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Let's start with an example. In Harry Potter, Hogwarts has 4 houses with students sorted into each house. The same happens on my website and I don't know how many users are in each house. It could be 20 in one house 50 in another and 100 in the third and fourth.
Now, each student can earn points on the website and at the end of the year, the house with the most points will win.
But it's not fair to "only" do a sum of the points, as the house with a 100 students will have a much higher chance to win, as they have more users to earn points. So I need to come up with an algorithm which is fair.
You can see an example here: https://worldofpotter.dk/points
What I do now is to sum all the points for a house, and then divide it by the number of users who have earned more than 10 points. This is still not fair, though.
Any ideas on how to make this calculation more fair?
Things we need to take into account:
* The percent of users earning points in each house
* Few users earning LOTS of points
* Many users earning FEW points (It's not bad earning few points. It still counts towards the total points of the house)
Link to MySQL dump(with users, houses and points): https://worldofpotter.dk/wop_points_example.sql
Link to CSV of points only: https://worldofpotter.dk/points.csv
I'd use something like Discounted Cumulative Gain which is used for measuring the effectiveness of search engines.
The concept is as it follows:
FUNCTION evalHouseScore (0_INDEXED_SORTED_ARRAY scores):
score = 0;
FOR (int i = 0; i < scores.length; i++):
score += scores[i]/log2(i);
END_FOR
RETURN score;
END_FUNCTION;
This must be somehow modified as this way of measuring focuses on the first result. As this is subjective you should decide on your the way you would modify it. Below I'll post the code which some constants which you should try with different values:
FUNCTION evalHouseScore (0_INDEXED_SORTED_ARRAY scores):
score = 0;
FOR (int i = 0; i < scores.length; i++):
score += scores[i]/log2(i+K);
END_FOR
RETURN L*score;
END_FUNCTION
Consider changing the logarithm.
Tests:
int[] g = new int[] {758,294,266,166,157,132,129,116,111,88,83,74,62,60,60,52,43,40,28,26,25,24,18,18,17,15,15,15,14,14,12,10,9,5,5,4,4,4,4,3,3,3,2,1,1,1,1,1};
int[] s = new int[] {612,324,301,273,201,182,176,139,130,121,119,114,113,113,106,86,77,76,65,62,60,58,57,54,54,42,42,40,36,35,34,29,28,23,22,19,17,16,14,14,13,11,11,9,9,8,8,7,7,7,6,4,4,3,3,3,3,2,2,2,2,2,2,2,1,1,1};
int[] h = new int[] {813,676,430,382,360,323,265,235,192,170,107,103,80,70,60,57,43,41,21,17,15,15,12,10,9,9,9,8,8,6,6,6,4,4,4,3,2,2,2,1,1,1};
int[] r = new int[] {1398,1009,443,339,242,215,210,205,177,168,164,144,144,92,85,82,71,61,58,47,44,33,21,19,18,17,12,11,11,9,8,7,7,6,5,4,3,3,3,3,2,2,2,1,1,1,1};
The output is for different offsets:
1182
1543
1847
2286
904
1231
1421
1735
813
1120
1272
1557
It sounds like some sort of constraint between the houses may need to be introduced. I might suggest finding the person that earned the most points out of all the houses and using it as the denominator when rolling up the scores. This will guarantee the max value of a user's contribution is 1, then all the scores for a house can be summed and then divided by the number of users to normalize the house's score. That should give you a reasonable comparison. It does introduce issues with low numbers of users in a house that are high achievers in which you may want to consider lower limits to the number of house members. Another technique may be to introduce handicap scores for users to balance the scales. The algorithm will most likely flex over time based on the data you receive. To keep it fair it will take some responsive action after the initial iteration. Players can come up with some creative ways to make scoring systems work for them. Here is some pseudo-code in PHP that you may use:
<?php
$mostPointsEarned; // Find the user that earned the most points
$houseScores = [];
foreach ($houses as $house) {
$numberOfUsers = 0;
$normalizedScores = [];
foreach ($house->getUsers() as $user) {
$normalizedScores[] = $user->getPoints() / $mostPointsEarned;
$numberOfUsers++;
}
$houseScores[] = array_sum($normalizedScores) / $numberOfUsers;
}
var_dump($houseScores);
You haven't given any examples on what should be preferred state, and what are situations against which you want to be immune. (3,2,1,1 compared to 5,2 etc.)
