There is an excellent discussion in the following thread:
http://groups-beta.google.com/group/rec.puzzles/browse_frm/thread/1bc83090d74d5643/e29f26e647c4303e?q=imdb+formula+minimum+votes+rating&rnum=4#e29f26e647c4303e
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Determining weighted rankings
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Carlos Moreno Nov 29 2002, 8:08 am show options
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From: Carlos Moreno <moreno_at_mochima_dot_...@xx.xxx> - Find messages
by this author
Date: Fri, 29 Nov 2002 11:07:29 -0500
Local: Fri,Nov 29 2002 8:07 am
Subject: Determining weighted rankings
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Here's an interesting puzzle -- at least I'm puzzled with this! :-)
I see the Top 250 movies in the IMDB (Internet Movie DataBase),
and I'm curious about the formula for the weighted ranking, which
they call a "true Bayesian estimate".
People vote for a movie, giving 1 to 10, and then they rank the
movies with this formula, that considers the average of votes
and also the number of votes:
v m
------- * R + ------- * c
v + m v + m
Where:
v = number of votes
m = minimum number of votes to be considered (currently 1250)
R = average "vote" given to the movie
c = average "vote" for all the movies (currently 6.9)
(when I put "currently", I mean that that is what they list in
their page, where they describe the formula)
I understand part of the "intent" of that formula, but I'm having
trouble understanding another part...
I see how it conveniently gives better ranking to a movie with
higher number of votes for the same average vote (if 10 people
say that a movie is 8/10, that's not as good as 1000 people
saying it is 8/10). So, the estimate gives certain weight to
the movie's ranking, but then it also considers the overall
average -- kind of an "in case of doubt, I'll assume you are
in the overall average" -- which translates to "in case of
insufficient information, I'll assume that probabilistically
speaking, you'll approach the overall average".
Obviously, the fraction that considers the movie's average vote
approaches 1 when the number of votes approaches infinity,
and the other fraction (the one that considers the overall
average) approaches zero when the number of votes approaches
infinity.
My doubt is: what is that magical number 1250? Is it arbitrary?
If it is, then how can they call this a "true Bayesian estimate"?
A true estimate tells me that they are estimating what is the
value with maximum likelihood to be the true value of the movie's
average vote, given that we don't count on an infinite number
of votes (which is what we would require to obtain the true
average vote). But this value, this formula is, the way I see
it, "tainted" by this mysterious 1250...
Let me put my doubt this way: We have to determine things like
the following: we have two movies: move 1 has 2000 votes and
they average 9/10; movie 2 has 4000 votes and they average 8/10;
which movie is better?? (defining "better" as the movie for
which an infinite number of voters would give a higher average).
Movie 2 may have less average, but it carries more weight, as I
have higher "confidence" in that value, since a lot of people
have voted, whereas movie 1 has higher average, but it is
"doubtful" -- not a lot of people have voted...
So you see that if I make m = 10, I obtain one value, but if I
make m = 2000, I obtain something very different. So, is that
m number really arbitrary? Or does it obey some mathematical
formula depending on the probability distribution of the votes?
Thanks,
Carlos
--
Mensanator Nov 29 2002, 1:00 pm show options
Newsgroups: rec.puzzles
From: mensana...@aol.com (Mensanator) - Find messages by this author
Date: 29 Nov 2002 21:00:19 GMT
Local: Fri,Nov 29 2002 1:00 pm
Subject: Re: Determining weighted rankings
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>Subject: Determining weighted rankings
>From: Carlos Moreno moreno_at_mochima_dot_...@xx.xxx
>Date: 11/29/2002 10:07 AM Central Standard Time
>Message-id: <3DE790C1.2030...@xx.xxx>
>Here's an interesting puzzle -- at least I'm puzzled with this! :-)
>I see the Top 250 movies in the IMDB (Internet Movie DataBase),
>and I'm curious about the formula for the weighted ranking, which
>they call a "true Bayesian estimate".
