I'm developing a tournament model for a virtual city commerce game (Urbien.com) and would love to get some algorithm suggestions. Here's the scenario and current "basic" implementation:
Scenario
Entries are paired up duel-style, like on the original Facemash or Pixoto.com.
The "player" is a judge, who gets a stream of dueling pairs and must choose a winner for each pair.
Tournaments never end, people can submit new entries at any time and winners of the day/week/month/millenium are chosen based on the data at that date.
Problems to be solved
Rating algorithm - how to rate tournament entries and how to adjust their ratings after each match?
Pairing algorithm - how to choose the next pair to feed the player?
Current solution
Rating algorithm - the Elo rating system currently used in chess and other tournaments.
Pairing algorithm - our current algorithm recognizes two imperatives:
Give more duels to entries that have had less duels so far
Match people with similar ratings with higher probability
Given:
N = total number of entries in the tournament
D = total number of duels played in the tournament so far by all players
Dx = how many duels player x has had so far
To choose players x and y to duel, we first choose player x with probability:
p(x) = (1 - (Dx / D)) / N
Then choose player y the following way:
Sort the players by rating
Let the probability of choosing player j at index jIdx in the sorted list be:
p(j) = ...
0, if (j == x)
n*r^abs(jIdx - xIdx) otherwise
where 0 < r < 1 is a coefficient to be chosen, and n is a normalization factor.
Basically the probabilities in either direction from x form a geometic series, normalized so they sum to 1.
Concerns
Maximize informational value of a duel - pairing the lowest rated entry against the highest rated entry is very unlikely to give you any useful information.
Speed - we don't want to do massive amounts of calculations just to choose one pair. One alternative is to use something like the Swiss pairing system and pair up all entries at once, instead of choosing new duels one at a time. This has the drawback (?) that all entries submitted in a given timeframe will experience roughly the same amount of duels, which may or may not be desirable.
Equilibrium - Pixoto's ImageDuel algorithm detects when entries are unlikely to further improve their rating and gives them less duels from then on. The benefits of such detection are debatable. On the one hand, you can save on computation if you "pause" half the entries. On the other hand, entries with established ratings may be the perfect matches for new entries, to establish the newbies' ratings.
Number of entries - if there are just a few entries, say 10, perhaps a simpler algorithm should be used.
Wins/Losses - how does the player's win/loss ratio affect the next pairing, if at all?
Storage - what to store about each entry and about the tournament itself? Currently stored:
Tournament Entry: # duels so far, # wins, # losses, rating
Tournament: # duels so far, # entries
instead of throwing in ELO and ad-hoc probability formulae, you could use a standard approach based on the maximum likelihood method.
The maximum likelihood method is a method for parameter estimation and it works like this (example). Every contestant (player) is assigned a parameter s[i] (1 <= i <= N where N is total number of contestants) that measures the strength or skill of that player. You pick a formula that maps the strengths of two players into a probability that the first player wins. For example,
P(i, j) = 1/(1 + exp(s[j] - s[i]))
which is the logistic curve (see http://en.wikipedia.org/wiki/Sigmoid_function). When you have then a table that shows the actual results between the users, you use global optimization (e.g. gradient descent) to find those strength parameters s[1] .. s[N] that maximize the probability of the actually observed match result. E.g. if you have three contestants and have observed two results:
Player 1 won over Player 2
Player 2 won over Player 3
then you find parameters s[1], s[2], s[3] that maximize the value of the product
P(1, 2) * P(2, 3)
Incidentally, it can be easier to maximize
log P(1, 2) + log P(2, 3)
Note that if you use something like the logistics curve, it is only the difference of the strength parameters that matters so you need to anchor the values somewhere, e.g. choose arbitrarily
s[1] = 0
In order to have more recent matches "weigh" more, you can adjust the importance of the match results based on their age. If t measures the time since a match took place (in some time units), you can maximize the value of the sum (using the example)
e^-t log P(1, 2) + e^-t' log P(2, 3)
where t and t' are the ages of the matches 1-2 and 2-3, so that those games that occurred more recently weigh more.