It's also a pity you haven't provided us the dataset in some nice way to play.
scala> val input = Map( // as seen on 2016-09-09 14:10 UTC on https://worldofpotter.dk/points
'G' -> Seq(758,294,266,166,157,132,129,116,111,88,83,74,62,60,60,52,43,40,28,26,25,24,18,18,17,15,15,15,14,14,12,10,9,5,5,4,4,4,4,3,3,3,2,1,1,1,1,1),
'S' -> Seq(612,324,301,273,201,182,176,139,130,121,119,114,113,113,106,86,77,76,65,62,60,58,57,54,54,42,42,40,36,35,34,29,28,23,22,19,17,16,14,14,13,11,11,9,9,8,8,7,7,7,6,4,4,3,3,3,3,2,2,2,2,2,2,2,1,1,1),
'H' -> Seq(813,676,430,382,360,323,265,235,192,170,107,103,80,70,60,57,43,41,21,17,15,15,12,10,9,9,9,8,8,6,6,6,4,4,4,3,2,2,2,1,1,1),
'R' -> Seq(1398,1009,443,339,242,215,210,205,177,168,164,144,144,92,85,82,71,61,58,47,44,33,21,19,18,17,12,11,11,9,8,7,7,6,5,4,3,3,3,3,2,2,2,1,1,1,1)
) // and the results on the website were: 1. R 1951, 2. H 1859, 3. S 990, 4. G 954
Here is what I thought of:
def singleValuedScore(individualScores: Seq[Int]) = individualScores
.sortBy(-_) // sort from most to least
.zipWithIndex // add indices e.g. (best, 0), (2nd best, 1), ...
.map { case (score, index) => score * (1 + index) } // here is the 'logic'
.max
input.mapValues(singleValuedScore)
res: scala.collection.immutable.Map[Char,Int] =
Map(G -> 1044,
S -> 1590,
H -> 1968,
R -> 2018)
The overall positions would be:
Ravenclaw with 2018 aggregated points
Hufflepuff with 1968
Slytherin with 1590
Gryffindor with 1044
Which corresponds to the ordering on that web: 1. R 1951, 2. H 1859, 3. S 990, 4. G 954.
The algorithms output is maximal product of score of user and rank of the user within a house.
This measure is not affected by "long-tail" of users having low score compared to the active ones.
There are no hand-set cutoffs or thresholds.
You could experiment with the rank attribution (score * index or score * Math.sqrt(index) or score / Math.log(index + 1) ...)
I take it that the fair measure is the number of points divided by the number of house members. Since you have the number of points, the exercise boils down to estimate the number of members.
We are in short supply of data here as the only hint we have on member counts is the answers on the website. This makes us vulnerable to manipulation, members can trick us into underestimating their numbers. If the suggested estimation method to "count respondents with points >10" would be known, houses would only encourage the best to do the test to hide members from our count. This is a real problem and the only thing I will do about it is to present a "manipulation indicator".
How could we then estimate member counts? Since we do not know anything other than test results, we have to infer the propensity to do the test from the actual results. And we have little other to assume than that we would have a symmetric result distribution (of the logarithm of the points) if all members tested. Now let's say the strong would-be respondents are more likely to actually test than weak would-be respondents. Then we could measure the extra dropout ratio for the weak by comparing the numbers of respondents in corresponding weak and strong test-point quantiles.
To be specific, of the 205 answers, there are 27 in the worst half of the overall weakest quartile, while 32 in the strongest half of the best quartile. So an extra 5 respondents of the very weakest have dropped out from an assumed all-testing symmetric population, and to adjust for this, we are going to estimate member count from this quantile by multiplying the number of responses in it by 32/27=about 1.2. Similarly, we have 29/26 for the next less-extreme half quartiles and 41/50 for the two mid quartiles.
