Mafia Chat~!

[Le King]

Wait lemme run the numbers.


ABCD

A kills B

B kills C

C kills D

D kills... A

Wait no. <********.

I did not think this through. burning_eyes  

[123jimmy321]

Le King

Vigoureux

Le King

Vigoureux

If A targets B:
• B targets A, and C targets A, we're on 2.
• B targets A, and C targets B, we're still on 2.
• B targets C, and C targets A, everyone dies.
If A targets C:
• B targets A, and C targets A, we're on 2.
• B targets A, and C targets B, everyone dies.
• B targets C, and C targets A, we're still on 2.

Yes

My brain broke a little but yup!

But if it's
"Only one winner. If 2 people remain, everyone loses."
then everyone loses in all scenarios.
Unless the point is to make everyone play until they notice. ninja

Not necessarily.

3 would make it unwinable.
2 will make it a loss.

Their goal should be to end up with at least 4 people in the final round.

The whole point of it was to make them talk to each other and figure out who's killing whom and who should die to have the correct numbers.

It's meant to be a numbers game.

But more importantly, it's meant to be hilarious for the host.

What if people band wagon kill people?  

[Le King]

Let's try with 5


ABCDE

A kills E
B kills E
C kills E
D kills A
E kills B


2 left



ABCDEF

A kills F
B kills F
C kills F
D kills A
E kills C
F.... ******** takes a nosedive off a cliff into a pool of lava with sharks.  

[Le King]

123jimmy321

Le King

Vigoureux

Le King

Vigoureux

If A targets B:
• B targets A, and C targets A, we're on 2.
• B targets A, and C targets B, we're still on 2.
• B targets C, and C targets A, everyone dies.
If A targets C:
• B targets A, and C targets A, we're on 2.
• B targets A, and C targets B, everyone dies.
• B targets C, and C targets A, we're still on 2.

Yes

My brain broke a little but yup!

But if it's
"Only one winner. If 2 people remain, everyone loses."
then everyone loses in all scenarios.
Unless the point is to make everyone play until they notice. ninja

Not necessarily.

3 would make it unwinable.
2 will make it a loss.

Their goal should be to end up with at least 4 people in the final round.

The whole point of it was to make them talk to each other and figure out who's killing whom and who should die to have the correct numbers.

It's meant to be a numbers game.

But more importantly, it's meant to be hilarious for the host.

What if people band wagon kill people?

That's the point.  

[Vigoureux]

Le King

Not necessarily.

3 would make it unwinable.
2 will make it a loss.

Their goal should be to end up with at least 4 people in the final round.

The whole point of it was to make them talk to each other and figure out who's killing whom and who should die to have the correct numbers.

It's meant to be a numbers game.

But more importantly, it's meant to be hilarious for the host.

Ok, I get it now.
Although I can already see the outcome:
"I'll betray everyone by killing an extra because I can!"
/ends with three players
(┛◉Д◉)┛彡┻━┻

 

[Le King]

Vigoureux

Le King

Not necessarily.

3 would make it unwinable.
2 will make it a loss.

Their goal should be to end up with at least 4 people in the final round.

The whole point of it was to make them talk to each other and figure out who's killing whom and who should die to have the correct numbers.

It's meant to be a numbers game.

But more importantly, it's meant to be hilarious for the host.

Ok, I get it now.
Although I can already see the outcome:
"I'll betray everyone by killing an extra because I can!"
/ends with three players
(┛◉Д◉)┛彡┻━┻

THAT WAS THE WHOLE HILARIOUS POINT UNTIL I REALIZED MY MATH WAS WRONG crying


screw the 2 people cancel each other rule.
there needs to be a cancel option, just apparently not this one. x-x  

[Vigoureux]

Le King

Vigoureux

THAT WAS THE WHOLE HILARIOUS POINT UNTIL I REALIZED MY MATH WAS WRONG crying

screw the 2 people cancel each other rule.
there needs to be a cancel option, just apparently not this one. x-x

Sorry! Sometimes I'm a party-pooper with my overthinking. sweatdrop

 

[Le King]

Vigoureux

Sorry! Sometimes I'm a party-pooper with my overthinking. sweatdrop

Hey you accepted the 4-people victory before I destroyed my own thing. qqqqqqq

Let's-team this nice party s**t. I have drugs. And by drugs I mean laxatives. crying  

[Vigoureux]

Le King

I have drugs. And by drugs I mean laxatives. crying

That would be a shitty way to die.
OD'ed on laxatives. lol

 

[Le King]

Vigoureux

Le King

I have drugs. And by drugs I mean laxatives. crying

That would be a shitty way to die.
OD'ed on laxatives. lol

Well, then.

