Shuffling a deck of cards "n" number of times to get an organized order
The scenario is that a fair deck of 52 playing cards is shuffled a certain number of times (52! times) . (a) What is the probability of shuffling a correctly ordered (Ace to 2) deck of cards? (b) What happens to the probability of shuffling a correctly ordered deck of cards as the number of shuffles increase?
Here's what I did so far:
1. Probability of shuffling once and getting the correctly ordered deck = 1/52!. (52! arrangements of 52 cards, 1 permutation that works)
2. Probability of shuffling twice and getting the correctly ordered deck:
P(2) = 1/52!^(2) so in general: P(n) = 1/52!^(n); We see that as the number of trials increase, P(n) approaches zero. So for (b) the probability of shuffling a correctly ordered deck decreases as "n" approaches infinity. (Counterintuitive here so I'm a bit skeptical that this is right)
3. P(52!) = 1/52!^(52!) = very close to 0.
My friend disagrees maintaining that:
The probability of it NOT being the right deck is (1- 1/52!) If we shuffle again and again, and we do NOT get the deck we are looking for, then the probability of NOT getting the right deck after n shuffles is (1-1/52!)^n. As n approaches infinity, this probability approaches 0, meaning that we WILL eventually break the chain of not getting the right deck.
He thinks that the probabilities do not change for any specific shuffle. However, they do if you consider n shuffles, and are looking for the probability of success or failure at least once in these shuffles.
He interprets it in reverse Q(n) as opposed to P(n) and it makes sense intuitively. But i'm confused as to why my friend's interpretation contrast with mines even though we seem to be modeling the same scenario albeit in terms of complements.