What is the probability that a five-card poker hand contains cards of five different kinds?

Solution:

A deck of 52 cards contains 4 suits: spades, club, hearts and diamonds, each containing 13 cards.

We have to find the probability of a five-card poker hand with cards of five different kinds.

From a total of 52 cards, 5 cards are selected.

So, \(^{52}C_{5} = \frac{52!}{5!(52-5)!}\)

\(^{52}C_{5} = \frac{52!}{5!47!}\)

\(^{52}C_{5} = \frac{52\times 51\times 50\times 49........\times 1}{(5\times 4\times 3\times 2\times 1)(47\times 46\times 45\times 44........\times 1)}\)

\(^{52}C_{5} = \frac{52\times 51\times 50\times 49\times 48}{5\times 4\times 3\times 2\times1}\)

\(^{52}C_{5} = \frac{311875200}{120}\)

\(^{52}C_{5} = 2598960\)

There are 4 suits each containing 13 cards.

So, \(^{13}C_{5} = \frac{13!}{5!(13-5)!}=\frac{13!}{5!8!}\)

\(^{13}C_{5} = \frac{13\times 12\times 11\times 10......\times 1}{(5\times 4\times 3\times 2\times 1)(8\times 7\times 6\times 5\times 4\times 3\times 2\times 1)}\)

\(^{13}C_{5} = \frac{13\times 12\times 11\times 10\times 9}{5\times 4\times 3\times 2\times 1}\)

\(^{13}C_{5} = 1287\)

Since, we need one of the four cards from each rank

\(^{4}C_{1} = \frac{4!}{1!3!}=\frac{4\times 3\times 2\times 1}{3\times 2\times 1}\)

\(^{4}C_{1} = 4\)

There are 4 ways to pick one card from the first selected rank, and 4 ways to pick one of the cards from the second selected rank, and so on. 

The total number of hands is 4⁵.

Now, the required probability is \(\frac{_{13}C_{5}4^{5}}{^{52}C_{5}}\)

= \(\frac{1287\times 1024}{2598960}\)

= 0.50708

Therefore, the required probability is 0.50708

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What is the probability that a five-card poker hand contains cards of five different kinds?

Summary:

The probability that a five-card poker hand contains cards of five different kinds is 0.50708