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Robijnvogel

Well-Known Member
May 8, 2013
533
421
89
If my reasoning is correct, we need to get a wolf lynched tonight. If there's still four wolves alive tomorrow, and they play their cards correctly, it becomes mathematically impossible for the village to win. Which, oddly enough, seems to argue in favor of lynching as many random people as possible in the hopes that one of them is a wolf.

Given that we do lynch a wolf tonight, how many more villagers can we afford to lynch?
0: Tomorrow, we'd have six villagers and three wolves. Definitely winnable, at least if we can figure out who they are.
1: Five villagers and three wolves tomorrow. Winnable.
2: Four villagers and three wolves. Winnable if we lynch two of those wolves tomorrow, and the last one the next day.
3: Three and three. Only winnable if the wolves mess up.

I propose that we lynch exactly three people today. That gives us our best shot at eliminating one wolf (which is critical), while still permitting a town victory, even if the wolves don't cooperate. Anyone who tries to thwart this plan will garner themselves a bucketload of wolfpoints in my eyes.

Here's a list of the currently living players:
  1. Fowltief
  2. fraction2
  3. lethosos
  4. ljfa
  5. pyure
  6. realKC
  7. robijnvogel
  8. shazam08
  9. Someone Else 37
  10. spwnX
  11. the_j
  12. victini
... uhh, that's twelve people. If four of those are wolves, then there are eight villagers left, not seven. @Robijnvogel, check your math.

A solo mislynch tonight would put us at six and four tomorrow. Might still be winnable, but we'd definitely need to lynch a wolf tomorrow, and with only six villagers, we have a much smaller margin of error. And probably not much more information to go on.

Let's see... if we lynch three people at random tonight, each lynch has approximately a 1 in 3 chance of hitting a wolf. More specifically, the first lynch has a 4 in 12 chance; if that hits a villager, the second lynch would have a 4 in 11 chance of hitting a wolf, and if that fails, the third lynch would have a 4 in 10 chance.

4/12 + 8/12 * (4/11 + 7/11 * (4/10)) = 0.75454545...
4/12 + 8/12 * (4/11 + 7/11 * (4/10 + 6/10 * (4/9))) = 0.858585...

If my math is correct, we'd have a 75% chance of hitting a wolf by lynching three people at random. 86% by lynching four. We can do better than that, but only if we get input from *everyone*, so any wolves trying to get villagers lynched would be drowned out.

... On the other hand, *I* don't have much idea on who looks wolfy, so maybe it would be best to choose unfortunate victims at random. Any ideas on a wholly impartial random number generator that doesn't rely on anyone who might be a dirty wolf messing with things?
I was under the impression that only 16 people were in this game. That is how I came to my previous calculation 16/4 = 4 wolves. Did someone join the game after that?

I think that your calculations are correct, however you fail to include that the villagers know with 100% certainty that they aren't wolves themselves. I think this improves our chances a little bit. Also, if the wolves know of our plan, I think they could easily misguide us into killing three citizens. (Maybe even killing a fourth while they're at it.)

I'll do some calculating, I guess.
 

VictiniX888

New Member
Jul 29, 2019
285
-2
0
So37's plan sounds... well the numbers are very convincing, but it runs a huge risk of being hijacked by wolves. The 75% is for random lynching, but if the wolves took over, they could probably get 3 villagers lynched instead.
That being said,
We can do better than that, but only if we get input from *everyone*, so any wolves trying to get villagers lynched would be drowned out.
How would getting input from everyone drown out wolves trying to lynch townies? They could just go for townies that look suspicious.

Right now I'm going to vote robin. I know yesterday I said he's too wolf to actually be wolf, but then he just seemed to want to get Pyure lynched so badly. He's also been very inconsistent with his vote, first saying "I will not change my vote" but will not give any more reason why Pyure is suspicious. Then near end of day he goes and hops on Fred, saying "I want to make sure Pyure and I can continue our discussion tomorrow" even though Pyure wouldn't get lynched even if he hadn't changed his vote. From what I see, that's a wolf giving up on trying to get Pyure lynched and trying to avoid suspicion by hopping on the wagon.

