ZFR: TTTTTTT (no B, can be ignored)
Aaargh. Big mistake here. The TTTTTTT of course can be ignored in the final calculation, but it can't be ignored when calculating weights, since there
is a possibility of it appearing.
The probability of it appearing is very small (requires 7 Ts) so it doesn't have that big of an effect on the final solution, but it should be taken into account nonetheless.
So:
TTTTTTT has (7! / 7! * 1!)) = 1 permutation (duh! all Ts).
TTTTTXX has (7! / (5! * 2!)) = 21 permutations
TTTXXXX has (7! / (3! * 4!)) = 35 permutations
TXXXXXX has (7! / (1! * 6!)) = 7 permutations
And the final answer becomes:
[(1 - 1) * 1/64 ] + (this is the TTTTTTT part that here resolves to 0 and can be ignored
here. The weights though have changed slightly)
[(1 - 0.9*0.9) * 21/64 ] +
[(1 - 0.9*0.9*0.9*0.9) * 35/64 ] +
[(1 - 0.9*0.9*0.9*0.9*0.9*0.9) * 7/64 ]
= 0 + 0.06234 + 0.18807 + 0.05125
= 0.3016627
Which is...
Almost exactly same as the original probability to the point that it makes me wonder whether these two values are actually same or just pretty close.
EDIT: Yup the value is actually exactly same.
Conclusion: Knowing that a Serial Killer exists, does not change the probability of B appearing. This actually can be common-sense confirmed by looking at the fact that Serial Killer appears in exactly half of the possible permutations. The ones where he does and ones where he doesn't are "symmetrical".
The probability of B appearing does change for Mafia, who know the exact letter setup, or for an RP, who know if 1 or more letters are taken. It also would have been different if SPF was the roleblocker instead of dedo, since in that case with SK and Mafia roleblocker revealed, only TTT or T combinations are possible.