-
We have updated our Community Code of Conduct. Please read through the new rules for the forum that are an integral part of Paradox Interactive’s User Agreement.
Day 1 Votecount - 3 hours 6 minutes to deadline (valid as of post #81)
Arkasas: 3
aedan777 (54)
alynkio (61)
Rovsea (47 Dedonus -> 64)
Dedonus: 3
CaesarCzech (49)
Xarkan (72)
Panzer Commader (76)
CaesarCzech: 2
Tus3 (50)
randakar (55)
Rovsea: 2
Arkasas (51)
Wagonlitz (67)
Tus3: 1
Dedonus (46)
Panzer Commader: 1
Graf Zeppelin (52)
Wagonlitz: 1
jeray2000 (70)
Caillean: 1
Capibara (73)
Claude LC: 1
Caillean (80)
Not voted: 1
@Claude LC
The odds of rolling both villager and wolf in consecutive role assignments may be 17.99%, but a villager has better odds of rolling a wolf in the next role assignment than that, and a wolf is definitely more than 50% likely to become a villager on the second roll. The math may work out that it's unlikely for a wolf to get rolled twice, but the odds of any one person in this game being a wolf, or villager, or seer are independent of the last game, and should be treated as such.
Do you really think I would make that same argument if I were a wolf?
This is a reminder to explain the math.
I have this feeling this is going to backfire on me, but fortune favors the brave, right?
UNVOTE Tus3
VOTE CaesarCzech
Day 1 Votecount - 1 hours 30 minutes to deadline (valid as of post #84)
Arkasas: 3
aedan777 (54)
alynkio (61)
Rovsea (47 Dedonus -> 64)
Dedonus: 3
CaesarCzech (49)
Xarkan (72)
Panzer Commader (76)
CaesarCzech: 3
Tus3 (50)
randakar (55)
Dedonus (46 Tus3 -> 84)
Rovsea: 2
Arkasas (51)
Wagonlitz (67)
Panzer Commader: 1
Graf Zeppelin (52)
Wagonlitz: 1
jeray2000 (70)
Caillean: 1
Capibara (73)
Claude LC: 1
Caillean (80)
Not voted: 1
@Claude LC
Unvote CaesarCzech
I don't think a three-way tie on the first day is a good idea.
This a rather poor case, but is there even a case against Arkasas?
Vote Dedonus
Maybe it works. If it doesn't, well I ended up death in all my previous games anyway, so it can't turn out worse than average.
I don't think a three-way tie on the first day is a good idea.
This a rather poor case, but is there even a case against Arkasas?
Vote Dedonus
Maybe it works. If it doesn't, well I ended up death in all my previous games anyway, so it can't turn out worse than average.
.png){kind=avatar}
I mean, it's day one. My Bohemian colleague and I are in the front because... reasons? I mean, if you look at it, I have one random vote, one vote to make a TIE, and one revenge vote. Caesar had a random vote before you changed it, randy's third voter rule vote, and Dedonus making a TIE. Compared to us, Dedonus has legitimately acted erratically*, and unless anyone has any better ideas, in my own self-interest I will vote him.
Unvote Rovsea
Vote Dedonus
*admittedly, Caesar hasn't been here since he voted, so he hasn't acted much at all
Day 1 Votecount - 29 minutes to deadline (valid as of post #87)
Dedonus: 5
CaesarCzech (49)
Xarkan (72)
Panzer Commader (76)
Tus3 (50 CaesarCzech -> 86)
Arkasas (51 Rovsea -> 87)
Arkasas: 3
aedan777 (54)
alynkio (61)
Rovsea (47 Dedonus -> 64)
CaesarCzech: 2
randakar (55)
Dedonus (46 Tus3 -> 84)
Rovsea: 1
Wagonlitz (67)
Panzer Commader: 1
Graf Zeppelin (52)
Wagonlitz: 1
jeray2000 (70)
Caillean: 1
Capibara (73)
Claude LC: 1
Caillean (80)
Not voted: 1
@Claude LC
Official Day 1 Votecount
Dedonus: 5
CaesarCzech (49)
Xarkan (72)
Panzer Commader (76)
Tus3 (50 CaesarCzech -> 86)
Arkasas (51 Rovsea -> 87)
Arkasas: 4
aedan777 (54)
alynkio (61)
Rovsea (47 Dedonus -> 64)
Caillean (80 Claude LC -> 89)
CaesarCzech: 2
randakar (55)
Dedonus (46 Tus3 -> 84)
Rovsea: 1
Wagonlitz (67)
Panzer Commader: 1
Graf Zeppelin (52)
Wagonlitz: 1
jeray2000 (70)
Caillean: 1
Capibara (73)
Not voted: 1
@Claude LC
Claude LC will be subbed
Any objections ?
Last edited by a moderator:
Day and Night 1
Day and Night 1
At least, this act has one clear benefit: it makes it obvious that the Gungeoneers are among you. Presumably all 4 of them, though this information you can not verify. Since you are not sure how many of them there are (a couple could've died on their way here), you decide to find the most obvious Gungeoneer among you and kill him.
Since you all are Tableists, you all ask the Great Table to illuminate your brains and help you find one of the Gungeoneers by reading some of the Holy Words, namely the Chapter 1 of the "Tabla Sutra":
"And when a table is masterfully flipped, what a most magnificent sight! Surely all who witness the flip of a true master can only stand transfixed, agog."
