[REVIEW] Open Setup Reviews

[Gamma Emerald]

Seems like a jungle game not branded as one

[Gamma Emerald]

Yeah, there should be 1 move free per day I’d say

[Gamma Emerald]

With 3 out of 10, the odds of being placed into 3 sequential slots is I believe 1/18 (2/9 * 2/8 = 4/72 = 1/18).

[Gamma Emerald]

another thought that came to mind (if we're tinkering with the mechanic itself instead of the town:scum ratio), would be to not allow scum to trapdoor players who have received it before (instead of just non-consecutively). this way, if scum are in slots #1, #2, #4, scum can't just bounce between #7 and #8. they'll have used up the furthest-away slots, forcing them to bring the trapdoor closer. this also introduces town strategies of possibly leaving certain slots alive if they want to influence parts of the playerlist where the trapdoor can't go next.

Definitely don’t use the “free move a day” idea if this is used, as town can technically do that repeatedly until no options remain, though I’d assume a fail safe exists
Also I almost tried to dive into the math for 3 in 4 then I realized it was kinda a trick question.

[Gamma Emerald]

It’s 100% since you only ever have 1 negative spot, which with the circle formation results in them always being adjacent

[Gamma Emerald]

Actually hold on I think my math was off since you can technically fill a gap with the third
Had you not clarified 3 in 4 that wouldn’t have clicked

[Gamma Emerald]

I’m gonna be posting unga bunga math post soon, covering both the updated logic for all three being together, as well as the chance of three being in a range of four adjacent seats with a space somewhere between.

[Gamma Emerald]

I'll start with covering the chance of all 3 scum being together. The stuff inside the spoiler= tag is me going through the minutiae of calculating the probability, skip it if you don’t care for it.

Spoiler: all 3 together

Let’s assume we start like this, starting person’s placement is where all probability starts, so to make things simple we’ll put them in a fixed location.

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ooooxooooo

Acceptable locations for the second person when going for all 3 together are (the newly placed person will be the one capital letter. These will be marked with unique letters for the next step

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A: ooXoxooooo
B: oooXxooooo
C: ooooxXoooo
D: ooooxoXooo

That ends up at 4/9 locations having the potential to have all 3 together. For the next step, take each valid outcome and calculate the outcomes that meet the final goal for each of those.

The letter notations will now be used to separate the outcomes that relate to each placement of 2, with each valid option having an extra letter added (so the first valid option for A will be AA)

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AA: ooxXxooooo
BA: ooXxxooooo
BB: oooxxXoooo
CA: oooXxxoooo
CB: ooooxxXooo
DA: ooooxXxooo

What is interesting to note is that we end up with 3 different distinct sets of locations (345, 456, and 567), each one just having 2 different orders of picking them, the commutative property (5, then 4, then 6 results in 456, while 5, then 6, then 4 achieves the same outcome). This results in 6 possible outcomes, assuming a fixed location for our first person, that result in the 3 subjects all being together.
This is taken out of the number of outcomes in general to find our true probability of this event. That is 9 times 8, which equals 72, because our second person would have 9 potential locations, while the third has 8. So we get 6/72, which we can simplify to get 1/12.

To put this simply (and provide an answer without needing to open the spoiler= tag), we have 3 valid arrangements assuming a fixed placement for our first person, and each arrangement can have the two variable spots filled in either order, resulting in 2 permutations of each valid arrangement. Thus there are 6 valid outcomes, with 72 possible outcomes, for a final probability of 1/12.
Let's put this in more tangible terms. Assume there is a circular table with 10 chairs around it, and someone is already seated. A second person comes in and sits down, and let's assume they know there's a third person coming and they all want to sit together. In this situation, the second person can either sit in one of the two seats directly next to person already seated, in which case the third person can sit in the seat next to the first or second person. The second person can also sit with a seat of space between them and the seated person, in which case the third person has to sit between them.

Now for the chance of 3 people being in a region of 4 adjacent spots with a gap in their placement, which isn't actually as complex as I initially thought it might be.]. By virtue of how this group is defined, no previously covered outcomes will show up again, which I'll be expanding on later.

Spoiler: 3 in 4 / xoxx

So we'll start with the same spot we did last time.

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ooooxooooo

Since we are now covering a range of 4 total spots with our arrangement, the possible ground to cover now becomes 7 spots instead of 5 from last time (the ground covered to cover is equal to the starting location plus the amount of possible distance away on each side). There's no condition regarding how the second person can be placed, they can be next to, one away, or two away from the first. Like so, once again using letters to make it easier to distinguish outcome branches in the next step.

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A: oXooxooooo
B: ooXoxooooo
C: oooXxooooo
D: ooooxXoooo
E: ooooxoXooo
F: ooooxooXoo

With the second person placed, the third person now has to be placed in a way that will satisfy the 3 in 4 with a gap. I will go over the logic for where the third person can go after the possible locations. As you probably guessed, these outcomes will be marked as a pair of letters with the first representing a placement of the second person and the second being representative of a viable placement of the third.