Ladies and gents and everything in between, we got a death scene.

Go home everyone.

We got this.  

[Vigoureux]

Le King

Hey you accepted the 4-people victory before I destroyed my own thing.

Off the top of my head,
at least if A, B, and C target D, the game should be winnable with four players,
since only one kill would cancel another kill.

 

[123jimmy321]

Vigoureux

Le King

Now what if you have it if the same 2 people target each other the kill doesn't go through or you RNG a 50/50 assigning numbers to each player for a battle to the death kinda thing?  

[Vigoureux]

123jimmy321

Now what if you have it if the same 2 people target each other the kill doesn't go through or you RNG a 50/50 assigning numbers to each player for a battle to the death kinda thing?

In the example that King gave, I was counting all kills.
So if they both targeted each other, they would both die.

 

[Le King]

Vigoureux

Le King

Hey you accepted the 4-people victory before I destroyed my own thing.

Off the top of my head,
at least if A, B, and C target D, the game should be winnable with four players,
since only one kill would cancel another kill.

No.
Way ahead of you, already proved this can't be done.
*ABCD have equal values but represent different sets that also have equal values. In this case, all their possible permutations are irrelevant. You put it in all possible combinations in your other post, but that's redundant as you are arguing combinatorics not algebra. You only need to test with one permutation because they are all interchangeable and have the same value.

So for shorthand I'll be testing with 1 set.

ABCD (4 people)
3 v 1
You always end up with having to have 3 people (ABC) target 1 (D) to get someone killed since even (AB, BA, CD) cancel eachother, and you will be left with = 2 players, Loss.

ABC target D, 3 left.

Hence no, 4 does not work. You then need to test with 3.

i. ABC
A target B
B target C
C target D

i. A kills B, B kills C, C kills A = 0 players left. No winner.


ii. ABC
Any odd permutation of 3 means either 1 or 3 required to make a kill.
1 required to make a kill: see i. everyone dies.
3 required to make a kill: requires a suicide.
A kills C, B kills C, C kills C = 2 players left. Loss.



This is true for any number of players, because in all cases 1 or 3 (or 5 if 5 are alive) players are required to make a kill.
1 for 1 kills equals to 1 for 1 ratio, and everyone dies, No winner.

3 for 1 equals there will always be two players left, which means Loss.



Further, let's consider the premise "if 2 killers target each other, they cancel each other out."

I already proved there can only be two possible scenarios (targeting in a loop with no repeat AB BC CA or AB BC BD DA or AB BC CD DE EA to infinity) a repeat a the very minimum has to occur within the least possible (3) or higher.

Let's test with 3.
AB
BA
CA

A cancels B
B cancels A
both live
C kills A

BC left. 2 people left. Loss.


Let's test for 4.
AB
BA
CD
DA
AB is canceled, CD is canceled. 4 left. Stalemate.


AB
BA
CA
DB

AB cancel eachother, but then C and D kill them. 2 people left. Loss.


Test with 5.
AB
BA
CA
DB
EC

A and B cancel.
C kills A
D kills B
E kills C

DE left. This is scenario ii. 2 people left. Loss.

i.
ii.
iii. we just formulated the final rule: with two cancels, you are left with 2 deaths that cannot happen and at the very minimum two players will remain. Loss.



Combinatiorics. emotion_yatta



When I say I'm bad at math I'm actually lying.  

[Le King]

Now if we put it into pure math, we are drawing a line through a circle by plotting two points on the ******** circle and trying to find the shortest line possible.

Shortest line through the circle THEORETICALLY should be 1 point. A 1-point line is a false premise. emotion_facepalm

The everyone-is-a-killer game is based on a false premise that there can be a 1-point line. emotion_facepalm emotion_facepalm emotion_facepalm emotion_facepalm emotion_facepalm