So we would estimate members by simply counting the number of respondents but multiplying the number of respondents in the weak quartiles mentioned above by 1.2, 1.1 and 0.8 respectively. If however any result distribution within a house would be conspicuously skewed, which is not the case now, we would have to suspect manipulation and re-design our member count.
For the sample at hand however, these adjustments to member counts are minor, and yields the same house ranks as from just counting the respondents without adjustments.
I got myself to amuse me a little bit with your question and some python programming with some random generated data. As some people mentioned in the comments you need to define what is fairness. If as you said you don't know the number of people in each of the houses, you can use the number of participations of each house, thus you motivate participation (it can be unfair depending on the number of people of each house, but as you said you don't have this data on the first place).
The important part of the code is the following.
import numpy as np
from numpy.random import randint # import random int
# initialize random seed
np.random.seed(4)
houses = ["Gryffindor","Slytherin", "Hufflepuff", "Ravenclaw"]
houses_points = []
# generate random data for each house
for _ in houses:
# houses_points.append(randint(0, 100, randint(60,100)))
houses_points.append(randint(0, 50, randint(2,10)))
# count participation
houses_participations = []
houses_total_points = []
for house_id in xrange(len(houses)):
houses_total_points.append(np.sum(houses_points[house_id]))
houses_participations.append(len(houses_points[house_id]))
# sum the total number of participations
total_participations = np.sum(houses_participations)
# proposed model with weighted total participation points
houses_partic_points = []
for house_id in xrange(len(houses)):
tmp = houses_total_points[house_id]*houses_participations[house_id]/total_participations
houses_partic_points.append(tmp)
The results of this method are the following:
House Points per Participant
Gryffindor: [46 5 1 40]
Slytherin: [ 8 9 39 45 30 40 36 44 38]
Hufflepuff: [42 3 0 21 21 9 38 38]
Ravenclaw: [ 2 46]
House Number of Participations per House
Gryffindor: 4
Slytherin: 9
Hufflepuff: 8
Ravenclaw: 2
House Total Points
Gryffindor: 92
Slytherin: 289
Hufflepuff: 172
Ravenclaw: 48
House Points weighted by a participation factor
Gryffindor: 16
Slytherin: 113
Hufflepuff: 59
Ravenclaw: 4
You'll find the complete file with printing results here (https://gist.github.com/silgon/5be78b1ea0b55a20d90d9ec3e7c515e5).
You should enter some more rules to define the fairness.
Idea 1
You could set up the rule that anyone has to earn at least 10 points to enter the competition.
Then you can calculate the average points for each house.
Positive: Everyone needs to show some motivation.
Idea 2
Another approach would be to set the rule that from each house only the 10 best students will count for the competition.
Positive: Easy rule to calculate the points.
Negative: Students might become uninterested if they see they can't reach the top 10 places of their house.
From my point of view, your problem is diveded in a few points:
The best thing to do would be to re - assignate the player in the different Houses so that each House has the same number of players. (as explain by #navid-vafaei)
If you don't want to do that because you believe that it may affect your game popularity with player whom are in House that they don't want because you can change the choice of the Sorting Hat at least in the movie or books.
In that case, you can sum the point of the student's house and divide by the number of students. You may just remove the number of student with a very low score. You may remove as well the student with a very low activity because students whom skip school might be fired.
The most important part for me n your algorithm is weather or not you give points for all valuables things:
In the Harry Potter's story, the students earn point on the differents subjects they chose at school and get point according to their score.
At the end of the year, there is a special award event. At that moment, the Director gave points for valuable things which cannot be evaluated in the subject at school suche as the qualites (bravery for example).
I'm trying to sort a bunch of products by customer ratings using a 5 star system. The site I'm setting this up for does not have a lot of ratings and continue to add new products so it will usually have a few products with a low number of ratings.
