Werewolf LXXXI: Under the Misty Mountains - The Barrow-Downs Discussion Forum

the phantom · 09-22-2010, 05:22 PM

If we have a single-lynch today, there will be ten of us tomorrow. Now, suppose that the Seer somehow has a way to deliver to you the names of three innocents (or his own name plus two innocents, still adding up to three).

That will leave 7 total unknowns, with 3 Elves (2 if we're lucky today) hidden in their midst.

Since there will be 10 total, we will be able to elect 5 reps. And with 5 reps, we can lynch 5 people.

So, the proven innocents come up with a voting plan, mandate who the reps should be and who should vote for them, ensuring that there are 5 total. Then everyone debates between the 7 non-provens and decides which 5 to lynch (this should be decided EARLY- start doing it in the first half of the day).

The proven innocents then tell the 2 reps who aren't proven to cast their lynch votes for one of the 5 candidates decided upon. Once they've voted and thus everything is safe, the 3 provens split their votes on the other 3, thus clinching a 5-way tie. With numbers like that, I'd say the odds would be favorable that we'd lynch the Elves, or at the least not have more than 1 left alive.

If 1 is left alive, then naturally he will slay one of the provens during the night, but the next day there will be 4 villagers left and 2 of them proven innocent. And so those two tell the remaining two to make them Reps by casting a vote for either of them, and then the 2 known innocents proceed to lynch the two other villagers to secure the victory. If the Elf refuses to cast his Rep vote the final day, then of course everyone will know for a fact precisely who he is and the 2 provens can vote for the other unknown and he can lynch the last Elf.

So... thoughts? I'd say it's a darn good plan.

09-22-2010, 05:22 PM	#21
the phantom Beloved Shadow Join Date: Jul 2002 Location: The Stadium Posts: 5,971	something to consider for tomorrow... If we have a single-lynch today, there will be ten of us tomorrow. Now, suppose that the Seer somehow has a way to deliver to you the names of three innocents (or his own name plus two innocents, still adding up to three). That will leave 7 total unknowns, with 3 Elves (2 if we're lucky today) hidden in their midst. Since there will be 10 total, we will be able to elect 5 reps. And with 5 reps, we can lynch 5 people. So, the proven innocents come up with a voting plan, mandate who the reps should be and who should vote for them, ensuring that there are 5 total. Then everyone debates between the 7 non-provens and decides which 5 to lynch (this should be decided EARLY- start doing it in the first half of the day). The proven innocents then tell the 2 reps who aren't proven to cast their lynch votes for one of the 5 candidates decided upon. Once they've voted and thus everything is safe, the 3 provens split their votes on the other 3, thus clinching a 5-way tie. With numbers like that, I'd say the odds would be favorable that we'd lynch the Elves, or at the least not have more than 1 left alive. If 1 is left alive, then naturally he will slay one of the provens during the night, but the next day there will be 4 villagers left and 2 of them proven innocent. And so those two tell the remaining two to make them Reps by casting a vote for either of them, and then the 2 known innocents proceed to lynch the two other villagers to secure the victory. If the Elf refuses to cast his Rep vote the final day, then of course everyone will know for a fact precisely who he is and the 2 provens can vote for the other unknown and he can lynch the last Elf. So... thoughts? I'd say it's a darn good plan. __________________ the phantom has posted. This thread is now important.