From a standard deck of 52 cards, 4 cards are chosen without replacement. What is the probability that all the 4 cards have different numerical value?
N.B.-It is assumed that Jacks,Queens, Kings and Aces have numerical values 11,12,13 and 1 respectively.
My solution-After choosing any 1 card, we can choose 48 cards(removing other cards of the same value) out of the remaining 51 cards and so on till we pick 4 cards.Therefore, the required probability is $\frac{52}{52}\times\frac{48}{51}\times\frac{44}{50}\times\frac{40}{49}$.
Kindly verify.

  • 4
    $\begingroup$ IMO the approach and the result are correct. $\endgroup$ – digital-Ink Jun 15 '14 at 14:08
  • $\begingroup$ your way includes order of picking; divide by 4!. $\endgroup$ – RE60K Jun 21 '14 at 14:43

Community wiki answer so the question can be marked as answered:

As remarked in a comment, your calculation is correct. The other comment that says that you should divide by $4!$ is in error.


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