>People vote for a movie, giving 1 to 10, and then they rank the
>movies with this formula, that considers the average of votes
>and also the number of votes:
> v m
>------- * R + ------- * c
> v + m v + m
>Where:
>v = number of votes
>m = minimum number of votes to be considered (currently 1250)
>R = average "vote" given to the movie
>c = average "vote" for all the movies (currently 6.9)
>(when I put "currently", I mean that that is what they list in
>their page, where they describe the formula)
>I understand part of the "intent" of that formula, but I'm having
>trouble understanding another part...
>I see how it conveniently gives better ranking to a movie with
>higher number of votes for the same average vote (if 10 people
>say that a movie is 8/10, that's not as good as 1000 people
>saying it is 8/10). So, the estimate gives certain weight to
>the movie's ranking, but then it also considers the overall
>average -- kind of an "in case of doubt, I'll assume you are
>in the overall average" -- which translates to "in case of
>insufficient information, I'll assume that probabilistically
>speaking, you'll approach the overall average".
>Obviously, the fraction that considers the movie's average vote
>approaches 1 when the number of votes approaches infinity,
>and the other fraction (the one that considers the overall
>average) approaches zero when the number of votes approaches
>infinity.
>My doubt is: what is that magical number 1250? Is it arbitrary?
>If it is, then how can they call this a "true Bayesian estimate"?
>A true estimate tells me that they are estimating what is the
>value with maximum likelihood to be the true value of the movie's
>average vote, given that we don't count on an infinite number
>of votes (which is what we would require to obtain the true
>average vote). But this value, this formula is, the way I see
>it, "tainted" by this mysterious 1250...
>Let me put my doubt this way: We have to determine things like
>the following: we have two movies: move 1 has 2000 votes and
>they average 9/10; movie 2 has 4000 votes and they average 8/10;
>which movie is better?? (defining "better" as the movie for
>which an infinite number of voters would give a higher average).
>Movie 2 may have less average, but it carries more weight, as I
>have higher "confidence" in that value, since a lot of people
>have voted, whereas movie 1 has higher average, but it is
>"doubtful" -- not a lot of people have voted...
By what criteria do you claim that not a "lot" of people voted for movie 1? It
seems to me that the function of m is to determine what constitutes a "lot".
In your example, 2000 > 1250, so why should it be doubtful?
Plugging in the numbers gives rankings of
v=2000 R=9 m=1250
8.192307692 for movie 1
and
v=4000 R=8 m=1250
7.738095238 for movie 2
so yes, movie 1 is indeed "better" than movie 2.
One way to look at m is to ask: how many votes of 9 does a movie need to be
considered "better" than movie 2?
v=831 R=9 m=1250
7.738587218
Thus, with only 831 votes required when we actually have 2000, there is no
reason to be doubtful that movie 1 is better.
Another question that can be asked is what if m were smaller, say 625 instead
of 1250? The ranking formula will result in a value that falls between c and R.
Higher values of v bring the result closer to R. The parameter m determines how
quickly it approaches R. Lower values of m mean greater signifigance to each
vote.
So we can ask: if m were 625, how many votes of 9 would a movie need to equal
4000 votes of 8?
v=4000 R=8 m=625
7.851351351
v=518 R=9 m=625
7.851706037
Note that with a lower value of m, we only need 518 votes instead of 831 to
equal movie 2. Thus, each vote of 9 carries more weight when m is lower.
>So you see that if I make m = 10, I obtain one value, but if I
>make m = 2000, I obtain something very different. So, is that
>m number really arbitrary? Or does it obey some mathematical
>formula depending on the probability distribution of the votes?
I doubt that m is arbitrary, but I do not how how they arrive at it.
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>Thanks,
>Carlos
>--
Eb Oesch Dec 2 2002, 12:20 pm show options
Newsgroups: rec.puzzles
From: ericboe...@hotmail.com (Eb Oesch) - Find messages by this author
Date: 2 Dec 2002 12:20:55 -0800
Local: Mon,Dec 2 2002 12:20 pm
Subject: Re: Determining weighted rankings
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Carlos Moreno <moreno_at_mochima_dot_...@xx.xxx> wrote in message
<news:3DE790C1.2030700@xx.xxx>...
