Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
add comment

3 Answers

up vote 2 down vote accepted

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

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

share|improve this answer
    
This answered my question. Thank you very much! –  Mario Jan 17 '13 at 17:31
add comment

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

share|improve this answer
add comment

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 $52-52({43\over52})^4$.

share|improve this answer
    
Thanks for the additional clarity! –  Mario Jan 17 '13 at 18:18
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.