The interesting thing in this approach is that when the strength parameters have values, the P(...) formula can be used immediately to calculate the win/lose probability for any future match. To pair contestants, you can pair those where the P(...) value is close to 0.5, and then prefer those contestants whose time-adjusted number of matches (sum of e^-t1 + e^-t2 + ...) for match ages t1, t2, ... is low. The best thing would be to calculate the total impact of a win or loss between two players globally and then prefer those matches that have the largest expected impact on the ratings, but that could require lots of calculations.
You don't need to run the maximum likelihood estimation / global optimization algorithm all the time; you can run it e.g. once a day as a batch run and use the results for the next day for matching people together. The time-adjusted match masses can be updated real time anyway.
On algorithm side, you can sort the players after the maximum likelihood run base on their s parameter, so it's very easy to find equal-strength players quickly.
Related
I'm wondering if there is a way to speed up the calculation of guaranteed promotion in football (soccer) for a given football league table. It seems like there is a lot of structure to the problem so the exhaustive solution is perhaps slower than necessary.
In the problem there is a schedule (a list of pairs of teams) that will play each other in the future and a table (map) of points each team has earned in games in the past. I've included a sample real life points table below. Each future game can be won, lost or tied and teams earn 3 points for a win and 1 for a tie. Points (Pts) is what ultimately matters for promotion and num_of_promoted_teams (positive integer, usually around 1-3) are promoted at the end of each season.
The problem is to determine which (if any) teams are currently guaranteed promotion. Where guaranteed promotion means that no matter the outcome of the final games the team must end up promoted.
def promoted(num_of_promoted_teams, table, schedule):
return guaranteed_promotions
I've been thinking about using depth first search (of the future game results) to eliminate teams which would lower the average but not the worst case performance. This certainly help early in the season, but the problem could become large in mid-season before shrinking again near the end. It seems like there might be a better way.
A constraint solver should be fast enough in practice thanks to clever pruning algorithms, and hard to screw up. Here’s some sample code with the OR-Tools CP-SAT solver.
from ortools.sat.python import cp_model
def promoted(num_promoted_teams, table, schedule):
for candidate in table.keys():
model = cp_model.CpModel()
final_table = table.copy()
for home, away in schedule:
home_win = model.NewBoolVar("")
draw = model.NewBoolVar("")
away_win = model.NewBoolVar("")
model.AddBoolOr([home_win, draw, away_win])
model.AddBoolOr([home_win.Not(), draw.Not()])
model.AddBoolOr([home_win.Not(), away_win.Not()])
model.AddBoolOr([draw.Not(), away_win.Not()])
final_table[home] += 3 * home_win + draw
final_table[away] += draw + 3 * away_win
candidate_points = final_table[candidate]
num_not_behind = 0
for team, team_points in final_table.items():
if team == candidate:
continue
is_behind = model.NewBoolVar("")
model.Add(team_points < candidate_points).OnlyEnforceIf(is_behind)
model.Add(candidate_points <= team_points).OnlyEnforceIf(is_behind.Not())
num_not_behind += is_behind.Not()
model.Add(num_promoted_teams <= num_not_behind)
solver = cp_model.CpSolver()
status = solver.Solve(model)
if status == cp_model.INFEASIBLE:
yield candidate
print(*promoted(2, {"A": 10, "B": 8, "C": 8}, [("B", "C")]))
Here’s an alternative solution that is less extensible and probably slower in exchange for being self-contained and predictable.
This solution consists of an algorithm to test whether a particular team can finish behind a particular set of other teams (assuming unfavorable tie-breaking), wrapped in a loop over pairs consisting of a top-k team ℓ and a set of k teams W that might or might not finish ahead of ℓ (where k is the number of promoted teams).
If there were no draws, then we could use bipartite matching. Mark ℓ as having lost its remaining matches and mark W as having won their matches against teams not in W. On one side of the bipartite graph, there are nodes corresponding to matches between members of W. On the other side, there are zero or more nodes for each team in W, corresponding to the number of matches that that team must win to pull ahead of ℓ. If there is a matching that completely matches the latter side, then W can finish collectively in front of ℓ, and ℓ is not guaranteed promotion.
