Here's how I would do this question: First, in a standard deck of cards, there are 13 different "kinds" and 4 of each "kind". The first card dealt to you is some "kind". There are then 51 cards left of which 3 cards are that same "kind" so the probability the second card dealt is also of that "kind" is 3/51. There are then 50 cards left of which 2 cards are of that same "kind" so the probability the third card deal is also of that same "kind" is 2/50= 1/25. There are then 49 card left of which 1 card is of that same "kind" so the probability the fourth card is of that same "kind" is 2/49. Finally there are no cards of that same "kind" so the fifth card dealt can be anything.

The probability four cards of the same "kind" are dealt and then a different card is (3/51)(2/50)(1/49)= (3*2*1)/(51*50*49). We can write 3*2*1 as 3!. We can write 1/(51*50*49) as (48*47*...*3*2*1)/(51*50*49*...*3*2*1)= 48!/51! so (3*2*1)/(51*50*49)= 3!(48!)/51!.

Now that is the probability that the four cards of the same "kind" are all dealt first, then the "odd" card. There are, in fact, 4 possible orders- "KKKKO", "KKKOK", "KKOKK", "KOKKK", and "OKKKK". It is easy to show that they all have that same probability so the probability of "four of a kind" is 4(3!)(48!)/51!.

As to why you got a wrong answer, please tell us exactly **what** you did and **why** you did it!