$n$ cards are randomly chosen from $52$ standard cards ($n<52$). One of the $n$ cards was checked, and it turned out to be an ace. This card was returned to the $n$ cards, after which the deck was shuffled. One card is chosen randomly again from these $n$ cards. What is the probability that this card is an ace?
I am approaching the problem this way: if choose the first ace again (with $\frac{1}{n}$ probability), we, of course, see this ace.
Otherwise, for any of the remaining $n-1$ cards, the probability to see an ace equals $\frac{3}{51}$. So the answer is $\frac{1+\frac{n-1}{17}}{n}$
Is that correct?
I was also trying to solve this problem in a more straightforward manner. The total probability for there to be an ace among $n$ cards equals:
$$ \frac{\binom{4}{1}\binom{48}{n-1}+\binom{4}{2}\binom{48}{n-2}+\binom{4}{3}\binom{48}{n-3}+\binom{4}{4}\binom{48}{n-4}}{\binom{52}{6}}$$
How do I take account of the fact that one ace is definitely among the $n$ cards?
Best Answer
The probability that there are $k$ aces among the rest $(n-1)$ cards is: $$ \frac{\binom3k\binom{48}{n-k-1}}{\binom{51}{n-1}}. $$ Thus the probability in question is: $$ \sum_{k=0}^3\frac{\binom3k\binom{48}{n-k-1}}{\binom{51}{n-1}}\frac{k+1}n. $$ The result coincides with your much simpler answer.