As at yesterday there were 18 town and 5 scum. Randomly pick 4 the chances they are all town is (18*17*16*15)/(23*22*21*20) = 34.6% [1dp]Balki Bartokomous wrote: ↑Tue Dec 01, 2020 10:34 amDamo, choose any group of four. Collectively, the group is very likely to contain at least one scum.damo666 wrote: ↑Tue Dec 01, 2020 10:25 amThe chances of 4 townies all voting town both days is (19/25)*(17/22)^4=11.9%.
It is therefore quite likely there is at least 1 scum in the four (another way of saying not ALL town). The chance an individual town randomly votes 2 town in a row is about 58.7% so not taking anything else into account any one individual is more likely to be town but collectively the group is very likely to contain AT LEAST ONE SCUM.
Yet again you choose to misconstrue what I say to throw shade at me. You need to detunnel.
That doesn't support your position here. You need to explain why the members of this particular foursome are more likely to flip scum than a random foursome. I don't believe you have made that case persuasively. But I'm listening.
The chance of 4 townies voting town both days 11.9%.
I think this is a significant enough difference to support my assertion.