Simplify the problem! If you do not consider the color, you can just look at four people and four A cards. Because all cards are of equal probability, you can consider other cards instead of.
Assume that all a is the same:
One person takes four A: Four (each person is a possibility) or C)
Three persons in one person, and one other person: 3*4 = 12 (C (3100) * C (1300), indicating the order of existence, that is, and are two different types, or C () * ))
One person and two others: 1 + 2 + 3 = 6 (C () * C (), which includes the order, divide By 2200 and. cards are the same.) or C)
One person and two others: 3*4 = 12 types (2110, C () * C (), divide by 2 to get 12 ), or C () * 3 = 12, 2 has only three types of positions. After 2 is completed, the remaining positions are determined.
A per person: 1 Type
So the probability that each person gets a is 1/(4 + 12 + 6 + 12 + 1) = 1/35
Send 52 cards to 4 people at random, with 13 cards for each person, and calculate the probability that each person will receive a "".