I want to find the probability of certain cases of picking 4 cards from 12 cards. Let's say the type of cards are A, B, and C, and there are 4 duplications of each type. The set of cards S={4.A,4.B,4.C} The cases are the following:
- All 4 cards are identical: I calculated it as 3((4*3*2*1)/(12*11*10*9))
- 3 are identical and one is different: 6((4*3*2*4)/(12*11*10*9))
- 2 identical of two different types: 3(4*3*4*3)/(12*11*10*9))
- 2 identical and two different: 3(4*3*4*4)/(12*11*10*9))
The thing is, 4 cards picked from the set cannot be any different from these 4 cases, it has to be one of them (If this is not clear intuitively, using combinations of objects with repeated items there are 15 different combinations, and since there are 3 ways in case 1, 6 in case 2,3 and 4 that sums up to 15, so these are the only possible case scenarios). However, when I sum these probabilities, they do not sum up to one(It adds up to 23/165), why? Did I calculate something wrong or am I missing something? I feel it is an easy problem but I am so confused.
The problem is that when you calculated the second case as $6\times\frac{4\times 3\times 2\times 4}{12\times 11\times10\times9}$, you counted that there are six different ways to choose which one appears three times and which appears once, but you haven't accounted for the fact that there are four different orders in which you can get them (if there are three As and a B then you could draw AAAB, AABA, ABAA or BAAA, and each of these has probability $\frac{4\times 3\times 2\times 4}{12\times 11\times10\times9}$). Similarly you need to take this into account in the third and fourth cases.