Next attempt.
There are 16 cards. All cards are dealt.
[comment deleted]
16C4*12C4*8C4*1 = 1820*495*70*1 = 63 063 000 ways to choose 4 groups of 4
12C3*9C3*6C3*1 ways to chose 4 groups of 3
12C3*9C3*6C3*1 *4*3*2*1 ways to chose 4 groups of three with no 1s and then add a 1 to each group.
\(\text{Prob(1 in each group)}=\frac{12C3*9C3*6C3*1 *4!}{16C4*12C4*8C4*1}\\ \text{Prob(1 in each group)}=\frac{12C3*9C3*6C3 *4!}{16C4*12C4*8C4}\\ \text{Prob(1 in each group)}=\frac{369600*4!}{63063000}\\ \text{Prob(1 in each group)}=\frac{3696*4!}{630630}\\ \text{Prob(1 in each group)}=\frac{616*4!}{105105}\\ \text{Prob(1 in each group)}=\frac{14784}{105105}\\ \text{Prob(1 in each group)}=\frac{4928}{35035}\\ \text{Prob(1 in each group)}\approx 0.14065\)
check
((nCr(12,3)*nCr(9,3)*nCr(6,3)*1*4!)/(nCr(16,4)*nCr(12,4)*nCr(8,4)*1) = 0.1406593406593407((nCr(12,3)*nCr(9,3)*nCr(6,3)*1*4!)/(nCr(16,4)*nCr(12,4)*nCr(8,4)*1 = 0.1406593406593407
So I have an answer. I have little idea about whether it is correct.