*** Coronavirus Mafia ***

All I’m going to say is you better hope I’m not the asymptomatic carrier because then the innos are in trouble with a runaway
agreed...it would be smart to spread votes out in case that's true.
 
7: Arob - (Cy$ 2:09pm) (Ctown 4:09pm) (Cleaver 6:24 pm) (Gordyo5 7:01 pm) (CloniesForLife 7:42 pm) (Vikes 7:46 pm) (michael 9:00 pm)

This is what I was looking for. If Gordy is innocent, is tomorrow's plan to look at someone else who voted for arob?

agreed...it would be smart to spread votes out in case that's true.

Keeping these two things in mind, I'm going to Vote Clonies to try and help michael out of a jam.
 
agreed...it would be smart to spread votes out in case that's true.
Hypothesis:
It doesn't matter statistically whether all the votes are on one person or spread out

Assumptions:
There are 23 remaining players and 1 of those is the AC
Everyone has the same chance to be the AC

Scenarios:
1) 22 votes for one person who happens to be the AC: 22 x (0.25) = 5.5 new mafia members - 1 old mafia member (AC) = 4.5 mafia added
There is a 1 in 23 chance that the above happens: 4.5 / 23 = 0.20 mafia added on average

2) 3 different players get 7 votes each, one of which is the AC: (7+1[assuming the AC votes for one of the other two]) x (0.25) = 2 new mafia members - (1 in 3 chance the AC is voted out) = 1.67 mafia add
There is a 3 in 23 chance the above happens: 1.67 x (3/23) = 0.22 mafia added on average

Other notes:
This isn't perfect because who current mafia votes for isn't accounted, but I don't have those numbers so I can't include it

Conclusion:
It doesn't matter whether the votes are spread out or all placed on one person.


Feel free to rip about my math during the peer review process.
 
Hypothesis:
It doesn't matter statistically whether all the votes are on one person or spread out

Assumptions:
There are 23 remaining players and 1 of those is the AC
Everyone has the same chance to be the AC

Scenarios:
1) 22 votes for one person who happens to be the AC: 22 x (0.25) = 5.5 new mafia members - 1 old mafia member (AC) = 4.5 mafia added
There is a 1 in 23 chance that the above happens: 4.5 / 23 = 0.20 mafia added on average

2) 3 different players get 7 votes each, one of which is the AC: (7+1[assuming the AC votes for one of the other two]) x (0.25) = 2 new mafia members - (1 in 3 chance the AC is voted out) = 1.67 mafia add
There is a 3 in 23 chance the above happens: 1.67 x (3/23) = 0.22 mafia added on average

Other notes:
This isn't perfect because who current mafia votes for isn't accounted, but I don't have those numbers so I can't include it

Conclusion:
It doesn't matter whether the votes are spread out or all placed on one person.


Feel free to rip about my math during the peer review process.
@PSYclone22
 
I'm going to continue on my war path and

vote Cleaver

for what seems like the 24th time this game.
 
Unofficial Vote Count

7 – Cleaver (cmjh 7:09 am) (Clonies 9:10 am) (dandy 9:31 am) (Fitzy 9:50 am) (MeowingCows 9:54 am) (CoKane 10:03 am) (andybernard 1:53 pm)
3 – Cy$ (mdk2isu 8:42 am) (Fitzy 8:42 am rescind 9:50 am) (cubs 8:58 am) (mj4cy 9:30 am)
2 – Clonies (Cleaver 9:45 am) (Ctown 1:19 pm)
1 – SCNCY (Cy$ 10:08 am)
 
I rated his post a winner.

It's really an analysis of what type of risk to take:

Spread votes out (we in the industry call it isolation) = increased chance ONE person gets the virus but limit chance of epidemic

Place all the votes on a few = reduced chance of having someone get infected, but increased chance of an epidemic.

All-in-all, the mathematic cost-benefit is equal, spiker was spot on.

End of ghost input.
 
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