I also have ljfa and SO37 at the top of my wolf list.

Ljfa - what the heck. I mean, well, does ljfa usually play like this? Voting against a bandwagon, going against the one lynch a day thing, stuff. He could have actually misread the "two votes" rule, but well, his voting just seems wolfy.

So37 - only been hopping on wagons. For so37 I find this kind of wolfy. Then he proposed an awfully flawed idea based solely on numbers and skewed in favour of the wolves. Might go with your idea if you could address the flaws I mentioned, but for now, ack.
 

Robijnvogel

Well-Known Member
May 8, 2013
533
421
89
I predict:
1. I will die today (unless I make a really good case for myself, which I doubt).

2.Three people will vote on three different other people. One of the people voted on is a wolf.
3. Three other people will second these votes. One of the wolves is seconding the vote on the other wolf.
4. Other people, including wolves will second the non-wolf votes.
5. Eventually the wolf from (3.) will switch his own vote to a third villager and try to convince others of following his example.

6. We lynch 3 townsfolk + the wolves kill one at night. (I will be one of those three.)

7. 4 town, 4 wolves.
8. Town kills two of the wolves, of which they now know the identity if they payed attention.
9. Wolves lynch 2 and maul 1 villager.
10. Two remaining wolves kill lynch the last villager.

As it stands, I agree with Victini, that Someone Else's idea is flawed. Regardless, we have to do something.

Still calculating.
 

Robijnvogel

Well-Known Member
May 8, 2013
533
421
89
Statistics
We have 8 villagers against 4 wolves.

Statistical chance to
-kill only townspeople: 44,44%
-hit at least one wolf: 55,55%
-hit only wolves: 11,11%
with two absolutely random lynches.

Statistical chance to
-kill only townspeople: 29,96%
-hit at least one wolf: 70,04%
-hit at least two wolves: 25,93%
-hit only wolves: 3,70%
with three absolutely random lynches.

Statistical chance to
-kill only townspeople: 19,75%
-hit at least one wolf: 80,25%
-hit at least two wolves: 40,74%
-hit at least three wolves: 11,11%
-hit only wolves: 1,23%
with four absolutely random lynches.

Of course the wolves would make this non-random. Them being with four, they could influence 4 lynches and set up 2 lynches by themselves.

Randomiser
However, I think there is a way we could make the votes semi-random though:

I'd need 5 people (at least 1 non-wolf) to each send me 3 integer numbers from 0 to 11 via pm, after they have publicly (in this thread) announced that they will be sending me those numbers.
I will only use the 15 numbers of the people who have first publicly announced their cooperation in this randomness system.

Once I have received all 5 sets of 3, I would publish all 15 numbers, asking all 5 submitters to check whether I am not lying about them (removing my influence over the outcome).

I would then take the three sums of all 15 numbers and take the modulo of 12 of them, resulting in three numbers between 0 and 11, which each of the 5 submitters has influenced equally (all of them could have changed the number in any of the outcomes, just by changing their own submission).

I would then pick those three numbers out of the following list:

0. Fowltief
1. fraction2
2. lethosos
3. ljfa
4. pyure
5. realKC
6. robijnvogel
7. shazam08
8. Someone Else 37
9. spwnX
10 the_j
11 victini

Example
I receive the following numbers
1, 6 and 9
3, 5 and 11
8, 9 and 10
4, 5 and 8
5, 3 and 2

Result:
21 % 12 = 9, 28 % 12 = 4 and 40 % 12 = 4
SpwnX and Pyure get lynched

In this case we'd have to do another round if we'd want to lynch a third person I guess. In this regard, maybe four votes may be wiser.

Feedback
If you have any ideas that would cripple this system, please share them. I think it's pretty solid, but I could be mistaken.

If we agree on doing this, effectively giving away all control, the people that still vote different than the outcome of this randomiser will definitely be considered wolves (even if they effectively kill a wolf).
So what do you (all) think? Should we do this, or not? If we should, then how many numbers should every submitter contribute?

