# Probability of cards without combinatorics

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

• 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. Dec 4 '12 at 23:34
• 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). Dec 4 '12 at 23:38
• 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. Dec 4 '12 at 23:41
• 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. Dec 4 '12 at 23:48
• 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. Dec 5 '12 at 0:01

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