5 cards are dealt from a well shuffled standard deck of cards. What is the probability of getting 2 kings and 2 jacks?

I started by evaluating each choice in turn. Probability that the first card is a king= 4/52. That the second card is a king=3/51, that the third card is a jack=4/50, that the fourth card is a jack=3/49 and that the last card is neither a jack nor a king =44/48.

I multiply these all together to get 6336/311875200. I know I am missing the factor to address the possible arrangements, but I am not sure how to determine this.

Any help would be appreciated

  • 2
    $\begingroup$ The factor for the possible arrangements is the number of distinct arrangements of the letters KKJJO. I could tell you how to do that, but I can't tell you how to do it without combinatorics. $\endgroup$ Dec 4 '12 at 23:34
  • $\begingroup$ please procede with your answer. I'd like to know why I would use 5C3 or 5C2 or a combo of them (I think the correct factor is 30, but I can't reason it out). $\endgroup$
    – Dan Dillon
    Dec 4 '12 at 23:38
  • $\begingroup$ When you say “without combinatorics”, what exactly do you mean? If you mean that you don’t have specific background in combinatorics (eg taking a course or similar), then let us know a little of what background you do have, so we know what kind of techniques to use. $\endgroup$ Dec 4 '12 at 23:41
  • $\begingroup$ I think I am missing something basic here. I can solve the problem as follows: 4C2*4C2*44C1/52C5, however, when I try to verify the results using the method shown above, I am off by a factor of 30. I took a stats and prob class in college a long time ago, and I am trying to refresh my memory. $\endgroup$
    – Dan Dillon
    Dec 4 '12 at 23:48
  • 1
    $\begingroup$ I think I may have answered my own question (after staring at Gerry Myerson's comment). The possible ways to arrange KKJJO are as follows: The Kings can be ordered 5C2 ways, the jacks can be ordered 3C2 ways, and the last card can be ordered 1C1 way. Which equals 30. Thank you Gerry. $\endgroup$
    – Dan Dillon
    Dec 5 '12 at 0:01

Write down all the possible patterns KKJJO such as JKOKJ.

Check they each have the same probability, so for example JKOKJ has probability $$\frac{4}{52}\times \frac{4}{51}\times \frac{44}{50}\times \frac{3}{49}\times \frac{3}{48}.$$

When you have the patterns and have checked none are repeated and none are missing then add up the probabilities of the distinct patterns. This the same as multiplying the probability of a single pattern by the number of distinct patterns.

If you are not multiplying by thirty then you have made an error.

If you want to use combinatorics to save time, you are multiplying by $\dfrac{5!}{2!\times 2! \times 1!}$ since there are two kings, two jacks, and one which is neither a king nor a jack.


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