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Nightless Expectation Rule: Difference between revisions
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==Proof== | ==Proof== | ||
This proof uses mathematical induction. | This proof uses mathematical induction. | ||
*Base case: In a 2 - 1 [[ | *Base case: In a 2 - 1 [[ELo]], Town will [[eliminate]] correctly 1/3 of the time at random. This is 1 - 2/3. | ||
*Induction Step: Given that P(Town win with ''S'' scum and ''N'' players) = (''N-2S'')/''N'', consider P(Town win with ''S'' scum and ''N+1'' players). | *Induction Step: Given that P(Town win with ''S'' scum and ''N'' players) = (''N-2S'')/''N'', consider P(Town win with ''S'' scum and ''N+1'' players). | ||
**P(Town win,''S'',''N+1'') = (''S''/(''N+1''))*P(Town win,''S-1,N'') + ((''N+1-S'')/(''N+1''))*P(Town win,''S'',''N) | **P(Town win,''S'',''N+1'') = (''S''/(''N+1''))*P(Town win,''S-1,N'') + ((''N+1-S'')/(''N+1''))*P(Town win,''S'',''N) |
Latest revision as of 07:00, 13 July 2020
The Nightless Expectation Rule is a mathematical rule stemming from The EV Project.
In Nightless games with no power roles, S scum alive, and N total players alive,
P(Town win) = 1 - 2S/N = (N-2S)/N
An obvious corollary would be:
P(Mafia win) = 2S/N
Proof
This proof uses mathematical induction.
- Base case: In a 2 - 1 ELo, Town will eliminate correctly 1/3 of the time at random. This is 1 - 2/3.
- Induction Step: Given that P(Town win with S scum and N players) = (N-2S)/N, consider P(Town win with S scum and N+1 players).
- P(Town win,S,N+1) = (S/(N+1))*P(Town win,S-1,N) + ((N+1-S)/(N+1))*P(Town win,S,N)
- =(S/(N+1))*(N-2(S-1))/N + ((N+1-S)/(N+1))*(N-2S)/N)
- =((SN-2S2+2S)+(N2-2SN+N-2S-SN+2S2))/(N*(N+1))
- =(N2+N-2SN)/(N*(N+1))
- =(N+1-2S)/(N+1)
...which is precisely the original formula, with (N+1) swapped in for N. Therefore the formula holds.
Balance
Letting P(Town win) = 1/2, this means:
- 1/2 = (N-2S)/N
- N/2 = N-2S
- N = 2N-4S
- 4S = N
Nightless games have a 50% win rate for either faction if scum make up a quarter of the initial population.