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I have a question.

I'm looking to calculate the probability of getting $4$ different suits cards in a $5$ card poker game using a standard $52$ card deck.

I think this is: $$\frac{\dbinom{4}{1}\dbinom{13}{2}\dbinom{13}{1}^{3}}{\dbinom{52}{5}}$$

What do you think? Thanks.

N. F. Taussig
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NM2
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  • Relevant: http://math.stackexchange.com/questions/211653/deal-4-cards-from-a-deck-what-is-the-probability-that-we-get-one-card-from-each – Vepir May 08 '16 at 07:54
  • Hey,I deal $5$cards. – NM2 May 08 '16 at 07:56
  • Do you have and idea why this answer isnt correct ?

    $\frac{52}{52}\cdot \frac{39}{51} \cdot \frac{26}{50} \cdot \frac{13}{49} \cdot \frac{48}{48}$

     – NM2 May 08 '16 at 08:01

1 Answers1

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The answer you have quoted in the body of the question is correct.

The answer you have posted in the comment is incorrect, because:

The first $4$ cards are permutations of $ABCD$ while the last is simply appended.

To make the numerator a true permutation, the last card could be placed at any of five spots,$\;(\;\bullet A \bullet B \bullet C \bullet D \bullet\;)$ and you would need to divide by $2$ to correct for two identical suits

ADDED

The link Matta has given is for four cards dealt, where the two methods would tally effortlessly:

$$\frac{13^4}{\binom{52}4} = \frac{52}{52}\cdot\frac{39}{51}\cdot\frac{26}{50}\cdot\frac{13}{49}$$

true blue anil
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