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AA: oxXoxooooo
AB: oxoXxooooo
BA: oXxoxooooo
BB: ooxoxXoooo
CA: oXoxxooooo
CB: oooxxoXooo
DA: ooXoxxoooo
DB: ooooxxoXoo
EA: oooXxoxooo
EB: ooooxoxXoo
FA: ooooxXoxoo
FB: ooooxoXxoo

Now, let's get into the meat of why the third has to be in the places it goes. Let's start with the cases of the first and second being adjacent (C and D). In this scenario there is no existing gap so it must be between the third and the person they're on the side of (the first for CB and DA, the second for CA and DB). Next up is the case of there being a one-spot gap between the first and second (B and E). In this case, the third now has to be in a location that is directly adjacent to either only the first (BB and EA) or the second (BA and EB). Finally, when we have two spots between the first and second (A and F), the third has to go somewhere in that in-between space, either being next to the first (AB and FA) or the second (AA and FB).

So we end up with 2 possible placements of the third for each placement of the second, which we had 6 of, giving us 12 placements that meet the "3 in 4 with a gap" criteria. Once again taking this out of 72, we have 12/72 which reduces to 1/6.
Now interestingly we once again have 2 possible permutations of each arrangement, which when you divide 12 by six, you get 6 unique arrangements. These are 235 (AA and BA), 245 (AB and CA), 356 (BB and DA), 457 (CB and EA), 568 (DB and FA), and 578 (EB and FB). If you're interested in a better visualization of these, I'll put it in a spoiler at the very end along with ones for the permutations for "all 3 together"

Again summing up for those who didn't read the complex math dive, the second person can now be one space farther away from the first person for a total of 6 spots the second person can be in, and the third will always have 2 spaces they can be for each possible location of the second, defined by being on the side of the first or second, for 12 viable "3 in 4 with a gap somewhere" outcomes, being taken out of the 72 overall possible outcomes for a total probability of 1/6. I won't be diving into the chair metaphor again, if you want to see me go into that deep logic without focusing on the full math behind it, I colored the text within the spoiler= tag that does so.

Now I'm going to flex my statistics muscles a bit. Let's say we want to find out the plain odds of the 3 being within 4 spaces at all with a fixed location of the first. This doesn't necessitate a gap, mean the first 6 are valid, and also doesn't require them all to be bunched together completely, allowing the other 12 that have that one extra bit of space. since the events are exclusive, you just add them, which is 6 + 12 = 18, and again take it out of your 72 overall outcomes for a 18 in 72, or 1/4, chance. That's about it for the math.

Spoiler: enhanced displays of the arrangements and permutations for those who care

For these, I'm using numbers to fill out the ten spots, o's becoming zeroes and the x's becoming 1, 2, or 3, showing the order they're placed. While the letter notations are gone the results will be in the same order as when they were there, for ease of reference.

"all 3 together":

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0023100000
0032100000
0002130000
0003120000
0000123000
0000132000

"3 in 4 with a gap":

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0230100000
0203100000
0320100000
0020130000
0302100000
0002103000
0030120000
0000120300
0003102000
0000102300
0000130200
0000103200

[Gamma Emerald]

3/4 is if you're doing straight row AND range of 4 with a gap
3 in 4 with a gap itself was 1/6

[Gamma Emerald]

I like the statistical mindset, but I think that specific solution goes against the original setup idea
The setup seems based around the static locations of the players. Taking that away creates a chaotic not-really-system that just seems less interesting.

[Gamma Emerald]

At any time, a Blue Mafia may submit a kill by PM. If the named player is a Mason, they will die. If the named player is a Mason, nothing will happen. If the named player is VT, the submitter will die.

:thonk:

[Gamma Emerald]

btw I think we should do a few runs of that trapdoor setup using the different ideas proposed during the upcoming marathon weekend

[Gamma Emerald]

probably save the seer shot for a mason recruit, that seems like a pretty good way to boost town winrate?

[Gamma Emerald]

confirmed town 1 shot true masonizer

doesn't mean masonizer should act only once fyi

[Gamma Emerald]

no, don't change the role, I was thinking about how it should be played

[Gamma Emerald]

How do the second row roles work? Kinda hard to share results if you’re dead.

[Gamma Emerald]

can the dead communicate?

[Gamma Emerald]

Tryna shit on my scumgame to my face do you wanna fuckn go right now Hectic I got a parking lot right here

...
I'm not afraid to say I think your scumgame needs work, if Haunted Village was any indication

[Gamma Emerald]

Who is gonna be asleep for 12+ hours???

[Gamma Emerald]

You have issues bro