Deck has 30 cards, out of which 5 cards have duplicates (20 cards are unique, 5 cards have 2 copies each).
If you draw X cards from the deck (without returning), what is the probability that there are at least one duplicate remaining in the deck?
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$\begingroup$ Also, by "there are at least one duplicate remaining in the deck", do you mean that at least $2$ copies of a card are remaining? $\endgroup$– true blue anilCommented May 10, 2016 at 7:52
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$\begingroup$ I have a B.Sc. degree in math, however, I finished my University back in 2001 so my memory is not fresh since I haven't worked with probability. But I can read math text, just started to read about hypergeometric distribution $\endgroup$– Mladen AdamovicCommented May 10, 2016 at 7:55
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$\begingroup$ To answer your second question "there are at least one duplicate remaining in the deck" means that there are 2 copies of the same card in the deck. This question is actually related to the game Hearthstone $\endgroup$– Mladen AdamovicCommented May 10, 2016 at 7:57
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$\begingroup$ Recursive methods at least give an answer: if $P(x,T,d)$ is the probability that a duplicate remains given $x$ draws from $T$ cards of which $2d$ have a duplicate, then $P(x,T,d)=\frac {2d}TP(x-1,T-1,d-1)+\frac {T-2d}T P(x-1,T-1,d)$ and $x<d\implies P(x,T,d)=1$, $P(x,T,0)=0$. I expect there is a closed formula, though. $\endgroup$– luluCommented May 10, 2016 at 9:57
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$\begingroup$ I'm not sure if this is correct. Assume that X is less than 28 (so there are at least 3 cards in the deck remaining), than possible combinations for remaining cards are $30 \choose 30-X$ Possibility to select $30-X$ cards with selected duplicate is $ {28 \choose 28-X} * 5$ . So it seems $$1 - \frac{{28 \choose 28-X} * 5}{{30 \choose 30 - X}}$$ $\endgroup$– Mladen AdamovicCommented May 10, 2016 at 10:00
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1 Answer
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Say there are $n$ cards, including $k$ pairs and $n-2k$ singletons, and we draw $m$ cards. The probability that a pair remains can be determined by inclusion-exclusion. There are $\binom kj$ ways to choose $j$ particular pairs, and the probability that they remain in the deck is $\frac{\binom{n-2j}m}{\binom nm}$. Thus the probability that no pair remains is
$$ \binom nm^{-1}\sum_{j=0}^k(-1)^j\binom kj\binom{n-2j}m\;, $$
and the probability that at least one pair remains is the complement,
$$ \binom nm^{-1}\sum_{j=1}^k(-1)^{j-1}\binom kj\binom{n-2j}m\;. $$
In your case, with $n=30$, $k=5$ and $m=X$, this is
$$ \binom{30}X^{-1}\sum_{j=1}^5(-1)^{j-1}\binom5j\binom{30-2j}X\\ =\binom{30}X^{-1}\left(5\binom{28}X-10\binom{26}X+10\binom{24}X-5\binom{22}X+\binom{20}X\right)\;. $$
Here's a table of the values for $0\le X\le30$.