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I am currently trying to figure out why my answer is not correct for that following question.

Q: What is the probability that 13 cards drawn from a standard deck has at least one card from each suit?

A: Clearly we know that there are 52C13 ways to draw 13 cards which will be the denominator for the final answer. Moreover, this is an inclusion and exclusion problem so the numerator is N(c1c2c3c4) where condition 1 is clubs, condition 2 is spades, condition 3 is hearts and finally condition 4 as diamonds. My logic is wrong here however...

I got N(c1c2c3c4) / 52C3 , however the answer key has N(-c1-c2-c3-c4)/52C3. I used the "-" to denote negation. Can someone explain to me why the book has it like this?

Visualizing this with a venn diagram, why would the numerator be the group of things that do not include conditions 1,2,3,4?

Belphegor
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  • It's hard to tell from the transcription, but presumably $-c1-c2-c3-c4$ means "(NOT $c1$) OR (NOT $c2$) OR (NOT $c3$) OR (NOT $c4$)", in which case applying de Morgan's Law leads to the inclusion-exclusion count you suggest. – Travis Willse Jan 28 '15 at 06:58

1 Answers1

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Call a hand bad if one or more suits are missing. We count the bad hands.

There are $\binom{39}{13}$ hands that have no $\spadesuit$. We get the same count for no $\heartsuit$, no $\diamondsuit$, no $\clubsuit$.

So our first estimate for the number of bad hands is $4\binom{39}{13}$.

However, this double-counts the hands that are missing both $\spadesuit$ and $\heartsuit$. There are $\binom{26}{13}$ of these, and the same number for the $5$ other choices of $2$ suits.

So our next estimate for the number of bad hands is $4\binom{39}{13}-6\binom{26}{13}$.

However, we have subtracted too much. For we have subtracted twice, for example, the hands that are missing all three of $\spadesuit$, $\heartsuit$, and $\diamondsuit$.

Our final count of bad hands is therefore $4\binom{39}{13}-6\binom{26}{13}+4\binom{13}{13}$.

André Nicolas
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  • So the inside of the Venn diagrams reprsent the bad hands and the outside is the good hands with at least 4 diff auits. Hence we find the outside not inside. – Belphegor Jan 28 '15 at 17:50
  • I chose to first count the bad hands, so the "outside," because I think it makes things clearer, We could have chosen to count the good hands, and then a similar argument yields $\binom{52}{13}-4\binom{39}{13}+6\binom{26}{13}-4\binom{13}{13}$. – André Nicolas Jan 28 '15 at 17:59