This could be extended easily if wins were 2 points instead of 3, but alas, 3 points causes the problem not to be convex, and we’re going to need some branching. The simplest branching method depends on the observation that it’s better for two teams to each win once and lose once against each other than draw twice. Hence, loop over all subsets of at most k choose 2 pairs of teams and run the algorithm above after marking each pair in the subset as having drawn once.
(I could propose improvements, but k is small, computers are cheap, programmers are expensive, and sports fans are relentless.)
Let's say we have N teams in a tournament and based on historical data we know what is the probability of each team winning any other team .Lets put all the probabilities in a matrix called P . P[a][b] is the probability team a winning team b. It is obvious that P[a][a] = 0 and P[a][b] = 1-P[b][a].
In this tournament at every round, two of teams compete against each other and the loser is eliminated. This two team are chosen randomly (with equal possibility of each team being picked). So at the first round we have n teams, next n-1 teams and so on until only one team remains and becomes the champion. What is the probability of each team becoming the champion? ( 1 <= N <= 18).
At first when I didn't know how to approach the problem but after some reading and search and keeping in mind that max n is 18 I figured at that using Dynamic programming and Bitmask is the way to go. How ever I couldn't figure at a solution. Here are my problems:
I have really hard time to figure at what are the sub problems and what sub problems should not be recomputed, basically I can't find a well defined recursive ( or not recursive) equation for the problem
In bitmask+dp problems we usually define something like dp[mask][n] or dp[n][mask]. I tried different approaches to define the mask but since the general solution is not clear to me there was no success
Some guidance on this two problems would be very helpful.
This is not really a dynamic programming problem.
If you have a vector V that gives the probability of each player being in the game after n rounds, then you can calculate the player probabilities for n+1 rounds by:
V'i = 2/((18-n)(17-n)) * sum over all j!=i of [ViVjPi,j]
That first factor is the probability that any given available match will be chosen, which depends on the number of previous rounds, because each successive round has fewer players to match up.
The second part is the probability of the players being available for each match, times the probability that the current player will win.
Just do this calculation 17 times to get the player probabilities after 17 rounds, which is the answer you're looking for. You can even drop that first factor, and fix it at the end by normalizing the vector so that the probabilities sum to 1.
I've been self-studying the Expectation Maximization lately, and grabbed myself some simple examples in the process:
http://cs.dartmouth.edu/~cs104/CS104_11.04.22.pdf
There are 3 coins 0, 1 and 2 with P0, P1 and P2 probability landing on Head when tossed. Toss coin 0, if the result is Head, toss coin 1 three times else toss coin 2 three times. The observed data produced by coin 1 and 2 is like this: HHH, TTT, HHH, TTT, HHH. The hidden data is coin 0's result. Estimate P0, P1 and P2.
http://ai.stanford.edu/~chuongdo/papers/em_tutorial.pdf
There are two coins A and B with PA and PB being the probability landing on Head when tossed. Each round, select one coin at random and toss it 10 times then record the results. The observed data is the toss results provided by these two coins. However, we don't know which coin was selected for a particular round. Estimate PA and PB.
While I can get the calculations, I can't relate the ways they are solved to the original EM theory. Specifically, during the M-Step of both examples, I don't see how they're maximizing anything. It just seems they are recalculating the parameters and somehow, the new parameters are better than the old ones. Moreover, the two E-Steps don't even look similar to each other, not to mention the original theory's E-Step.
So how exactly do these example work?
The second PDF won't download for me, but I also visited the wikipedia page http://en.wikipedia.org/wiki/Expectation%E2%80%93maximization_algorithm which has more information. http://melodi.ee.washington.edu/people/bilmes/mypapers/em.pdf (which claims to be a gentle introduction) might be worth a look too.
