Odds of guessing a sequence of cards in order

Question

For a deck of 52 cards, find the number m such that
$P$(by random guessing we get more than m correct guesses) < $1/10000.$
I was thinking along the lines of - if there is m correct guesses, there are $(52-m)!$ possible arrangements with a correct guess.
So the probability of m correct guesses is
$(52-m)!/52!=k$ and the solution is given by $k<1/10000$.
Is my reasoning correct? If not how to solve this problem?

Answer 1

Three formulations

1. Suppose you guess the sequence of cards (either entirely at the start or just before each card is turned over). You want more than $m$ of your guesses to be correct before you give a wrong answer. This has probability $\frac{(51-m)!}{52!}$ - almost what you wrote, but the question says more than - and will be less than $\frac1{1000}$ when $m \ge 1$, since $\frac{1}{52 \times 51}=\frac{1}{2652}$ is the probability you get at least two, i.e. more than one, correct before making an error, and is small enough.

2. You guess the full sequence of cards (a permutation of the deck) at the start and you want more than $m$ of any of your answers to be correct matches, not caring how many errors you make or when. This is related to rencontres numbers and the probability will be less than $\frac1{1000}$ when $m \ge 5$

3. You guess each card before it is turned over, after taking into account the cards previously turned over, and you want more than $m$ of any of your answers to be correct matches. You will probably do better in this formulation as, for example, your $52$nd guess should be correct by elimination and your $51$st has a probability of $\frac12$. The probability will be less than $\frac1{1000}$ when $m \ge 11$