NOTE
I will accept integers outside the range of 0 to 11, however submitting higher or lower numbers will not affect the outcome any more than the numbers 0 to 11, so please don't bother. If you submit more numbers than the limit,
 
Last edited:

Pyure

Not Totally Useless
Aug 14, 2013
8,334
7,191
383
Waterloo, Ontario
This is all getting silly.

I'm just going to go with my original analysis at this time and vote lynch spwnx. I should probably keep gunning for robij but he made a half-decent case for himself.

I can probably also be persuaded to lynch anyone who isn't participating. srsly.
 
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Robijnvogel

Well-Known Member
May 8, 2013
533
421
89
For now, I'll vote Ljfa, (<- killing vote) because Victini makes a good case on him, I guess...

Also, like Pyure, I could probably be persuaded to lynch anyone who isn't participating.
 

fowltief

New Member
Jul 29, 2019
42
0
1
vote the J

so this is the part of the movie where everyone points gunz at each other but doesn't know who to aim at and no one shoots and the main characters talk it out,right?
 

fowltief

New Member
Jul 29, 2019
42
0
1
I sure do like my red shirt. What do main characters wear, anyway?
that depends.. usually they are always good looking no matter the situation they are in. usually pay fan service so probably part sexualized and the camera and light is always on them.
 

Robijnvogel

Well-Known Member
May 8, 2013
533
421
89
I'm having some appointments tonight, which start about now, so I won't be there to witness my lynchment. Have fun!
 

Lethosos

New Member
Jul 29, 2019
898
-7
0
Numbers are a good way to confuse people. So, yeah, SOE and Robin are up there on the wolf quotient. However, I honestly agree that more peeps need to be lynched to have a chance against the wolves. So, I'll continue the vote on Spawnty in hopes that he really is a wolf.

Sent from my Puzzle Box of Yogg-Saron using Tapatalk 2
 

Someone Else 37

Forum Addict
Feb 10, 2013
1,876
1,440
168
Statistics
We have 8 villagers against 4 wolves.

Statistical chance to
-kill only townspeople: 44,44%
-hit at least one wolf: 55,55%
-hit only wolves: 11,11%
with two absolutely random lynches.

Statistical chance to
-kill only townspeople: 29,96%
-hit at least one wolf: 70,04%
-hit at least two wolves: 25,93%
-hit only wolves: 3,70%
with three absolutely random lynches.

Statistical chance to
-kill only townspeople: 19,75%
-hit at least one wolf: 80,25%
-hit at least two wolves: 40,74%
-hit at least three wolves: 11,11%
-hit only wolves: 1,23%
with four absolutely random lynches.

Of course the wolves would make this non-random. Them being with four, they could influence 4 lynches and set up 2 lynches by themselves.

Randomiser
However, I think there is a way we could make the votes semi-random though:

I'd need 5 people (at least 1 non-wolf) to each send me 3 integer numbers from 0 to 11 via pm, after they have publicly (in this thread) announced that they will be sending me those numbers.
I will only use the 15 numbers of the people who have first publicly announced their cooperation in this randomness system.

Once I have received all 5 sets of 3, I would publish all 15 numbers, asking all 5 submitters to check whether I am not lying about them (removing my influence over the outcome).

I would then take the three sums of all 15 numbers and take the modulo of 12 of them, resulting in three numbers between 0 and 11, which each of the 5 submitters has influenced equally (all of them could have changed the number in any of the outcomes, just by changing their own submission).

I would then pick those three numbers out of the following list:

0. Fowltief
1. fraction2
2. lethosos
3. ljfa
4. pyure
5. realKC
6. robijnvogel
7. shazam08
8. Someone Else 37
9. spwnX
10 the_j
11 victini

Example
I receive the following numbers
1, 6 and 9
3, 5 and 11
8, 9 and 10
4, 5 and 8
5, 3 and 2

Result:
21 % 12 = 9, 28 % 12 = 4 and 40 % 12 = 4
SpwnX and Pyure get lynched

In this case we'd have to do another round if we'd want to lynch a third person I guess. In this regard, maybe four votes may be wiser.

Feedback
If you have any ideas that would cripple this system, please share them. I think it's pretty solid, but I could be mistaken.

