# Question on overlapping sets

Let's say that four of us (persons A, B, C, and I) are each handed our own, complete deck of cards. All four decks are identical, meaning that an Ace of Hearts in B's deck is equivalent to (and therefore, not unique from) an Ace of Hearts in mine.

We each draw nine cards at random from our own decks, and set them aside in our own, separate piles. To clarify, these cards are drawn without replacements, meaning that once we've each drawn our nine cards, we are each left with two piles: a stack of nine cards, and the original deck of cards, which now has forty-three cards remaining.

We then take these four piles of nine cards, and combine them into a single stack of thirty-six cards. After removing duplicates from the new stack, how many unique cards are we likely to have?

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I don't know a standard mathematical interpretation for "how many unique cards are we likely to have?", so I'll interpret it as "what is the expected number of unique cards?".

The probability that at least one of the cards is an Ace of Hearts is

$$1-(1-9/52)^4=\frac{3892815}{7311616}\approx0.5324\;.$$

Thus the expected number of unique cards is

$$52\cdot\frac{3892815}{7311616}=\frac{3892815}{140608}\approx27.69\;.$$

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This answered my question. Thank you very much! – Mario Jan 17 '13 at 17:31

The number of unique cards will always be greater than 8 and less than 37.

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This is the same as Joriki's answer, with a bit more detail:

Label the cards of a deck numerically, $1$ through $52$.

For $i=1,2,\ldots,52$, let $X_i$ equal $1$ if no one chose the $i$'th card and $0$ otherwise. Let $X$ be the number of cards not chosen by anyone. We have $X=\sum\limits_{i=1}^{52} X_i$. The expected number of cards that were not chosen by anyone is $$\Bbb E(X)= \sum_{1=1}^{52} \Bbb E(X_i) =\sum_{i=1}^{52} P[X_i=1]= \sum_{i=1}^{52}\textstyle ({51\cdot50\cdots 43\over 52\cdot51\cdots 44})^4=52\cdot({43\over 52})^4.$$

The expected number of cards remaining in the stack is then $52-52({43\over52})^4$.

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Thanks for the additional clarity! – Mario Jan 17 '13 at 18:18