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I have two decks of 52 cards that are randomly shuffled. I draw five cards from the first deck. What is the probability that if I draw five cards from the second deck, I will draw at least two of the cards (same suit and number) I drew from the first deck?

I understand that my denominator will be ${52 \choose 5}$ but I'm struggling to determine the numerator.

I have an intuition about the probability of drawing all five of the same cards. Here the numerator will be 5! or the number of different orders that the five cards could appear in. So the probability is $\frac{120}{2598960}=0.000046$. However, I'm neither sure this is correct nor where to go from here to determine the probability of drawing at least 2.

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The number of ways of matching exactly $k$ of the five cards selected from the first deck is $$\binom{5}{k}\binom{47}{5 - k}$$ since if $k$ cards match, the remaining $5 - k$ cards must be selected from among the other $47$ cards in the deck.

Therefore, the number of ways of selecting at least two matching cards is $$\binom{5}{2}\binom{47}{3} + \binom{5}{3}\binom{47}{2} + \binom{5}{4}\binom{47}{1} + \binom{5}{5}\binom{47}{0}$$ Alternatively, we can subtract the number of hands with fewer than two matching cards from the total number of hands. Hence, the number of favorable hands is $$\binom{52}{5} - \binom{5}{0}\binom{47}{5} - \binom{5}{1}\binom{47}{4}$$ In your attempt, you counted sequences in the numerator and subsets in the denominator. There is only one subset in which all five cards match. If you want to use sequences in the numerator, you also have to use them in the denominator.

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The Hypergeometric distribution would be most useful here. When you draw from the second deck 5 cards will be successes (matches) and 47 cards will be failures.

Let $X$ represent the number of matched cards...

$$ P(X \ge 2)=1- P(X =0)-P(X =1) \\=1- \dfrac{ \binom {47}5+ \binom {47}4\binom {5}1}{\binom {52}5}$$

Your calculation for 5 matches is flawed because when you use $\binom{52}5$ for your denominator you are counting combinations, not permutations, so the numerator should also be a combination.

so $P(X =5) =\dfrac 1{\binom{52}5} $

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