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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).
Im trying to determine the "difficultly" of a quiz object.
My ultimate goal is to be able to create a "difficulty score" (DS) for any quiz. This would allow me to compare one quiz to another accurately, despite being made up of different questions/answers.
When creating my quiz object, I assign each question a "difficulty index" (DI), which is number on a scale from 1-15.
15 = most difficult
1 = least difficult
Now a strait forward way to measure this "difficulty score" could be to add up each question's "difficulty index" then divide by maximum possible "difficulty index" for the quiz. ( ex. 16/30 = 53.3% Difficulty )
However, I also have multiple "weighting" properties associated to each question. These weights are again one a scale of 1-5.
5 = most impact
1 = least impact
The reason I have (2) instead of the more common (1) is so I can accommodate a scenario as follows...
If presenting the student with a very difficult question (DI=15) and the student answers "incorrect", don't have it hurt their score so much BUT if they get it "correct" have it improve their score greatly. I call these my "positive" (PW) and "negative" (NW) weights.
Quiz Example A:
Question 1: DI = 1 | PW = 3 | NW = 3
Question 2: DI = 1 | PW = 3 | NW = 3
Question 3: DI = 1 | PW = 3 | NW = 3
Question 4: DI = 15 | PW = 5 | NW = 1
Quiz Example B:
Question 1: DI = 1 | PW = 3 | NW = 3
Question 2: DI = 1 | PW = 3 | NW = 3
Question 3: DI = 1 | PW = 3 | NW = 3
Question 4: DI = 15 | PW = 1 | NW = 5
Technically the above two quizzes are very similar BUT Quiz B should be more "difficult" because the hardest question will have the greatest impact on your score if you get it wrong.
My question now becomes how can I accurately determine the "difficulty score" when considering the complex weighting system?
Any help is greatly appreciated!
The challenge of course is to determine the difficulty score for each single question.
I suggest the following model:
Hardness (H): Define a hard question such that chances of answering it correctly are lower. The hardest question is such that (1) the chance of answering it correctly are equal to random choice (because it is inherently very hard), and (2) it has the largest number of possible answers. We'll define such question as (H = 15). On the other end of the scale, we'll define (H = 0) for a question where the chances of answering it correctly are 100% (because it is trivial) (I know - such question will never appear). Now - define the hardness of each question by subjective extrapolation (remember that one can always guess between the given options). For example, if a (H = 15) question has 4 answers, and another question with similar inherent hardness has 2 answers - it would be (H = 7.5). Another example: If you believe that an average student has 62.5% of answering a question correctly - it would also be a (H = 7.5) question (this is because a H = 15 has 25% of correct answer, while H = 0 has 100%. The average is 62.5%)
Effect (E): Now, we'll measure the effect of PW and NW. For questions with 50% chance of answering correctly - the effect is E = 0.5*PW - 0.5*NW. For questions with 25% chance of answering correctly - the effect is E = 0.25*PW - 0.75*NW. For trivial question NW doesn't matter so the effect is E = PW.
Difficulty (DI): The last step is to integrate the hardness and the effect - and call it difficulty. I suggest DI = H - c*E, where c is some positive constant. You may want to normalize again.
Edit: Alternatively, you may try the following formula: DI = H * (1 - c*E), where the effect magnitude is not absolute, but relative to the question's hardness.
Clarification:
The teacher needs to estimate only one parameter about each question: What is the probability that an average student would answer this question correctly. His estimation, e, will be in the range [1/k, 1], where k is the number of answers.
The hardness, H, is a linear function of e such that 1/k is mapped to 15 and 1 is mapped to 0. The function is: H = 15 * k / (k-1) * (1-e)
The effect E depends on e, PW and NW. The formula is E = e*PW - (1-e)*NW
Example based on OP comments:
Question 1:
k = 4, e = 0.25 (hardest). Therefore H = 15
PW = 1, NW = 5, e = 0.25. Therefore E = 0.25*1 - 0.75*5 = -3.5
c = 5. DI = 15 - 5*(-3.5) = 32.5
Question 2:
k = 4, e = 0.95 (very easy). Therefore H = 1
PW = 1, NW = 5, e = 0.95. Therefore E = 0.95*1 - 0.05*5 = 0.7
c = 5. DI = 1 - 5*(0.7) = -2.5
I'd say the core of the problem is that mathematically your example quizzes A and B are identical, except that quiz A awards the student 4 gratuitous bonus points (or, equivalently, quiz B arbitrarily takes 4 points away from them). If the same students take both of them, the score distribution will be the same, except shifted by 4 points. So while the two quizzes may feel different psychologically (because, let's face it, getting extra points feels good, and losing points feels bad, even if you technically did nothing to deserve it), finding an objective way to distinguish them seems tricky.
