Kleene Onigiri wrote:Isn't is simply 4 out of 7 people?
List is 6 long and 2 people are fixed (detective+informant), so 4 variable slots. And at that time it was 7 people that could put into those slots (not counting yourself, since I think you can't appear on your own list, and not counting Jecka as the informant).
I might have used the wrong formula tho XD
If you eliminate the detective, the informant and the owner of the list, there are 10 players left who can fill those 4 slots.
The total possible number of lists at this point is a 4-combination of 10, which is 210.
We need to calculate the probability that at least to civilians have the same list (two out of 10 civilians).
This is where it gets complicated (and I'm not 100% sure of my logic), but a *aproximatte* formula is (1 + 2 + ... + 9) / 210.
Let's use this notation for event = a generation of a list in the context described above.
If we have only one event, we'll generate one list => the condition isn't met (we would need at least two lists).
For two events: we shall consider the first civilian will receive any list, and the condition is met if the second civilian receives the same list, so the probability in this case is 1/210.
For three events: the above case plus the probability that the third civilian receives same list as either the second or the first civilian, therefore 1/210 + 2/210.
For ten events: 1/210 + 2/210 + .. + 9/210 = (1 + 2 + ... + 9) / 210.
There's a small flaw in this logic though. If you take the three-events case, the probability for the third civilian to have either the second or the first civilian list isn't 2/210 <- it's a little smaller than that (if 2nd and 1st civilian have the same list already).
The easy solution I found was to use a binomial probability calculator (http://stattrek.com/online-calculator/binomial.aspx). The probability computed in both ways was pretty close (~0.1% difference).
Sorry if I don't make any sense.
*will stop saying stuff that might be relevant to the game now*