The whole point of the EM algorithm is to find parameters which maximize the likelihood of the observed data. This is the only bullet point on page 8 of the first PDF, the equation for capital Theta subscript ML.
The EM algorithm comes in handy where there is hidden data which would make the problem easy if you knew it. In the three coins example this is the result of tossing coin 0. If you knew the outcome of that you could (of course) produce an estimate for the probability of coin 0 turning up heads. You would also know whether coin 1 or coin 2 was tossed three times in the next stage, which would allow you to make estimates for the probabilities of coin 1 and coin 2 turning up heads. These estimates would be justified by saying that they maximized the likelihood of the observed data, which would include not only the results that you are given, but also the hidden data that you are not - the results from coin 0. For a coin that gets A heads and B tails you find that the maximum likelihood for the probability of A heads is A/(A+B) - it might be worth you working this out in detail, because it is the building block for the M step.
In the EM algorithm you say that although you don't know the hidden data, you come in with probability estimates which allow you to write down a probability distribution for it. For each possible value of the hidden data you could find the parameter values which would optimize the log likelihood of the data including the hidden data, and this almost always turns out to mean calculating some sort of weighted average (if it doesn't the EM step may be too difficult to be practical).
What the EM algorithm asks you to do is to find the parameters maximizing the weighted sum of log likelihoods given by all the possible hidden data values, where the weights are given by the probability of the associated hidden data given the observations using the parameters at the start of the EM step. This is what almost everybody, including the Wikipedia algorithm, calls the Q-function. The proof behind the EM algorithm, given in the Wikipedia article, says that if you change the parameters so as to increase the Q-function (which is only a means to an end), you will also have changed them so as to increase the likelihood of the observed data (which you do care about). What you tend to find in practice is that you can maximize the Q-function using a variation of what you would do if you know the hidden data, but using the probabilities of the hidden data, given the estimates at the start of the EM-step, to weight the observations in some way.
In your example it means totting up the number of heads and tails produced by each coin. In the PDF they work out P(Y=H|X=) = 0.6967. This means that you use weight 0.6967 for the case Y=H, which means that you increment the counts for Y=H by 0.6967 and increment the counts for X=H in coin 1 by 3*0.6967, and you increment the counts for Y=T by 0.3033 and increment the counts for X=H in coin 2 by 3*0.3033. If you have a detailed justification for why A/(A+B) is a maximum likelihood of coin probabilities in the standard case, you should be ready to turn it into a justification for why this weighted updating scheme maximizes the Q-function.
Finally, the log likelihood of the observed data (the thing you are maximizing) gives you a very useful check. It should increase with every EM step, at least until you get so close to convergence that rounding error comes in, in which case you may have a very small decrease, signalling convergence. If it decreases dramatically, you have a bug in your program or your maths.
As luck would have it, I have been struggling with this material recently as well. Here is how I have come to think of it:
Consider a related, but distinct algorithm called the classify-maximize algorithm, which we might use as a solution technique for a mixture model problem. A mixture model problem is one where we have a sequence of data that may be produced by any of N different processes, of which we know the general form (e.g., Gaussian) but we do not know the parameters of the processes (e.g., the means and/or variances) and may not even know the relative likelihood of the processes. (Typically we do at least know the number of the processes. Without that, we are into so-called "non-parametric" territory.) In a sense, the process which generates each data is the "missing" or "hidden" data of the problem.
Now, what this related classify-maximize algorithm does is start with some arbitrary guesses at the process parameters. Each data point is evaluated according to each one of those parameter processes, and a set of probabilities is generated-- the probability that the data point was generated by the first process, the second process, etc, up to the final Nth process. Then each data point is classified according to the most likely process.
At this point, we have our data separated into N different classes. So, for each class of data, we can, with some relatively simple calculus, optimize the parameters of that cluster with a maximum likelihood technique. (If we tried to do this on the whole data set prior to classifying, it is usually analytically intractable.)
Then we update our parameter guesses, re-classify, update our parameters, re-classify, etc, until convergence.