That said, one reasonable measure of "psychological difficulty" could simply be the average score (per question) that a randomly chosen student would be expected to get from the quiz. Of course, that's not something you can reliably calculate in advance, although you could estimate it from actual quiz results after the fact.
However, if you could somehow relate your (presumably arbitrary) difficulty ratings to the fraction of students likely to answer the question correctly, then you could use that to estimate the expected average score. So, for example, we might simply assume a linear relationship with the question difficulty as the success rate, with difficulty 1 corresponding to a 100% expected success rate, and difficulty 15 corresponding to a 0% expected success rate. Then the expected average score S per question for the quiz could be calculated as:
S = avg(PW × X − NW × (1 − X))
where the average is taken over all questions in the quiz, and where PW and NW are the point weights for a correct and an incorrect answer respectively, DI below is the difficulty rating for the question, and X = (15 − DI) / 14 is the estimated success rate.
Of course, we might want to also account for the fact that, even if a student doesn't know the answer to a question, they can still guess. Basically this means that the estimated success rate X should not range from 0 to 1, but from 1/N to 1, where N is the number of options for the question. So, taking that into account, we can adjust the formula for X to be:
X = (1 + (N − 1) × (15 − DI) / 14) / N
One problem with this estimated average score S as a difficulty measure is that it isn't bounded in either direction, and provides no obvious scale to indicate what counts as an "easy" quiz or a "hard" one. The fundamental problem here is that you haven't specified any limits for the question weights, so there's technically nothing to stop someone from making a question with, say, a positive or negative weight of one million points.
That said, if you do impose some reasonable limits on the weights (even if they're only recommendations), then you should be able to also establish reasonable thresholds on S for a quiz to be considered e.g. easy, moderate or hard. And even if you don't, you can still at least use it to rank quizzes relative to each other by difficulty.
Ps. One way to present the expected score in a UI might be to multiply it by the number of questions in the quiz, and display the result as "par" for the quiz. That way, students could roughly judge their own performance against the difficulty of the quiz by seeing whether they scored above or below par.
I have an application (Node/Angular) that I'm creating where I'm trying to rank users based on overall performance across two metrics. There are two metrics used to track the users we are using are the following:
Units Produced (ranges between 0 - 6000)
Rate of production = [ Units Produced ] / [ Labor Hours ] (ranges between 0 - 100)
However, ranking users explicitly by either of these variables doesn't make sense, because it creates some strange incentives/behaviors.
For instance, it is possible to have a really high Rate of Production, but a super low number of total number of units produced by working really hard over a short period of time. Alternatively, you can have a very high number of Units Produced, but it may be due to the fact that they worked overtime, and thus were able to produce more units than anyone else just due to the fact that they had longer to work, and they could have a low Rate of Production.
Does anyone have experience designing these types of scoring systems? How have you handled it?
First, I would recommend to bring them on the same scale. E.g. divide Units produced by 60.
Then, if you are fine with equal weights, there are three common simple choices:
Add the scores
Multiply the scores (equal to adding logs of each)
Take the minimum of the two scores
Which of the ones is best, depends on to what extent you want it to be a measure of combined good results. In your case, I would recommend you to multiply and put a scale on the resulting product.
If you want to go a little more complex and weigh or play around with how much to reward separate vs joint scores, you can use the following formula:
V = alpha * log_b[Units Produced / 60] + (1-alpha) * log_b[Rate of Production],
where alpha determines the weighting of one vs the other and the base of the logarithmic function determines to what extent a joint success is rewarded.