I tried using average star rating but that algorithm fails when there is a small number of ratings.
Example a product that has 3x 5 star ratings would show up better than a product that has 100x 5 star ratings and 2x 2 star ratings.
Shouldn't the second product show up higher because it is statistically more trustworthy because of the larger number of ratings?
Prior to 2015, the Internet Movie Database (IMDb) publicly listed the formula used to rank their Top 250 movies list. To quote:
The formula for calculating the Top Rated 250 Titles gives a true Bayesian estimate:
weighted rating (WR) = (v ÷ (v+m)) × R + (m ÷ (v+m)) × C
where:
R = average for the movie (mean)
v = number of votes for the movie
m = minimum votes required to be listed in the Top 250 (currently 25000)
C = the mean vote across the whole report (currently 7.0)
For the Top 250, only votes from regular voters are considered.
It's not so hard to understand. The formula is:
rating = (v / (v + m)) * R +
(m / (v + m)) * C;
Which can be mathematically simplified to:
rating = (R * v + C * m) / (v + m);
The variables are:
R – The item's own rating. R is the average of the item's votes. (For example, if an item has no votes, its R is 0. If someone gives it 5 stars, R becomes 5. If someone else gives it 1 star, R becomes 3, the average of [1, 5]. And so on.)
C – The average item's rating. Find the R of every single item in the database, including the current one, and take the average of them; that is C. (Suppose there are 4 items in the database, and their ratings are [2, 3, 5, 5]. C is 3.75, the average of those numbers.)
v – The number of votes for an item. (To given another example, if 5 people have cast votes on an item, v is 5.)
m – The tuneable parameter. The amount of "smoothing" applied to the rating is based on the number of votes (v) in relation to m. Adjust m until the results satisfy you. And don't misinterpret IMDb's description of m as "minimum votes required to be listed" – this system is perfectly capable of ranking items with less votes than m.
All the formula does is: add m imaginary votes, each with a value of C, before calculating the average. In the beginning, when there isn't enough data (i.e. the number of votes is dramatically less than m), this causes the blanks to be filled in with average data. However, as votes accumulates, eventually the imaginary votes will be drowned out by real ones.
In this system, votes don't cause the rating to fluctuate wildly. Instead, they merely perturb it a bit in some direction.
When there are zero votes, only imaginary votes exist, and all of them are C. Thus, each item begins with a rating of C.
See also:
A demo. Click "Solve".
Another explanation of IMDb's system.
An explanation of a similar Bayesian star-rating system.
Evan Miller shows a Bayesian approach to ranking 5-star ratings:
where
nk is the number of k-star ratings,
sk is the "worth" (in points) of k stars,
N is the total number of votes
K is the maximum number of stars (e.g. K=5, in a 5-star rating system)
z_alpha/2 is the 1 - alpha/2 quantile of a normal distribution. If you want 95% confidence (based on the Bayesian posterior distribution) that the actual sort criterion is at least as big as the computed sort criterion, choose z_alpha/2 = 1.65.
In Python, the sorting criterion can be calculated with
def starsort(ns):
"""
http://www.evanmiller.org/ranking-items-with-star-ratings.html
"""
N = sum(ns)
K = len(ns)
s = list(range(K,0,-1))
s2 = [sk**2 for sk in s]
z = 1.65
def f(s, ns):
N = sum(ns)
K = len(ns)
return sum(sk*(nk+1) for sk, nk in zip(s,ns)) / (N+K)
fsns = f(s, ns)
return fsns - z*math.sqrt((f(s2, ns)- fsns**2)/(N+K+1))
For example, if an item has 60 five-stars, 80 four-stars, 75 three-stars, 20 two-stars and 25 one-stars, then its overall star rating would be about 3.4:
x = (60, 80, 75, 20, 25)
starsort(x)
# 3.3686975120774694
and you can sort a list of 5-star ratings with
sorted([(60, 80, 75, 20, 25), (10,0,0,0,0), (5,0,0,0,0)], key=starsort, reverse=True)
# [(10, 0, 0, 0, 0), (60, 80, 75, 20, 25), (5, 0, 0, 0, 0)]
This shows the effect that more ratings can have upon the overall star value.
