And now for more probabilities...
The odds of a single die coming up "B" is 5%.
Seven dice were rolled.
So the odds of no "B" coming up are 0.95*0.95*0.95*... = 69.8%. Which means that there's a 30% chance that there's at least one town roleblocker in the game.
This of course is true for a virgin pre-game draw. For the stage the game was in however, you could look at it differently, since you had more information.
For example, Mafia knew the setup was TTTXXXX, where X is a non-T letter. The probability of X being B is 10%.
So for them, the probability of a town roleblocker being present in game was:
1 - (0.9*0.9*0.9*0.9) = 34.4%
Of course since they knew GameRager was Town, they had a strong reason to believe he wasn't lying, so the probability was close to 100%, but still...
Town on the other hand, knew there was a Serial Killer, so the possible combinations were:
TTTTTTT (no B, can be ignored) (EDIT: this is not entirely true, and the remainder of the calculations in this post are slightly wrong because of this. See post 1113
The latter 3 combinations didn't have an equal chance of appearing. There was for example a bigger chance of TTTXXXX appearing than TXXXXXX.
So we have to find the probability of each combination appearing, and using that as a weight to multiply the probability of B appearing in that setup. Basically
P(setup appearing) * P(B appearing in given setup) for all three setups, and then summing them up.
So: T and X have both equal chance of appearing, therefore all permutations have equal chance of appearing.
TTTTTXX has (7! / (5! * 2!)) = 21 permutations
TTTXXXX has (7! / (3! * 4!)) = 35 permutations
TXXXXXX has (7! / (1! * 6!)) = 7 permutations
If we divide all by 7, we get the first combination has a weight of 3/9, the second has 5/9 and the last one has 1/9
So P(B appearing, given a Serial Killer) =
[(1 - 0.9*0.9) * 3/9 ] +
[(1 - 0.9*0.9*0.9*0.9) * 5/9 ] +
[(1 - 0.9*0.9*0.9*0.9*0.9*0.9) * 1/9 ]
= 0.06333 + 0.19106 + 0.05206
(Yeah, I know it's also around 30%, but it's compeltely different).
The above are even more different for a PR who already knows one letter (e.g. trent would know one X is taken as D), and yet more different for a PR who knows more than one letter (e.g. Joe knows two Xs are taken as DD). And finally, to someone who's a town roleblocker himself, the probability is different and equal to 100%.