I did something very similar I found it valuable to break them into leagues or tiers, for example using Units Produced as a base.
Novice = 100 Units Produced
Beginner = 500 Units Produced
Advanced = 2000 Units Produced
Expert = 4000 Units Produced
Putting this into a useable object
var levels = [
{id: 1, name: "Novice", minUnits: 100, maxUnits: 599 },
{id: 2, name: "Beginner", minUnits: 500, maxUnits: 1999 },
{id: 3, name: "Advanced", minUnits: 2000, maxUnits: 3999 },
{id: 4, name: "Expert", minUnits: 4000, maxUnits: 6000 }
]
You can then use your Rate of production to multiply by a weighted value inside the levels, you can determine what this is. You can play with the values to make it as hard or as easy as you want.
You can do a combination with
SCORE = 200/( K_1/x_1 + K_2/x_2 )
// x_1 : Score 1
// x_2 : Score 2
// K_1 : Maximum of Score 1
// K_2 : Maximum of Score 2
Of course be carefull when dividing by zero. If either x_1 or x_2 are zero then SCORE=0. If x_1=K_1 and x_2=K_2 then SCORE=100 (maximum)
Otherwise the score is somewhere in between. If x_1/K_1 = x_2/K_2 = z then SCORE = 100*z
This weighs the lower score more such that you get rewarded when raising one of the two scores (unlike a minimum of the two scenarios) but not as much as raising both.
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I am finding it hard to understand the process of Naive Bayes, and I was wondering if someone could explain it with a simple step by step process in English. I understand it takes comparisons by times occurred as a probability, but I have no idea how the training data is related to the actual dataset.
Please give me an explanation of what role the training set plays. I am giving a very simple example for fruits here, like banana for example
training set---
round-red
round-orange
oblong-yellow
round-red
dataset----
round-red
round-orange
round-red
round-orange
oblong-yellow
round-red
round-orange
oblong-yellow
oblong-yellow
round-red
The accepted answer has many elements of k-NN (k-nearest neighbors), a different algorithm.
Both k-NN and NaiveBayes are classification algorithms. Conceptually, k-NN uses the idea of "nearness" to classify new entities. In k-NN 'nearness' is modeled with ideas such as Euclidean Distance or Cosine Distance. By contrast, in NaiveBayes, the concept of 'probability' is used to classify new entities.
Since the question is about Naive Bayes, here's how I'd describe the ideas and steps to someone. I'll try to do it with as few equations and in plain English as much as possible.
First, Conditional Probability & Bayes' Rule
Before someone can understand and appreciate the nuances of Naive Bayes', they need to know a couple of related concepts first, namely, the idea of Conditional Probability, and Bayes' Rule. (If you are familiar with these concepts, skip to the section titled Getting to Naive Bayes')
Conditional Probability in plain English: What is the probability that something will happen, given that something else has already happened.
Let's say that there is some Outcome O. And some Evidence E. From the way these probabilities are defined: The Probability of having both the Outcome O and Evidence E is:
(Probability of O occurring) multiplied by the (Prob of E given that O happened)
One Example to understand Conditional Probability:
Let say we have a collection of US Senators. Senators could be Democrats or Republicans. They are also either male or female.
If we select one senator completely randomly, what is the probability that this person is a female Democrat? Conditional Probability can help us answer that.
Probability of (Democrat and Female Senator)= Prob(Senator is Democrat) multiplied by Conditional Probability of Being Female given that they are a Democrat.
P(Democrat & Female) = P(Democrat) * P(Female | Democrat)
We could compute the exact same thing, the reverse way:
P(Democrat & Female) = P(Female) * P(Democrat | Female)
Understanding Bayes Rule
Conceptually, this is a way to go from P(Evidence| Known Outcome) to P(Outcome|Known Evidence). Often, we know how frequently some particular evidence is observed, given a known outcome. We have to use this known fact to compute the reverse, to compute the chance of that outcome happening, given the evidence.
P(Outcome given that we know some Evidence) = P(Evidence given that we know the Outcome) times Prob(Outcome), scaled by the P(Evidence)
The classic example to understand Bayes' Rule:
Probability of Disease D given Test-positive =
P(Test is positive|Disease) * P(Disease)
_______________________________________________________________
(scaled by) P(Testing Positive, with or without the disease)
Now, all this was just preamble, to get to Naive Bayes.