You'll find that this formula tends to give an overall rating which is a bit
lower than the overall rating reported by sites such as Amazon, Ebay or Wal-mart
particularly when there are few votes (say, less than 300). This reflects the
higher uncertainy that comes with fewer votes. As the number of votes increases
(into the thousands) all overall these rating formulas should tend to the
(weighted) average rating.
Since the formula only depends on the frequency distribution of 5-star ratings
for the item itself, it is easy to combine reviews from multiple sources (or,
update the overall rating in light of new votes) by simply adding the frequency
distributions together.
Unlike the IMDb formula, this formula does not depend on the average score
across all items, nor an artificial minimum number of votes cutoff value.
Moreover, this formula makes use of the full frequency distribution -- not just
the average number of stars and the number of votes. And it makes sense that it
should since an item with ten 5-stars and ten 1-stars should be treated as
having more uncertainty than (and therefore not rated as highly as) an item with
twenty 3-star ratings:
In [78]: starsort((10,0,0,0,10))
Out[78]: 2.386028063783418
In [79]: starsort((0,0,20,0,0))
Out[79]: 2.795342687927806
The IMDb formula does not take this into account.
See this page for a good analysis of star-based rating systems, and this one for a good analysis of upvote-/downvote- based systems.
For up and down voting you want to estimate the probability that, given the ratings you have, the "real" score (if you had infinite ratings) is greater than some quantity (like, say, the similar number for some other item you're sorting against).
See the second article for the answer, but the conclusion is you want to use the Wilson confidence. The article gives the equation and sample Ruby code (easily translated to another language).
Well, depending on how complex you want to make it, you could have ratings additionally be weighted based on how many ratings the person has made, and what those ratings are. If the person has only made one rating, it could be a shill rating, and might count for less. Or if the person has rated many things in category a, but few in category b, and has an average rating of 1.3 out of 5 stars, it sounds like category a may be artificially weighed down by the low average score of this user, and should be adjusted.
But enough of making it complex. Let’s make it simple.
Assuming we’re working with just two values, ReviewCount and AverageRating, for a particular item, it would make sense to me to look ReviewCount as essentially being the “reliability” value. But we don’t just want to bring scores down for low ReviewCount items: a single one-star rating is probably as unreliable as a single 5 star rating. So what we want to do is probably average towards the middle: 3.
So, basically, I’m thinking of an equation something like X * AverageRating + Y * 3 = the-rating-we-want. In order to make this value come out right we need X+Y to equal 1. Also we need X to increase in value as ReviewCount increases...with a review count of 0, x should be 0 (giving us an equation of “3”), and with an infinite review count X should be 1 (which makes the equation = AverageRating).
So what are X and Y equations? For the X equation want the dependent variable to asymptotically approach 1 as the independent variable approaches infinity. A good set of equations is something like:
Y = 1/(factor^RatingCount)
and (utilizing the fact that X must be equal to 1-Y)
X = 1 – (1/(factor^RatingCount)
Then we can adjust "factor" to fit the range that we're looking for.
I used this simple C# program to try a few factors:
// We can adjust this factor to adjust our curve.
double factor = 1.5;
// Here's some sample data
double RatingAverage1 = 5;
double RatingCount1 = 1;
double RatingAverage2 = 4.5;
double RatingCount2 = 5;
double RatingAverage3 = 3.5;
double RatingCount3 = 50000; // 50000 is not infinite, but it's probably plenty to closely simulate it.