What the expectation-maximization algorithm does is similar, but more general: Instead of a hard classification of data points into class 1, class 2, ... through class N, we are now using a soft classification, where each data point belongs to each process with some probability. (Obviously, the probabilities for each point need to sum to one, so there is some normalization going on.) I think we might also think of this as each process/guess having a certain amount of "explanatory power" for each of the data points.
So now, instead of optimizing the guesses with respect to points that absolutely belong to each class (ignoring the points that absolutely do not), we re-optimize the guesses in the context of those soft classifications, or those explanatory powers. And it so happens that, if you write the expressions in the correct way, what you're maximizing is a function that is an expectation in its form.
With that said, there are some caveats:
1) This sounds easy. It is not, at least to me. The literature is littered with a hodge-podge of special tricks and techniques-- using likelihood expressions instead of probability expressions, transforming to log-likelihoods, using indicator variables, putting them in basis vector form and putting them in the exponents, etc.
These are probably more helpful once you have the general idea, but they can also obfuscate the core ideas.
2) Whatever constraints you have on the problem can be tricky to incorporate into the framework. In particular, if you know the probabilities of each of the processes, you're probably in good shape. If not, you're also estimating those, and the sum of the probabilities of the processes must be one; they must live on a probability simplex. It is not always obvious how to keep those constraints intact.
3) This is a sufficiently general technique that I don't know how I would go about writing code that is general. The applications go far beyond simple clustering and extend to many situations where you are actually missing data, or where the assumption of missing data may help you. There is a fiendish ingenuity at work here, for many applications.
4) This technique is proven to converge, but the convergence is not necessarily to the global maximum; be wary.
I found the following link helpful in coming up with the interpretation above: Statistical learning slides
And the following write-up goes into great detail of some painful mathematical details: Michael Collins' write-up
I wrote the below code in Python which explains the example given in your second example paper by Do and Batzoglou.
I recommend that you read this link first for a clear explanation of how and why the 'weightA' and 'weightB' in the code below are obtained.
Disclaimer : The code does work but I am certain that it is not coded optimally. I am not a Python coder normally and have started using it two weeks ago.
import numpy as np
import math
#### E-M Coin Toss Example as given in the EM tutorial paper by Do and Batzoglou* ####
def get_mn_log_likelihood(obs,probs):
""" Return the (log)likelihood of obs, given the probs"""
# Multinomial Distribution Log PMF
# ln (pdf) = multinomial coeff * product of probabilities
# ln[f(x|n, p)] = [ln(n!) - (ln(x1!)+ln(x2!)+...+ln(xk!))] + [x1*ln(p1)+x2*ln(p2)+...+xk*ln(pk)]
multinomial_coeff_denom= 0
prod_probs = 0
for x in range(0,len(obs)): # loop through state counts in each observation
multinomial_coeff_denom = multinomial_coeff_denom + math.log(math.factorial(obs[x]))
prod_probs = prod_probs + obs[x]*math.log(probs[x])
multinomial_coeff = math.log(math.factorial(sum(obs))) - multinomial_coeff_denom
likelihood = multinomial_coeff + prod_probs
return likelihood
# 1st: Coin B, {HTTTHHTHTH}, 5H,5T
# 2nd: Coin A, {HHHHTHHHHH}, 9H,1T
# 3rd: Coin A, {HTHHHHHTHH}, 8H,2T
# 4th: Coin B, {HTHTTTHHTT}, 4H,6T
# 5th: Coin A, {THHHTHHHTH}, 7H,3T
# so, from MLE: pA(heads) = 0.80 and pB(heads)=0.45
# represent the experiments
head_counts = np.array([5,9,8,4,7])
tail_counts = 10-head_counts
experiments = zip(head_counts,tail_counts)