Getting to Naive Bayes'
So far, we have talked only about one piece of evidence. In reality, we have to predict an outcome given multiple evidence. In that case, the math gets very complicated. To get around that complication, one approach is to 'uncouple' multiple pieces of evidence, and to treat each of piece of evidence as independent. This approach is why this is called naive Bayes.
P(Outcome|Multiple Evidence) =
P(Evidence1|Outcome) * P(Evidence2|outcome) * ... * P(EvidenceN|outcome) * P(Outcome)
scaled by P(Multiple Evidence)
Many people choose to remember this as:
P(Likelihood of Evidence) * Prior prob of outcome
P(outcome|evidence) = _________________________________________________
P(Evidence)
Notice a few things about this equation:
If the Prob(evidence|outcome) is 1, then we are just multiplying by 1.
If the Prob(some particular evidence|outcome) is 0, then the whole prob. becomes 0. If you see contradicting evidence, we can rule out that outcome.
Since we divide everything by P(Evidence), we can even get away without calculating it.
The intuition behind multiplying by the prior is so that we give high probability to more common outcomes, and low probabilities to unlikely outcomes. These are also called base rates and they are a way to scale our predicted probabilities.
How to Apply NaiveBayes to Predict an Outcome?
Just run the formula above for each possible outcome. Since we are trying to classify, each outcome is called a class and it has a class label. Our job is to look at the evidence, to consider how likely it is to be this class or that class, and assign a label to each entity.
Again, we take a very simple approach: The class that has the highest probability is declared the "winner" and that class label gets assigned to that combination of evidences.
Fruit Example
Let's try it out on an example to increase our understanding: The OP asked for a 'fruit' identification example.
Let's say that we have data on 1000 pieces of fruit. They happen to be Banana, Orange or some Other Fruit.
We know 3 characteristics about each fruit:
Whether it is Long
Whether it is Sweet and
If its color is Yellow.
This is our 'training set.' We will use this to predict the type of any new fruit we encounter.
Type Long | Not Long || Sweet | Not Sweet || Yellow |Not Yellow|Total
___________________________________________________________________
Banana | 400 | 100 || 350 | 150 || 450 | 50 | 500
Orange | 0 | 300 || 150 | 150 || 300 | 0 | 300
Other Fruit | 100 | 100 || 150 | 50 || 50 | 150 | 200
____________________________________________________________________
Total | 500 | 500 || 650 | 350 || 800 | 200 | 1000
___________________________________________________________________
We can pre-compute a lot of things about our fruit collection.
The so-called "Prior" probabilities. (If we didn't know any of the fruit attributes, this would be our guess.) These are our base rates.
P(Banana) = 0.5 (500/1000)
P(Orange) = 0.3
P(Other Fruit) = 0.2
Probability of "Evidence"
p(Long) = 0.5
P(Sweet) = 0.65
P(Yellow) = 0.8
Probability of "Likelihood"
P(Long|Banana) = 0.8
P(Long|Orange) = 0 [Oranges are never long in all the fruit we have seen.]
....
P(Yellow|Other Fruit) = 50/200 = 0.25
P(Not Yellow|Other Fruit) = 0.75
Given a Fruit, how to classify it?
Let's say that we are given the properties of an unknown fruit, and asked to classify it. We are told that the fruit is Long, Sweet and Yellow. Is it a Banana? Is it an Orange? Or Is it some Other Fruit?
We can simply run the numbers for each of the 3 outcomes, one by one. Then we choose the highest probability and 'classify' our unknown fruit as belonging to the class that had the highest probability based on our prior evidence (our 1000 fruit training set):
P(Banana|Long, Sweet and Yellow)
P(Long|Banana) * P(Sweet|Banana) * P(Yellow|Banana) * P(banana)
= _______________________________________________________________
P(Long) * P(Sweet) * P(Yellow)
= 0.8 * 0.7 * 0.9 * 0.5 / P(evidence)
= 0.252 / P(evidence)
P(Orange|Long, Sweet and Yellow) = 0
P(Other Fruit|Long, Sweet and Yellow)
P(Long|Other fruit) * P(Sweet|Other fruit) * P(Yellow|Other fruit) * P(Other Fruit)
= ____________________________________________________________________________________
P(evidence)
= (100/200 * 150/200 * 50/200 * 200/1000) / P(evidence)
= 0.01875 / P(evidence)
By an overwhelming margin (0.252 >> 0.01875), we classify this Sweet/Long/Yellow fruit as likely to be a Banana.