After hearing the revelation, all of you stay silent. You are all thinking, counting on the Great Table to designate one of the Gungeoneers. Unfortunately, the ways of the Great Table are mysterious, and you can hardly agree on one candidate. Clans form, voting for various candidates, before finally a majority manages to make a name stand out of the rank: Dedonus. Yes, Dedonus is clearly the most suspicious person here. The Great Table proclaims so ! He must die.
You execute him rather brutally - it's more of a murder than an execution. Finally, when scraps are what is left of Dedonus, you can only stand, bewildered, in front of the small heap of powder in the middle of the room. Dedonus was actually a Gundead. Who knew ?
Since it is already late, you decide to go to sleep again. On top of that, you figure, killing an innocent after reciting the Holy Words has surely angered the Great Table. Trying your luck again today would surely be most unwise.
In the morning, you are sadly not surprised to see another one of your kin dead. Rovsea is inside a wall, having being fired from a gun which, you hope, you will never face.
Dedonus the Gundead is lynched
Rovsea the Gundead is hunted
I recommend everyone who voted for me not go to Vegas anytime soon. You guys wagered against some really bad odds (~1% = (4/17)^3) .
Okay here we go. So first let us introduce some notation.
Let P(A) be the Probability P that some event A happens. Then P(A∩B) is the probability that both A AND B happen. P(A/B) is the probability that A happens, given that B has already happened.
Now the general formula relating all these is: P(A∩B) = P(A/B) * P(B);
This makes sense if you think of it logically. For both A and B to happen, first one event must happen, and then the other must happen given that the first one did.
In this case, let event B be: A player is a werewolf in the first role distribution. And event A: A player is a werewolf in the second role distribution.
There are two ways of looking at it. A priori or before the fact: P(A∩B) is indeed very low as you said(1/16), but that is not what we are looking for. We want P(A/B), or the probability of a player being a werewolf now given what we know: That they were a werewolf before. Using the forumla above we get:
P(A/B) = P(A∩B) / P(B) = (1/16) / (4/16) = 1 / 4 = 4 / 16, same as everyone else.
If this doesn't seem intuitively obvious to you, you can apply the theorem after-the-fact and try to find P(A∩B) with all the information that we know.
P(A∩B) = P(A/B) * P(B);
P(A/B) is, of course, still 4/16 or 1/4, that is a constant. But P(B) = 1, obviously, since we know for sure that the player was a werewolf the first time.
Therefore: P(A∩B) = 1/4 * 1 = 1/4 or 4/16, same as everyone else.
In other words, a player is unlikely to be a werewolf twice in a row for the simple reason that they must pass the 1/4 test twice. But anyone who was a werewolf the first time has already passed that once, so it is irrelevant now. This confirms what we intuitively know to be obvious: The chance of someone being a werewolf the second time is unrelated to whether or not they were a werewolf the first time.
Let P(A) be the Probability P that some event A happens. Then P(A∩B) is the probability that both A AND B happen. P(A/B) is the probability that A happens, given that B has already happened.
Now the general formula relating all these is: P(A∩B) = P(A/B) * P(B);
This makes sense if you think of it logically. For both A and B to happen, first one event must happen, and then the other must happen given that the first one did.
In this case, let event B be: A player is a werewolf in the first role distribution. And event A: A player is a werewolf in the second role distribution.
There are two ways of looking at it. A priori or before the fact: P(A∩B) is indeed very low as you said(1/16), but that is not what we are looking for. We want P(A/B), or the probability of a player being a werewolf now given what we know: That they were a werewolf before. Using the forumla above we get:
P(A/B) = P(A∩B) / P(B) = (1/16) / (4/16) = 1 / 4 = 4 / 16, same as everyone else.
If this doesn't seem intuitively obvious to you, you can apply the theorem after-the-fact and try to find P(A∩B) with all the information that we know.
P(A∩B) = P(A/B) * P(B);
P(A/B) is, of course, still 4/16 or 1/4, that is a constant. But P(B) = 1, obviously, since we know for sure that the player was a werewolf the first time.
Therefore: P(A∩B) = 1/4 * 1 = 1/4 or 4/16, same as everyone else.
In other words, a player is unlikely to be a werewolf twice in a row for the simple reason that they must pass the 1/4 test twice. But anyone who was a werewolf the first time has already passed that once, so it is irrelevant now. This confirms what we intuitively know to be obvious: The chance of someone being a werewolf the second time is unrelated to whether or not they were a werewolf the first time.
Meh, it's some rather straightforward conditional probability. Yes, the odds of being a wolf twice in a row is (4/17)^2, but given that you've already been made a wolf in game 1, the probability of being a wolf in game 2 is 4/17. Suggesting it's lower than that is just your basic Gambler's Fallacy.
EDIT: Of course Panzer posts the full drawn-out mathematical explanation moments before I casually dismiss it as trivial. Oh well.
Official announcement
@Dedonus II: Electric Boogaloo will be subbing in for Claude LC. Welcome back !
Sub me out please
Something came up that takes all my attention, can't really play right now.
Something came up that takes all my attention, can't really play right now.