# Probability of drawing at least one card from each suit

Suppose you have a deck of 16 cards, 4 from each suit. You shuffle them and deal out eight cards face up. What is the probability that you will have at least one card from each suit showing?

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## 1 Answer

Let $C(n,k)$ be the binomial coefficient, then the probability of a particular suit NOT appearing is $$\frac{{C(16 - 4,8)}}{{C(16,8)}}.$$ Similarly, the probability of two suits NOT appearing is $$\frac{{C(16 - 8,8)}}{{C(16,8)}}$$ and the probability of three suits NOT appearing is $$\frac{{C(16 - 12,8)}}{{C(16,8)}}$$ and the probability of four suits not appearing is $$\frac{{C(16 - 16,8)}}{{C(16,8)}} = 0.$$ Therefore, the probability that at least one of the suits does NOT appear is $$\frac{{4C(16 - 4,8) - 6C(16 - 8,8) + 4C(16 - 12,8)}}{{C(16,8)}}$$ so $$1 - \frac{{4C(16 - 4,8) - 6C(16 - 8,8) + 4C(16 - 12,8)}}{{C(16,8)}}$$ is the probability we seek (4,-6, and 4 come from the inclusion-exclusion principle). Watch this YouTube video if you do not understand the inclusion-exclusion principle.

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