Mafia Statistics (Super Cool Table)

[Callid]

To your first equation:
Yeah it's basically the same as mine, but I just don't see much of a need for the +/-/sqrt on the # games played. The sqrt I guess, to make # of games not matter as much after a point, but I don't really see the point of the +3-3. It's not like anyone's going to complain about data being skewed at the bottom of the table.

To your second equation:
Yeah admittedly, the chance to win as a townie vs the chance to win as a BO is different, however this is largely inconsequential imo. Not to mention, this equation fails to take into account (as it's really impossible for it to) the effectiveness of town makeups when people win, or whatever variation on the ruleset the GM decides to play. Maybe all the times I play town, the BO is loaded with worthless roles and the town is loaded with good ones, while when you play the BO has gin/vodka/vermouth/tequila/chianti, and the town just has detective boys, kazuha, eri, and a bunch of other "less useful" roles? You would certainly have to play as a "better townie" than I would, but this owuldn't be taken into account. Certainly it's an impressive formula, but I just think it's overly complicated for what small benefit (if any) it would give.

First Equation:
The point is simply that the players would NOT end up at the end of the list, but somewhere around rank 5, because one win out of one game is 100%. However, now they get a error, and finish the list.

Second Equation:
You're right, and my equation doesn't make sense if you've only played 5 or 6 games. However, after 20-30 games such ruleset varieties will fade (not to mention that things might - might! - get more stable). And my system adjusts itself to the overall likeliness, so that if we'd get a stable system now which would grant the BO much easier wins, my system would weight a BO-win less. Also, how complicated it is doesn't really matter to Excel, so...
Not to mention that it can be simplified via algebra a lot, but I'll explain that tomorrow.

[Akonyl]

I fail to see how someone with 1 game played, 1 game won (which would come out to a score of 1, without using the -3+3) would be around rank 5.

Also, after 20 or 30 games, people will begin to have the same amount of times as BO/town so I'm not exactly sure how much it would matter. Not to mention, some people may just be better at playing one side than the other.

[blurfbreg]

((((W_T/P_T)*(P_T/P_∑)/(G_T/G_∑)) + ((W_B/P_B)*(P_B/P_∑)/(G_B/G_∑)) + ((W_L/P_L)*(P_L/P_∑)/(G_L/G_∑)))/3)² * (((P_∑+P_GM)-3)^0.5+3)

May I suggest the second factor of the weighted winning to be: (log₂(P_∑+P_GM)+1)
This equation gets rid of the problem with people who played less than three games. You can always change the base of the logarithmic part if necessary.

Your equation works regardless of how many times a person played as a townie, BO or lover.

I'll look at it in detail with some numbers later. I'm getting tired.

[Callid]

I fail to see how someone with 1 game played, 1 game won (which would come out to a score of 1, without using the -3+3) would be around rank 5.

Also, after 20 or 30 games, people will begin to have the same amount of times as BO/town so I'm not exactly sure how much it would matter. Not to mention, some people may just be better at playing one side than the other.

Considering 5:
He played as a lover and won the only game the lovers won out of 15. So the formula is 0 for both the BO and town part, as he has never won as BO or town (and therefore W_T=W^B=0).
So we get:
(0+0+((1/1)*(1/1)/(1/15)))/3 = (1*1*(15/1))/3 = 15/3 = 5 = 500%.

Also, the players will not get the same amount of town and BO plays, cause there are always less people in the BO than in the town, so P_T>P_B>P_L.

((((W_T/P_T)*(P_T/P_∑)/(G_T/G_∑)) + ((W_B/P_B)*(P_B/P_∑)/(G_B/G_∑)) + ((W_L/P_L)*(P_L/P_∑)/(G_L/G_∑)))/3)² * (((P_∑+P_GM)-3)^0.5+3)

May I suggest the second factor of the weighted winning to be: (log₂(P_∑+P_GM)+1)
This equation gets rid of the problem with people who played less than three games. You can always change the base of the logarithmic part if necessary.

Your equation works regardless of how many times a person played as a townie, BO or lover.

I'll look at it in detail with some numbers later. I'm getting tired.

That wouldn't work. Your formula (with +2 instead of +1) is optimal for more than 3 games, it's values for less than 4 are way off (they have to be so low because of what I explained to Akonyl above):


Number of games   Optimal Value   My Value   Your Value   Your Value +1
1<0.1-12
2<0.1-23
30.1<X<332.583.58
43<X<4434
5~4.334.413.324.32
6~4.64.733.584.58
7~4.8353.814.81
8~55.2445
9~5.155.444.175.17
...............

Not to mention that it can be simplified via algebra a lot, but I'll explain that tomorrow.