// Do the calculations
double modfactor = Math.Pow(factor, RatingCount1);
double modRating1 = (3 / modfactor)
+ (RatingAverage1 * (1 - 1 / modfactor));
double modfactor2 = Math.Pow(factor, RatingCount2);
double modRating2 = (3 / modfactor2)
+ (RatingAverage2 * (1 - 1 / modfactor2));
double modfactor3 = Math.Pow(factor, RatingCount3);
double modRating3 = (3 / modfactor3)
+ (RatingAverage3 * (1 - 1 / modfactor3));
Console.WriteLine(String.Format("RatingAverage: {0}, RatingCount: {1}, Adjusted Rating: {2:0.00}",
RatingAverage1, RatingCount1, modRating1));
Console.WriteLine(String.Format("RatingAverage: {0}, RatingCount: {1}, Adjusted Rating: {2:0.00}",
RatingAverage2, RatingCount2, modRating2));
Console.WriteLine(String.Format("RatingAverage: {0}, RatingCount: {1}, Adjusted Rating: {2:0.00}",
RatingAverage3, RatingCount3, modRating3));
// Hold up for the user to read the data.
Console.ReadLine();
So you don’t bother copying it in, it gives this output:
RatingAverage: 5, RatingCount: 1, Adjusted Rating: 3.67
RatingAverage: 4.5, RatingCount: 5, Adjusted Rating: 4.30
RatingAverage: 3.5, RatingCount: 50000, Adjusted Rating: 3.50
Something like that? You could obviously adjust the "factor" value as needed to get the kind of weighting you want.
You could sort by median instead of arithmetic mean. In this case both examples have a median of 5, so both would have the same weight in a sorting algorithm.
You could use a mode to the same effect, but median is probably a better idea.
If you want to assign additional weight to the product with 100 5-star ratings, you'll probably want to go with some kind of weighted mode, assigning more weight to ratings with the same median, but with more overall votes.
If you just need a fast and cheap solution that will mostly work without using a lot of computation here's one option (assuming a 1-5 rating scale)
SELECT Products.id, Products.title, avg(Ratings.score), etc
FROM
Products INNER JOIN Ratings ON Products.id=Ratings.product_id
GROUP BY
Products.id, Products.title
ORDER BY (SUM(Ratings.score)+25.0)/(COUNT(Ratings.id)+20.0) DESC, COUNT(Ratings.id) DESC
By adding in 25 and dividing by the total ratings + 20 you're basically adding 10 worst scores and 10 best scores to the total ratings and then sorting accordingly.
This does have known issues. For example, it unfairly rewards low-scoring products with few ratings (as this graph demonstrates, products with an average score of 1 and just one rating score a 1.2 while products with an average score of 1 and 1k+ ratings score closer to 1.05). You could also argue it unfairly punishes high-quality products with few ratings.
This chart shows what happens for all 5 ratings over 1-1000 ratings:
http://www.wolframalpha.com/input/?i=Plot3D%5B%2825%2Bxy%29/%2820%2Bx%29%2C%7Bx%2C1%2C1000%7D%2C%7By%2C0%2C6%7D%5D
You can see the dip upwards at the very bottom ratings, but overall it's a fair ranking, I think. You can also look at it this way:
http://www.wolframalpha.com/input/?i=Plot3D%5B6-%28%2825%2Bxy%29/%2820%2Bx%29%29%2C%7Bx%2C1%2C1000%7D%2C%7By%2C0%2C6%7D%5D
If you drop a marble on most places in this graph, it will automatically roll towards products with both higher scores and higher ratings.
Obviously, the low number of ratings puts this problem at a statistical handicap. Never the less...
A key element to improving the quality of an aggregate rating is to "rate the rater", i.e. to keep tabs of the ratings each particular "rater" has supplied (relative to others). This allows weighing their votes during the aggregation process.
Another solution, more of a cope out, is to supply the end-users with a count (or a range indication thereof) of votes for the underlying item.
One option is something like Microsoft's TrueSkill system, where the score is given by mean - 3*stddev, where the constants can be tweaked.
After look for a while, I choose the Bayesian system.
If someone is using Ruby, here a gem for it:
https://github.com/wbotelhos/rating
I'd highly recommend the book Programming Collective Intelligence by Toby Segaran (OReilly) ISBN 978-0-596-52932-1 which discusses how to extract meaningful data from crowd behaviour. The examples are in Python, but its easy enough to convert.