What is the probability that a five-card poker hand has no pairs (possibly straight or flush)?

Question

I know that the total number of five-card poker hands is $C(52,5)$. I am having trouble figuring our the total number of hands with no pairs, possibly a straight or a flush.

Any hints or help/guidance is appreciated.

Answer 1

HINT

We have that

• for the first card we have $p_1=\frac{52}{52}$

• for the second card we have $p_2=\frac{48}{51}$

• for the third card we have $p_3=\frac{44}{50}$

• for the fourth card we have $p_4=\frac{40}{49}$

• for the fifth card we have $p_5=\frac{36}{48}$

Answer 2

Think about it this way. You can construct the no-pair hand in two steps:

1. Choose the numbers in the hand ($13$ options, you must choose $5$)
2. Order the cards from smallest to largest
3. Pick a color for the first card. Then for the second. Then for the third, and so on.

The middle step is there to ensure you aren't counting anything twice, the number of ways you can perform last step is independent of the first step, so you can just multiply the number of times you can perform the first and last step to get the result.

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Alternatively, you can use the approach by gimusi in his answer (and for practice, try both approaches which should give you the same answer!)