((((W_T/P_T)*(P_T/P_∑)/(G_T/G_∑)) + ((W_B/P_B)*(P_B/P_∑)/(G_B/G_∑)) + ((W_L/P_L)*(P_L/P_∑)/(G_L/G_∑)))/3)² * (((P_∑+P_GM)-3)^0.5+3) =
((((W_T/P_∑)/(G_T/G_∑)) + ((W_B/P_∑)/(G_B/G_∑)) + ((W_L/P_∑)/(G_L/G_∑)))/3)² * (((P_∑+P_GM)-3)^0.5+3) =
(((W_T*G_∑/P_∑*G_T) + (W_B*G_∑/P_∑*G_B) + (W_L*G_∑/P_∑*G_L))/3)² * (((P_∑+P_GM)-3)^0.5+3) =
((W_T/G_T + W_B/G_B + W_L/G_L)*G_∑/(3*P_∑))² * (((P_∑+P_GM)-3)^0.5+3)

[Akonyl]

I fail to see how someone with 1 game played, 1 game won (which would come out to a score of 1, without using the -3+3) would be around rank 5.

Also, after 20 or 30 games, people will begin to have the same amount of times as BO/town so I'm not exactly sure how much it would matter. Not to mention, some people may just be better at playing one side than the other.

Considering 5:
He played as a lover and won the only game the lovers won out of 15. So the formula is 0 for both the BO and town part, as he has never won as BO or town (and therefore W_T=W^B=0).
So we get:
(0+0+((1/1)*(1/1)/(1/15)))/3 = (1*1*(15/1))/3 = 15/3 = 5 = 500%.

Also, the players will not get the same amount of town and BO plays, cause there are always less people in the BO than in the town, so P_T>P_B>P_L.

that's not even the formula I was referring to >_>

The first comment was made about your formula including the -3+3, not your huge one. The +3-3 formula you proposed doesn't have anything to do with special roles.

[blurfbreg]

May I suggest the second factor of the weighted winning to be: (log₂(P_∑+P_GM)+1)
This equation gets rid of the problem with people who played less than three games. You can always change the base of the logarithmic part if necessary.

That wouldn't work. Your formula (with +2 instead of +1) is optimal for more than 3 games, it's values for less than 4 are way off (they have to be so low because of what I explained to Akonyl above):


Number of games   Optimal Value   My Value   Your Value   Your Value +1
1<0.1-12
2<0.1-23
30.1<X<332.583.58
43<X<4434
5~4.334.413.324.32
6~4.64.733.584.58
7~4.8353.814.81
8~55.2445
9~5.155.444.175.17
...............

1) According to the table, the logarithmic function fits better than yours (just didn't know it should be +1 or +2 for the second factor). In a way, I kind of agree with the people below 3 having less experience (since it won't be accurate to see people's abilities below 3 games anyways). If the equation is that much of a problem below 3, then make a second equation (using exponential function for the part we're talking about) for those who have played 3 games or less.
This isn't impossible for the formulas on Excel if you use the logic functions.

2) How in the world did you come up with the optimal values for each game? I just thought it would be a nice weighting index (judging by the rate of increase) depending on the function we're figuring out to use. That's why I chose log₂(x). It's not like log_1.75(x) (or its other variations) would be bad if you don't want to count less than 3 games.

3) If you want to know how well people really play, you'd have to play with them to know first-hand.

[Callid]

1) According to the table, the logarithmic function fits better than yours (just didn't know it should be +1 or +2 for the second factor). In a way, I kind of agree with the people below 3 having less experience (since it won't be accurate to see people's abilities below 3 games anyways). If the equation is that much of a problem below 3, then make a second equation (using exponential function for the part we're talking about) for those who have played 3 games or less.
This isn't impossible for the formulas on Excel if you use the logic functions.

2) How in the world did you come up with the optimal values for each game? I just thought it would be a nice weighting index (judging by the rate of increase) depending on the function we're figuring out to use. That's why I chose log₂(x). It's not like log_1.75(x) (or its other variations) would be bad if you don't want to count less than 3 games.

3) If you want to know how well people really play, you'd have to play with them to know first-hand.

1) ... or we could simply say you have to played four games before you an be rated Actually, most rating systems work this way, you also won't get an ELO number with just one game.

2) Well, I started at 4 (for 4 games), and then I went on with +0.33; +0.28; +0.23; +0.18; +0.13 etc. and I "adjusted" the numbers a little bit to more comfortable values, like 4.6; 5; 5.15. It should simply show that the difference between one game more or less is decreasing over time. See also my reply for Akonyl below.

3) Of course, that's always the best way. 

that's not even the formula I was referring to >_>

The first comment was made about your formula including the -3+3, not your huge one. The +3-3 formula you proposed doesn't have anything to do with special roles.

Oh, I see.
Well, that was basically because it down-rates the number of games a little, cause the change of the result of the root is greatly changed by one game, but if 3 is added to this change, it becomes much less important, because it's not like 10% anymore, but only around 3%. This was necessary because, without +/-3, a player who has won 8 of 10 games would rank higher than a player who won all 6 of 6 games. And if you take GMing into account, even more twisted scenarios are possible.

[CTU]

When will the new round info be added?

[Conia]

When will the new round info be added?

Since Sebolains has took a trip to Europe, we will have to wait till he comes back.