If we agree on doing this, effectively giving away all control, the people that still vote different than the outcome of this randomiser will definitely be considered wolves (even if they effectively kill a wolf).
So what do you (all) think? Should we do this, or not? If we should, then how many numbers should every submitter contribute?

NOTE
I will accept integers outside the range of 0 to 11, however submitting higher or lower numbers will not affect the outcome any more than the numbers 0 to 11, so please don't bother. If you submit more numbers than the limit,
Problem: If you are a wolf and one of the people who PM'd you numbers is a wolf, you could select numbers for that wolf's vote such that the lynches fall upon any players you wish, and when you publish the numbers, that wolf would just say "Yep."

Nonetheless, I like where you're going with this. In order to be completely foolproof, five players would have to send copies of the same three random numbers to five other players. Heck, since we have twelve peeps, make it six. Even-numbered players in your zero-indexed list would each send their three (or four? Five, just in case?) random numbers to all six odd-numbered players, who would then either publish the numbers they got or OK the published list. That way, no one can do any messing with the numbers without a villager catching them.

... That sounds like a logistical nightmare that could never be resolved in the few hours left in the day. That said, as I am #8 in that list, I will send some random numbers to Fraction, ljfa, RealKC, Shazam, SpwnX, and Victini momentarily in the off chance that anyone actually decides to go through with this plan.
 

Fraction2

New Member
Jul 29, 2019
27
0
0
Sorry, I've been incredibly busy, and werewolf falls below "not-failing-school. " I realize the day is almost over, so I'll place a Vote shazam08 , because I didn't see a vote on him earlier. If my reading skills suck in this regard, please inform me.
 

Pyure

Not Totally Useless
Aug 14, 2013
8,334
7,191
383
Waterloo, Ontario
Would anyone like to help me double-up on Fraction2? I read his vote as a rather blatant attempt to "wrap up the game for wolves today"

Vote fraction2.
 

Someone Else 37

Forum Addict
Feb 10, 2013
1,876
1,440
168
If nothing else, my crazy ideas have generated some discussion. Time for me to do some analysis myself!

Sounds good to me, SE.

Lynch ljfa for the crime of posting just before me.
This seems contradictory to me. You say you like my ideas, then make no effort to implement them. You could argue that voting for the last person to post is about the randomest thing you could do, but I could argue that if you were a wolf who knows that ljfa isn't, you'd do the same thing.

How would getting input from everyone drown out wolves trying to lynch townies? They could just go for townies that look suspicious.
If the wolves try to lynch townies, and the townies just kinda follow along, then you're right, we'd lynch townies and be doomed. If all of the townies do their own analysis and come up with their own conclusions not influenced by the opinions of others, then that's eight honest votes from townies and only four dishonest votes from wolves. They won't be able to control the top three votes. Probably.

So37 - only been hopping on wagons. For so37 I find this kind of wolfy. Then he proposed an awfully flawed idea based solely on numbers and skewed in favour of the wolves. Might go with your idea if you could address the flaws I mentioned, but for now, ack.
A reasonable position for an arithmophobe and an accurate description of the plan as it was at that time. Now that the worst of these flaws have been addressed, what do you think?

Statistics
We have 8 villagers against 4 wolves.

Statistical chance to
-kill only townspeople: 44,44%
-hit at least one wolf: 55,55%
-hit only wolves: 11,11%
with two absolutely random lynches.

Statistical chance to
-kill only townspeople: 29,96%
-hit at least one wolf: 70,04%
-hit at least two wolves: 25,93%
-hit only wolves: 3,70%
with three absolutely random lynches.

Statistical chance to
-kill only townspeople: 19,75%
-hit at least one wolf: 80,25%
-hit at least two wolves: 40,74%
-hit at least three wolves: 11,11%
-hit only wolves: 1,23%
with four absolutely random lynches.

Of course the wolves would make this non-random. Them being with four, they could influence 4 lynches and set up 2 lynches by themselves.
How did you come up with those percentages? Keep in mind that if we're going to lynch three different people, we're going to be drawing without replacement, to use the technical term. The drawings are not independent.

This is all getting silly.

