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$n$ 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?

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  • $\begingroup$ A priori, there are $5$ cases to consider according to the number of aces amongst the $n$ cards. We can easily compute the, a priori, probabilities of each case. Now, having selected an ace from the $n$, these probabilities change. For example, the probability that the $n$ contain no aces is now $0$. Use Bayes' to compute the a posteriori probabilities. $\endgroup$
    – lulu
    Commented Jun 23, 2019 at 11:26
  • $\begingroup$ Note: you say that the ace is returned to "the deck" but this is ambiguous. Do you mean that it is returned to the special $n$ cards or that it is returned to the remainder of the deck? $\endgroup$
    – lulu
    Commented Jun 23, 2019 at 11:27
  • $\begingroup$ @lulu, it's returned to the special $n$ cards $\endgroup$
    – Don Draper
    Commented Jun 23, 2019 at 11:28
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    $\begingroup$ That's what I figured, but I think you should clarify that in the post. $\endgroup$
    – lulu
    Commented Jun 23, 2019 at 11:28
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    $\begingroup$ That's the way to go. I'd run some sanity checks at the end...say $n=1$, or $n=52$ for instance. Just because the arithmetic is so messy. $\endgroup$
    – lulu
    Commented Jun 23, 2019 at 12:24

1 Answer 1

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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.

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  • $\begingroup$ Thank you. This is a great approach! Can you please take a look at my last comment under the OP? I have recalculated a posteriori probabilities. Now I can just sum the products of each probability $P(A_{k}|B)$ and $\frac{k}{n}$. Is that correct? $\endgroup$
    – Don Draper
    Commented Jun 23, 2019 at 12:08

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