5
$\begingroup$

How many five-card hands dealt from a standard deck of $52$ playing cards are all of the same suit? If a random hand is dealt, what is the probability that it will have this property?

Would the probability be: $$\frac{\dbinom{13}{5}*\dbinom{4}{1}}{\dbinom{52}{5}}$$

$\endgroup$
4
$\begingroup$

It would be $$\frac{\binom41_{\text{color}} \cdot \binom{13}5_{\text{cards of this color}} \cdot \binom{52-13}0_{\text{other cards}}}{\binom{52}{5}_{\text{total}}} = \frac{\binom41 \cdot \binom{13}5}{\binom{52}5} = \frac{33}{16660}$$ So you are correct. See here for some more on the number.

$\endgroup$
0
$\begingroup$

The first card can be any suit. For the second card there are 12 left of that suit out of 51 cards. For the third card there are 11 left of that suit out of 50 cards. For the fourth card there are 10 left of that suit out of 49 cards. For the fifth card there are 9 left of that suit out of 48 cards. So $ \frac{12}{51} \cdot \frac{11}{50} \cdot \frac{10}{49} \cdot \frac{9}{48} = .001981 $ (same answer as another solution).

$\endgroup$
  • $\begingroup$ Users will benefit more from your answer if you write a complete answer. $\endgroup$ – random123 Feb 17 '18 at 5:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.