So, probabilities.
We need 30 points, from 5 players. This can be reached via different means.
1) Play 12,9,6,4. Sum is 31 > 30
2) Play 12,9,6,2,1 Sum is 30
So for every roll, we have four outcomes:
A) A die we absolutely need (12,9,6) is rolled.
B) 4 is rolled AND not A
C) either 2 or 1 is rolled AND not (A or B)
D) not (A or B or C)
Let n be the number of dice each players rolls, k_A the number of dice in A we already have, k_B the number of dice in B we already have, k_C the number of dice in C we already have. The probabilities for A - D are:
p(A) = sum(i=0 to n-1) of ((9+k_A)/12)^i * (k_A/12)
p(B) = sum(i=0 to n-1) of ((8+k_A+k_B)/12)^i * (k_B/12)
p(C) = sum(i=0 to n-1) of ((6+k_A+k_B+k_C)/12)^i * (k_C/12)
p(D) = 1 - p(A) - p(B) - p(C)
Scenarios of success are:
I) Three players play 12,9,6, one player plays 4
II) Three players play 12,9,6, two players play 1,2
I fear I won't have time to complete this right now. Should be able to calculate it soonish though.
PPE: I guess WW's roll makes things easier.