# initialise the pA(heads) and pB(heads)
pA_heads = np.zeros(100); pA_heads[0] = 0.60
pB_heads = np.zeros(100); pB_heads[0] = 0.50
# E-M begins!
delta = 0.001
j = 0 # iteration counter
improvement = float('inf')
while (improvement>delta):
expectation_A = np.zeros((5,2), dtype=float)
expectation_B = np.zeros((5,2), dtype=float)
for i in range(0,len(experiments)):
e = experiments[i] # i'th experiment
ll_A = get_mn_log_likelihood(e,np.array([pA_heads[j],1-pA_heads[j]])) # loglikelihood of e given coin A
ll_B = get_mn_log_likelihood(e,np.array([pB_heads[j],1-pB_heads[j]])) # loglikelihood of e given coin B
weightA = math.exp(ll_A) / ( math.exp(ll_A) + math.exp(ll_B) ) # corresponding weight of A proportional to likelihood of A
weightB = math.exp(ll_B) / ( math.exp(ll_A) + math.exp(ll_B) ) # corresponding weight of B proportional to likelihood of B
expectation_A[i] = np.dot(weightA, e)
expectation_B[i] = np.dot(weightB, e)
pA_heads[j+1] = sum(expectation_A)[0] / sum(sum(expectation_A));
pB_heads[j+1] = sum(expectation_B)[0] / sum(sum(expectation_B));
improvement = max( abs(np.array([pA_heads[j+1],pB_heads[j+1]]) - np.array([pA_heads[j],pB_heads[j]]) ))
j = j+1
The key to understanding this is knowing what the auxiliary variables are that make estimation trivial. I will explain the first example quickly, the second follows a similar pattern.
Augment each sequence of heads/tails with two binary variables, which indicate whether coin 1 was used or coin 2. Now our data looks like the following:
c_11 c_12
c_21 c_22
c_31 c_32
...
For each i, either c_i1=1 or c_i2=1, with the other being 0. If we knew the values these variables took in our sample, estimation of parameters would be trivial: p1 would be the proportion of heads in samples where c_i1=1, likewise for c_i2, and \lambda would be the mean of the c_i1s.
However, we don't know the values of these binary variables. So, what we basically do is guess them (in reality, take their expectation), and then update the parameters in our model assuming our guesses were correct. So the E step is to take the expectation of the c_i1s and c_i2s. The M step is to take maximum likelihood estimates of p_1, p_2 and \lambda given these cs.
Does that make a bit more sense? I can write out the updates for the E and M step if you prefer. EM then just guarantees that by following this procedure, likelihood will never decrease as iterations increase.
So I've been working on a problem in my spare time and I'm stuck. Here is where I'm at. I have a number 40. It represents players. I've been given other numbers 39, 38, .... 10. These represent the scores of the first 30 players (1 -30). The rest of the players (31-40) have some unknown score. What I would like to do is find how many combinations of the scores are consistent with the given data.
So for a simpler example: if you have 3 players. One has a score of 1. Then the number of possible combinations of the scores is 3 (0,2; 2,0; 1,1), where (a,b) stands for the number of wins for player one and player two, respectively. A combination of (3,0) wouldn't work because no person can have 3 wins. Nor would (0,0) work because we need a total of 3 wins (and wouldn't get it with 0,0).
I've found the total possible number of games. This is the total number of games played, which means it is the total number of wins. (There are no ties.) Finally, I have a variable for the max wins per player (which is one less than the total number of players. No player can have more than that.)
I've tried finding the number of unique combinations by spreading out N wins to each player and then subtracting combinations that don't fit the criteria. E.g., to figure out many ways to give 10 victories to 5 people with no more than 4 victories to each person, you would use:
C(14,4) - C(5,1)*C(9,4) + C(5,2)*C(4,4) = 381. C(14,4) comes from the formula C(n+k-1, k-1) (google bars and strips, I believe). The next is picking off the the ones with the 5 (not allowed), but adding in the ones we subtracted twice.
Yeah, there has got to be an easier way. Lastly, the numbers get so big that I'm not sure that my computer can adequately handle them. We're talking about C(780, 39), which is 1.15495183 × 10^66. Regardless, there should be a better way of doing this.
To recap, you have 40 people. The scores of the first 30 people are 10 - 39. The last ten people have unknown scores. How many scores can you generate that are meet the criteria: all the scores add up to total possible wins and each player gets no more 39 wins.