I'm just going to go with my original analysis at this time and vote lynch spwnx. I should probably keep gunning for robij but he made a half-decent case for himself.

I can probably also be persuaded to lynch anyone who isn't participating. srsly.
Oh boy, another arithmophobe. Lynching quiet peeps is probably the best we can do at this point- some of them are bound to be wolves.

Numbers are a good way to confuse people. So, yeah, SOE and Robin are up there on the wolf quotient. However, I honestly agree that more peeps need to be lynched to have a chance against the wolves. So, I'll continue the vote on Spawnty in hopes that he really is a wolf.

Sent from my Puzzle Box of Yogg-Saron using Tapatalk 2
While I have been known to post objectivish numericky analyses as a wolf (because they look townish and I stink at making actual reads), I don't usually put this much effort into them.

Would anyone like to help me double-up on Fraction2? I read his vote as a rather blatant attempt to "wrap up the game for wolves today"

Vote fraction2.
Sure. Vote Fraction2.
God I hope this isn't a trap.

More math coming up in a bit.
 

Pyure

Not Totally Useless
Aug 14, 2013
8,334
7,191
383
Waterloo, Ontario
Full disclosure: When I read fraction2's comment that the day was over, I thought he meant 1PM EDT (5 minutes from now, 20 minutes from when he posted). The day doesn't actually end for another 2 hours. That makes his comment considerably less alarming. So feel free to change your vote.
 

Pyure

Not Totally Useless
Aug 14, 2013
8,334
7,191
383
Waterloo, Ontario
fwiw I apologize if I'm phoning this game in a bit. Its hockey draft season and I write the drafting software for a lot of the leagues in north america. Wrapped up the OHL draft on saturday, working on CCHL this week :\
 

Someone Else 37

Forum Addict
Feb 10, 2013
1,876
1,440
168
Probability lesson!

If there are v villagers and w wolves, then the probability that a randomly-chosen lynch victim will hit a villager is v / (v+w), and the probability that it will hit a wolf is w / (v+w). This leaves (v-1) villagers or (w-1) wolves, respectively, available for the next lynch.

At the moment, we have eight villagers and four wolves. v = 8 and w = 4. The best way to enumerate all possible outcomes is with a decision tree:
  • Hit a villager on the first lynch: P = 8/12, v = 7, w = 4
    • Hit a villager on the second lynch: P = 7/11, v = 6, w = 4
      • Hit a villager on the third lynch: P = 6/10, v = 5, w = 4
      • Hit a wolf on the third lynch: P = 4/10, v = 6, w = 3
    • Hit a wolf on the second lynch: P = 4/11, v = 7, w = 3
      • Hit a villager on the third lynch: P = 7/10, v = 6, w = 3
      • Hit a wolf on the third lynch: P = 3/10, v = 7, w = 2
  • Hit a wolf on the first lynch: P = 4/12, v = 8, w = 3
    • Hit a villager on the second lynch: P = 8/11, v = 7, w = 3
      • Hit a villager on the third lynch: P = 7/10, v = 6, w = 3
      • Hit a wolf on the third lynch: P = 3/10, v = 7, w = 2
    • Hit a wolf on the second lynch: P = 3/11, v = 8, w = 2
      • Hit a villager on the third lynch: P = 8/10, v = 7, w = 2
      • Hit a wolf on the third lynch: P = 2/10, v = 8, w = 1
To find the probability of any node, simply multiply together the probabilities all along the path leading up to it. Also, since the nodes on each level represent mutually exclusive and exhaustive events, the probability of any one of a subset of branches taking place is simply the sum of those branches' individual probabilities. We can't lynch a villager AND a wolf with the same lynch, so there's no need to account for that possibility.

As an example, the probability of hitting no wolves at all with three random lynches is the probability of the node on line 3 of the above tree.
P = 8/12 * 7/11 * 6/10 = 0.2545 ~= 25%
So (because these events are exhaustive) the probability of hitting at least one wolf with three random lynches is about 75%.

The real question is, is that better or worse than the probability of lynching at least one wolf at the rate we're going? I don't have the faintest idea.
 
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