Why is Bayes Classifier so popular?
Look at what it eventually comes down to. Just some counting and multiplication. We can pre-compute all these terms, and so classifying becomes easy, quick and efficient.
Let z = 1 / P(evidence). Now we quickly compute the following three quantities.
P(Banana|evidence) = z * Prob(Banana) * Prob(Evidence1|Banana) * Prob(Evidence2|Banana) ...
P(Orange|Evidence) = z * Prob(Orange) * Prob(Evidence1|Orange) * Prob(Evidence2|Orange) ...
P(Other|Evidence) = z * Prob(Other) * Prob(Evidence1|Other) * Prob(Evidence2|Other) ...
Assign the class label of whichever is the highest number, and you are done.
Despite the name, Naive Bayes turns out to be excellent in certain applications. Text classification is one area where it really shines.
Your question as I understand it is divided in two parts, part one being you need a better understanding of the Naive Bayes classifier & part two being the confusion surrounding Training set.
In general all of Machine Learning Algorithms need to be trained for supervised learning tasks like classification, prediction etc. or for unsupervised learning tasks like clustering.
During the training step, the algorithms are taught with a particular input dataset (training set) so that later on we may test them for unknown inputs (which they have never seen before) for which they may classify or predict etc (in case of supervised learning) based on their learning. This is what most of the Machine Learning techniques like Neural Networks, SVM, Bayesian etc. are based upon.
So in a general Machine Learning project basically you have to divide your input set to a Development Set (Training Set + Dev-Test Set) & a Test Set (or Evaluation set). Remember your basic objective would be that your system learns and classifies new inputs which they have never seen before in either Dev set or test set.
The test set typically has the same format as the training set. However, it is very important that the test set be distinct from the training corpus: if we simply
reused the training set as the test set, then a model that simply memorized its input, without learning how to generalize to new examples, would receive misleadingly high scores.
In general, for an example, 70% of our data can be used as training set cases. Also remember to partition the original set into the training and test sets randomly.
Now I come to your other question about Naive Bayes.
To demonstrate the concept of Naïve Bayes Classification, consider the example given below:
As indicated, the objects can be classified as either GREEN or RED. Our task is to classify new cases as they arrive, i.e., decide to which class label they belong, based on the currently existing objects.
Since there are twice as many GREEN objects as RED, it is reasonable to believe that a new case (which hasn't been observed yet) is twice as likely to have membership GREEN rather than RED. In the Bayesian analysis, this belief is known as the prior probability. Prior probabilities are based on previous experience, in this case the percentage of GREEN and RED objects, and often used to predict outcomes before they actually happen.
Thus, we can write:
Prior Probability of GREEN: number of GREEN objects / total number of objects
Prior Probability of RED: number of RED objects / total number of objects
Since there is a total of 60 objects, 40 of which are GREEN and 20 RED, our prior probabilities for class membership are:
Prior Probability for GREEN: 40 / 60
Prior Probability for RED: 20 / 60
Having formulated our prior probability, we are now ready to classify a new object (WHITE circle in the diagram below). Since the objects are well clustered, it is reasonable to assume that the more GREEN (or RED) objects in the vicinity of X, the more likely that the new cases belong to that particular color. To measure this likelihood, we draw a circle around X which encompasses a number (to be chosen a priori) of points irrespective of their class labels. Then we calculate the number of points in the circle belonging to each class label. From this we calculate the likelihood:
From the illustration above, it is clear that Likelihood of X given GREEN is smaller than Likelihood of X given RED, since the circle encompasses 1 GREEN object and 3 RED ones. Thus:
Although the prior probabilities indicate that X may belong to GREEN (given that there are twice as many GREEN compared to RED) the likelihood indicates otherwise; that the class membership of X is RED (given that there are more RED objects in the vicinity of X than GREEN). In the Bayesian analysis, the final classification is produced by combining both sources of information, i.e., the prior and the likelihood, to form a posterior probability using the so-called Bayes' rule (named after Rev. Thomas Bayes 1702-1761).
