At least one suit missing from a 7-card hand

Question

Seven cards are dealt at random from a standard 52-card deck (four suits of 13 cards each). What is the probability that at least one suit is completely missing from the hand?

Accepted Answer

Let As be the event that suit s is absent, and apply inclusion-exclusion over the four suits. Counts are over \binom{52}{7} = 133{,}784{,}560 hands. - One suit absent: \binom{39}{7} = 15{,}380{,}937, with \binom{4}{1} = 4 suits. - Two suits absent: \binom{26}{7} = 657{,}800, with \binom{4}{2} = 6 pairs. - Three suits absent: \binom{13}{7} = 1{,}716, with \binom{4}{3} = 4 triples. - Four suits absent is impossible. Alternating signs: 4(15{,}380{,}937) - 6(657{,}800) + 4(1{,}716) = 61{,}523{,}748 - 3{,}946{,}800 + 6{,}864 = 57{,}583{,}812. So P(	ext{some suit missing}) = \frac{57{,}583{,}812}{…

At least one suit missing from a 7-card hand

Your answer

Discussion

Learn the concepts

Related questions