RE: Murder By The Book - Day 1: Murder on the Disoriented Express
03-17-2014, 12:21 AM
I haven't been to many parties like this one, but not attempting to suss out who the killer is only means that they have time to act unimpeded.
If there's 21 of us, and n of us are Undercover (or whatever moral degenerates are using them as a cover), then we can make ceiling((21 - 2n)/2) - 1 = 10 - n mistakes before the game is lost. Costly mistakes, yes, but the alternative is the death of all of us, as Undercover's only possible escape is to step over our corpses.
If we choose wrong, and they wipe another of us out, we'll have one less mistake before we are doomed, 9 - n. If we choose right, then either we have scrubbed Undercover from our midst or the remainder will strike again, leaving us with ceiling((19 - 2(n-1))/2) -1 = 10 - n mistakes.
If we choose to do nothing, then we will be offering one of our innocents on a silver platter, reducing our number to 20 while leaving n Undercover agents. That limits us to ceiling((20 - 2n)/2) - 1 = 9 - n.
Assuming no information, the chance of any of us who aren't Undercover making it out of this alive is maintained or improved by choosing a lynch over no lynch. The expected value of the decision is (n-N)/N, compared to an expected value of no lynch of -1. Unless there's some further information you have that modifies this calculus?
If there's 21 of us, and n of us are Undercover (or whatever moral degenerates are using them as a cover), then we can make ceiling((21 - 2n)/2) - 1 = 10 - n mistakes before the game is lost. Costly mistakes, yes, but the alternative is the death of all of us, as Undercover's only possible escape is to step over our corpses.
If we choose wrong, and they wipe another of us out, we'll have one less mistake before we are doomed, 9 - n. If we choose right, then either we have scrubbed Undercover from our midst or the remainder will strike again, leaving us with ceiling((19 - 2(n-1))/2) -1 = 10 - n mistakes.
If we choose to do nothing, then we will be offering one of our innocents on a silver platter, reducing our number to 20 while leaving n Undercover agents. That limits us to ceiling((20 - 2n)/2) - 1 = 9 - n.
Assuming no information, the chance of any of us who aren't Undercover making it out of this alive is maintained or improved by choosing a lynch over no lynch. The expected value of the decision is (n-N)/N, compared to an expected value of no lynch of -1. Unless there's some further information you have that modifies this calculus?