Thoughts?
Generating functions:
Since the question is more about math, but still on a programming QA site, let me give you a partial solution that works for many of these problems using a symbolic algebra (like Maple of Mathematica). I highly recommend that you grab an intro combinatorics book, these kind of questions are answered there.
First of all the first 30 players who score 10-39 (with a total score of 735) are a bit of a red herring - what we would like to do is solve the other problem, the remaining 10 players whose score could be in the range of (0...39).
If we think of the possible scores of the players as the polynomial:
f(x) = x^0 + x^1 + x^2 + ... x^39
Where a value of x^2 is the score of 2 for example, consider what this looks like
f(x)^10
This represents the combined score of all 10 players, ie. the coefficent of x^385 is 2002, which represents the fact that there are 2002 ways for the 10 players to score 385. Wolfram Alpha (a programming lanuage IMO) can evaluate this for us.
If you'd like to know how many possible ways of doing this, just substitute in x=1 in the expression giving 8,140,406,085,191,601, which just happens to be 39^10 (no surprise!)
Why is this useful?
While I know it may seem silly to some to set up all this machinery for a simple problem that can be solved on paper - the approach of generating functions is useful when the problem gets messy (and asymptotic analysis is possible). Consider the same problem, but now we restrict the players to only score prime numbers (2,3,5,7,11,...). How many possible ways can the 10 of them score a specific number, say 344? Just modify your f(x):
f(x) = x^2 + x^3 + x^5 + x^7 + x^11 ...
and repeat the process! (I get [x^344]f(x)^10 = 1390).
For a school project, we'll have to implement a ranking system. However, we figured that a dumb rank average would suck: something that one user ranked 5 stars would have a better average that something 188 users ranked 4 stars, and that's just stupid.
So I'm wondering if any of you have an example algorithm of "smart" ranking. It only needs to take in account the rankings given and the number of rankings.
Thanks!
You can use a method inspired by Bayesian probability. The gist of the approach is to have an initial belief about the true rating of an item, and use users' ratings to update your belief.
This approach requires two parameters:
What do you think is the true "default" rating of an item, if you have no ratings at all for the item? Call this number R, the "initial belief".
How much weight do you give to the initial belief, compared to the user ratings? Call this W, where the initial belief is "worth" W user ratings of that value.
With the parameters R and W, computing the new rating is simple: assume you have W ratings of value R along with any user ratings, and compute the average. For example, if R = 2 and W = 3, we compute the final score for various scenarios below:
100 (user) ratings of 4: (3*2 + 100*4) / (3 + 100) = 3.94
3 ratings of 5 and 1 rating of 4: (3*2 + 3*5 + 1*4) / (3 + 3 + 1) = 3.57
10 ratings of 4: (3*2 + 10*4) / (3 + 10) = 3.54
1 rating of 5: (3*2 + 1*5) / (3 + 1) = 2.75
No user ratings: (3*2 + 0) / (3 + 0) = 2
1 rating of 1: (3*2 + 1*1) / (3 + 1) = 1.75
This computation takes into consideration the number of user ratings, and the values of those ratings. As a result, the final score roughly corresponds to how happy one can expect to be about a particular item, given the data.
Choosing R
When you choose R, think about what value you would be comfortable assuming for an item with no ratings. Is the typical no-rating item actually 2.4 out of 5, if you were to instantly have everyone rate it? If so, R = 2.4 would be a reasonable choice.
You should not use the minimum value on the rating scale for this parameter, since an item rated extremely poorly by users should end up "worse" than a default item with no ratings.
If you want to pick R using data rather than just intuition, you can use the following method:
Consider all items with at least some threshold of user ratings (so you can be confident that the average user rating is reasonably accurate).
For each item, assume its "true score" is the average user rating.
Choose R to be the median of those scores.
If you want to be slightly more optimistic or pessimistic about a no-rating item, you can choose R to be a different percentile of the scores, for instance the 60th percentile (optimistic) or 40th percentile (pessimistic).