Finally, we classify X as RED since its class membership achieves the largest posterior probability.
Naive Bayes comes under supervising machine learning which used to make classifications of data sets.
It is used to predict things based on its prior knowledge and independence assumptions.
They call it naive because it’s assumptions (it assumes that all of the features in the dataset are equally important and independent) are really optimistic and rarely true in most real-world applications.
It is classification algorithm which makes the decision for the unknown data set. It is based on Bayes Theorem which describe the probability of an event based on its prior knowledge.
Below diagram shows how naive Bayes works
Formula to predict NB:
How to use Naive Bayes Algorithm ?
Let's take an example of how N.B woks
Step 1: First we find out Likelihood of table which shows the probability of yes or no in below diagram.
Step 2: Find the posterior probability of each class.
Problem: Find out the possibility of whether the player plays in Rainy condition?
P(Yes|Rainy) = P(Rainy|Yes) * P(Yes) / P(Rainy)
P(Rainy|Yes) = 2/9 = 0.222
P(Yes) = 9/14 = 0.64
P(Rainy) = 5/14 = 0.36
Now, P(Yes|Rainy) = 0.222*0.64/0.36 = 0.39 which is lower probability which means chances of the match played is low.
For more reference refer these blog.
Refer GitHub Repository Naive-Bayes-Examples
Ram Narasimhan explained the concept very nicely here below is an alternative explanation through the code example of Naive Bayes in action
It uses an example problem from this book on page 351
This is the data set that we will be using
In the above dataset if we give the hypothesis = {"Age":'<=30', "Income":"medium", "Student":'yes' , "Creadit_Rating":'fair'} then what is the probability that he will buy or will not buy a computer.
The code below exactly answers that question.
Just create a file called named new_dataset.csv and paste the following content.
Age,Income,Student,Creadit_Rating,Buys_Computer
<=30,high,no,fair,no
<=30,high,no,excellent,no
31-40,high,no,fair,yes
>40,medium,no,fair,yes
>40,low,yes,fair,yes
>40,low,yes,excellent,no
31-40,low,yes,excellent,yes
<=30,medium,no,fair,no
<=30,low,yes,fair,yes
>40,medium,yes,fair,yes
<=30,medium,yes,excellent,yes
31-40,medium,no,excellent,yes
31-40,high,yes,fair,yes
>40,medium,no,excellent,no
Here is the code the comments explains everything we are doing here! [python]
import pandas as pd
import pprint
class Classifier():
data = None
class_attr = None
priori = {}
cp = {}
hypothesis = None
def __init__(self,filename=None, class_attr=None ):
self.data = pd.read_csv(filename, sep=',', header =(0))
self.class_attr = class_attr
'''
probability(class) = How many times it appears in cloumn
__________________________________________
count of all class attribute
'''
def calculate_priori(self):
class_values = list(set(self.data[self.class_attr]))
class_data = list(self.data[self.class_attr])
for i in class_values:
self.priori[i] = class_data.count(i)/float(len(class_data))
print "Priori Values: ", self.priori
'''
Here we calculate the individual probabilites
P(outcome|evidence) = P(Likelihood of Evidence) x Prior prob of outcome
___________________________________________
P(Evidence)
'''
def get_cp(self, attr, attr_type, class_value):
data_attr = list(self.data[attr])
class_data = list(self.data[self.class_attr])
total =1
for i in range(0, len(data_attr)):
if class_data[i] == class_value and data_attr[i] == attr_type:
total+=1
return total/float(class_data.count(class_value))
'''
Here we calculate Likelihood of Evidence and multiple all individual probabilities with priori
(Outcome|Multiple Evidence) = P(Evidence1|Outcome) x P(Evidence2|outcome) x ... x P(EvidenceN|outcome) x P(Outcome)
scaled by P(Multiple Evidence)
'''
def calculate_conditional_probabilities(self, hypothesis):
for i in self.priori:
self.cp[i] = {}
for j in hypothesis:
self.cp[i].update({ hypothesis[j]: self.get_cp(j, hypothesis[j], i)})
print "\nCalculated Conditional Probabilities: \n"
