=\frac{1}{n}\cdot \frac{(n-1)!}{n!}=\frac{1}{n^2}$.
Which I somehow don't believe, especially after studying here that the probability that the probability that exactly one card stays on the same position is $\frac{1}{n!}\binom{n}{k}\lfloor\frac{(n-k)!}{n!}\rfloor$ which is kind of more complicated than my solution which I guess is kind of related.
Any help and hint is much appreciated!