Choosing W
The choice of W should depend on how many ratings a typical item has, and how consistent ratings are. W can be higher if items naturally obtain many ratings, and W should be higher if you have less confidence in user ratings (e.g., if you have high spammer activity). Note that W does not have to be an integer, and can be less than 1.
Choosing W is a more subjective matter than choosing R. However, here are some guidelines:
If a typical item obtains C ratings, then W should not exceed C, or else the final score will be more dependent on R than on the actual user ratings. Instead, W should be close to a fraction of C, perhaps between C/20 and C/5 (depending on how noisy or "spammy" ratings are).
If historical ratings are usually consistent (for an individual item), then W should be relatively small. On the other hand, if ratings for an item vary wildly, then W should be relatively large. You can think of this algorithm as "absorbing" W ratings that are abnormally high or low, turning those ratings into more moderate ones.
In the extreme, setting W = 0 is equivalent to using only the average of user ratings. Setting W = infinity is equivalent to proclaiming that every item has a true rating of R, regardless of the user ratings. Clearly, neither of these extremes are appropriate.
Setting W too large can have the effect of favoring an item with many moderately-high ratings over an item with slightly fewer exceptionally-high ratings.
I appreciated the top answer at the time of posting, so here it is codified as JavaScript:
const defaultR = 2;
const defaultW = 3; // should not exceed typicalNumberOfRatingsPerAnswers 0 is equivalent to using only average of ratings
function getSortAlgoValue(ratings) {
const allRatings = ratings.reduce((sum, r) => sum + r, 0);
return (defaultR * defaultW + allRatings) / (defaultW + ratings.length);
}
Only listed as a separate answer because the formatting of the code block as a reply wasn't very
Since you've stated that the machine would only be given the rankings and the number of rankings, I would argue that it may be negligent to attempt a calculated weighting method.
First, there are two many unknowns to confirm the proposition that in enough circumstances a larger quantity of ratings are a better indication of quality than a smaller number of ratings. One example is how long have rankings been given? Has there been equal collection duration (equal attention) given to different items ranked with this same method? Others are, which markets have had access to this item and, of course, who specifically ranked it?
Secondly, you've stated in a comment below the question that this is not for front-end use but rather "the ratings are generated by machines, for machines," as a response to my comment that "it's not necessarily only statistical. One person might consider 50 ratings enough, where that might not be enough for another. And some raters' profiles might look more reliable to one person than to another. When that's transparent, it lets the user make a more informed assessment."
Why would that be any different for machines? :)
In any case, if this is about machine-to-machine rankings, the question needs greater detail in order for us to understand how different machines might generate and use the rankings.
Can a ranking generated by a machine be flawed (so as to suggest that more rankings may somehow compensate for those "flawed" rankings? What does that even mean - is it a machine error? Or is it because the item has no use to this particular machine, for example? There are many issues here we might first want to unpack, including if we have access to how the machines are generating the ranking, on some level we may already know the meaning this item may have for this machine, making the aggregated ranking superfluous.
What you can find on different plattforms is the blanking of ratings without enough votings: "This item does not have enough votings"
The problem is you can't do it in an easy formula to calculate a ranking.
I would suggest a hiding of ranking with less than minimum votings but caclulate intern a moving average. I always prefer moving average against total average as it prefers votings from the last time against very old votings which might be given for totaly different circumstances.
Additionally you do not need to have too add a list of all votings. you just have the calculated average and the next voting just changes this value.
newAverage = weight * newVoting + (1-weight) * oldAverage
with a weight about 0.05 for a preference of the last 20 values. (just experiment with this weight)
Additionally I would start with these conditions:
no votings = medium range value (1-5 stars => start with 3 stars)
the average will not be shown if less than 10 votings were given.
A simple solution might be a weighted average:
sum(votes) / number_of_votes
That way, 3 people voting 1 star, and one person voting 5 would give a weighted average of (1+1+1+5)/4 = 2 stars.
Simple, effective, and probably sufficient for your purposes.