pprint.pprint(self.cp)
def classify(self):
print "Result: "
for i in self.cp:
print i, " ==> ", reduce(lambda x, y: x*y, self.cp[i].values())*self.priori[i]
if __name__ == "__main__":
c = Classifier(filename="new_dataset.csv", class_attr="Buys_Computer" )
c.calculate_priori()
c.hypothesis = {"Age":'<=30', "Income":"medium", "Student":'yes' , "Creadit_Rating":'fair'}
c.calculate_conditional_probabilities(c.hypothesis)
c.classify()
output:
Priori Values: {'yes': 0.6428571428571429, 'no': 0.35714285714285715}
Calculated Conditional Probabilities:
{
'no': {
'<=30': 0.8,
'fair': 0.6,
'medium': 0.6,
'yes': 0.4
},
'yes': {
'<=30': 0.3333333333333333,
'fair': 0.7777777777777778,
'medium': 0.5555555555555556,
'yes': 0.7777777777777778
}
}
Result:
yes ==> 0.0720164609053
no ==> 0.0411428571429
I try to explain the Bayes rule with an example.
What is the chance that a random person selected from the society is a smoker?
You may reply 10%, and let's assume that's right.
Now, what if I say that the random person is a man and is 15 years old?
You may say 15 or 20%, but why?.
In fact, we try to update our initial guess with new pieces of evidence ( P(smoker) vs. P(smoker | evidence) ). The Bayes rule is a way to relate these two probabilities.
P(smoker | evidence) = P(smoker)* p(evidence | smoker)/P(evidence)
Each evidence may increase or decrease this chance. For example, this fact that he is a man may increase the chance provided that this percentage (being a man) among non-smokers is lower.
In the other words, being a man must be an indicator of being a smoker rather than a non-smoker. Therefore, if an evidence is an indicator of something, it increases the chance.
But how do we know that this is an indicator?
For each feature, you can compare the commonness (probability) of that feature under the given conditions with its commonness alone. (P(f | x) vs. P(f)).
P(smoker | evidence) / P(smoker) = P(evidence | smoker)/P(evidence)
For example, if we know that 90% of smokers are men, it's not still enough to say whether being a man is an indicator of being smoker or not. For example if the probability of being a man in the society is also 90%, then knowing that someone is a man doesn't help us ((90% / 90%) = 1. But if men contribute to 40% of the society, but 90% of the smokers, then knowing that someone is a man increases the chance of being a smoker (90% / 40%) = 2.25, so it increases the initial guess (10%) by 2.25 resulting 22.5%.
However, if the probability of being a man was 95% in the society, then regardless of the fact that the percentage of men among smokers is high (90%)! the evidence that someone is a man decreases the chance of him being a smoker! (90% / 95%) = 0.95).
So we have:
P(smoker | f1, f2, f3,... ) = P(smoker) * contribution of f1* contribution of f2 *...
=
P(smoker)*
(P(being a man | smoker)/P(being a man))*
(P(under 20 | smoker)/ P(under 20))
Note that in this formula we assumed that being a man and being under 20 are independent features so we multiplied them, it means that knowing that someone is under 20 has no effect on guessing that he is man or woman. But it may not be true, for example maybe most adolescence in a society are men...
To use this formula in a classifier
The classifier is given with some features (being a man and being under 20) and it must decide if he is an smoker or not (these are two classes). It uses the above formula to calculate the probability of each class under the evidence (features), and it assigns the class with the highest probability to the input. To provide the required probabilities (90%, 10%, 80%...) it uses the training set. For example, it counts the people in the training set that are smokers and find they contribute 10% of the sample. Then for smokers checks how many of them are men or women .... how many are above 20 or under 20....In the other words, it tries to build the probability distribution of the features for each class based on the training data.
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.