Probability no two pairs are grouped together twice in a row? There is a room with 48 people divided into 16 groups of 3 in round 1. In round 2, the group is again randomly divided into 16 groups. What is the probability that no two groupmates in round one are in the same group again in round two?
Progress
I thought at first that if you label two arbitrary people A and B then the probability that they will not be in the same group in round two if they were in round one is $1/21$ (from $3/63$). From there I thought maybe do $1 - (1/21)*$the number of possible pairs, but that will yield a negative answer.
 A: I am assuming that the goal in round 2 is that each person is grouped with two new people, as opposed to "I was in group 7 in round 1 and I'm in group 7 again in round 2." I am proceeding according to this assumption. Also, I am not new to Mathematics or Probability, but I am new to entering formulas on this website. Please pardon my lack of LaTeX skills, and feel free to edit this or re-post with the fractions properly presented. Thank you.
Round 1 can be represented by a regular 48-gon. Suppose P1 is one vertex of the polygon, and P2 and P3 are its neighbors (that is, an edge of the polygon connects P1 to P2, and another edge connects P1 to P3.) Then, these two edges, plus the diagonal between P2 and P3, forms a triangle. P1, P2, and P3 represent the members of one of the 16 groups. Continue this pattern for the rest of the vertices of the polygon, and you will have 16 triangles, representing the 16 groups of three people.
In round 2, then, brand new triangles will be formed, apparently at random, using any of the edges and diagonals of the polygon. There are 48*47/2 = 1128 total edges and diagonals in the polygon. 48 of them will be chosen to use in the triangles. So there are 1128 C 48 different possibilities of grouping the people. However, there are 48 lines that we do not want to draw in round 2: those that were used in round 1. We can draw any of the other lines in round 2, so there are (1128 - 48) C 48 possibilities for groups in round 2, without using any of the pairings from round 1.
Thus, the probability that each person will be grouped with two new people in round 2 is 1080 C 48 divided by 1128 C 48 = 0.11848.