> Here's an interesting puzzle -- at least I'm puzzled with this! :-)
> I see the Top 250 movies in the IMDB (Internet Movie DataBase),
> and I'm curious about the formula for the weighted ranking, which
> they call a "true Bayesian estimate".
> People vote for a movie, giving 1 to 10, and then they rank the
> movies with this formula, that considers the average of votes
> and also the number of votes:
> v m
> ------- * R + ------- * c
> v + m v + m
> Where:
> v = number of votes
> m = minimum number of votes to be considered (currently 1250)
> R = average "vote" given to the movie
> c = average "vote" for all the movies (currently 6.9)
> (when I put "currently", I mean that that is what they list in
> their page, where they describe the formula)
> I understand part of the "intent" of that formula, but I'm having
> trouble understanding another part...
> I see how it conveniently gives better ranking to a movie with
> higher number of votes for the same average vote (if 10 people
> say that a movie is 8/10, that's not as good as 1000 people
> saying it is 8/10). So, the estimate gives certain weight to
> the movie's ranking, but then it also considers the overall
> average -- kind of an "in case of doubt, I'll assume you are
> in the overall average" -- which translates to "in case of
> insufficient information, I'll assume that probabilistically
> speaking, you'll approach the overall average".
> Obviously, the fraction that considers the movie's average vote
> approaches 1 when the number of votes approaches infinity,
> and the other fraction (the one that considers the overall
> average) approaches zero when the number of votes approaches
> infinity.
> My doubt is: what is that magical number 1250? Is it arbitrary?
> If it is, then how can they call this a "true Bayesian estimate"?
I agree with you. "A true Bayesian estimate" implies to me that the
estimate assumes a plausible model of the population of voters and
uses that model to derive the best possible estimate. I can't guess
what model that would be.
- Hide quoted text -
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> A true estimate tells me that they are estimating what is the
> value with maximum likelihood to be the true value of the movie's
> average vote, given that we don't count on an infinite number
> of votes (which is what we would require to obtain the true
> average vote). But this value, this formula is, the way I see
> it, "tainted" by this mysterious 1250...
> Let me put my doubt this way: We have to determine things like
> the following: we have two movies: move 1 has 2000 votes and
> they average 9/10; movie 2 has 4000 votes and they average 8/10;
> which movie is better?? (defining "better" as the movie for
> which an infinite number of voters would give a higher average).
> Movie 2 may have less average, but it carries more weight, as I
> have higher "confidence" in that value, since a lot of people
> have voted, whereas movie 1 has higher average, but it is
> "doubtful" -- not a lot of people have voted...
The 1250 value is unrelated to the usual statistical confidence
levels. With population sizes of 2000 and 4000 and all ratings
limited to the range 1/10..10/10, we can be very confident in the
significance of the difference between the two means of 9/10 and 8/10
(we're talking on the order of 10 standard deviations).
Maybe the rating formula is based on the assumption that the
moviegoers are voting with their feet. You could consider the
decision not to see a movie as a tentative thumbs-down on that movie
based on its reputation, or lack of same. But the formula isn't
directly based on this consideration, either, since the formula is
nonlinear in the number of voters -- it treats the difference between
2000 and 10000 votes as being much more important than the difference
between 10000 and 80000 votes.
You could rightly object that a linear model based on the
voting-with-their-feet assumption is too simplistic. A person might
prefer to see movie 1 but ends up seeing movie 2 instead just because
movie 2 happened to be the one showing at the local theater. Any
model you choose will be incomplete, since you can't model all the
elements that determine how many people will end up rating a given
movie on IMDB.
So the actual formula used by IMDB seems like a reasonable hack. Such
hacks would probably offend an anti-Bayesian, and if the enemy of your
enemy is your friend, maybe that makes it a Bayesian hack. Or maybe
the writer just thought that "a true Bayesian estimate" sounded more
scientificky than "a weighted average". |