Odds of guessing a sequence of cards in order 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?
 A: Three formulations


*

*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.

*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 $m \ge 5$ 

*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 $m \ge 11$ 
