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Math question
glithert
The richest duck in the worldDuckburg, CalisotaRegistered User regular
The richest duck in the worldDuckburg, CalisotaRegistered User regular
If I have a scale that goes from 1-10, with 5 being the average, and my friend has a scale that goes from 1-10 with 7 being the average, how would I go about converting a score from my scale to his scale and vice-versa?
I suck at math and I have no idea how to even begin tackling this.
I suck at math and I have no idea how to even begin tackling this.
glithert on
Posts
I'd start by equating 5 to 7, and creating two separate conversion factors, one for below the average, one for above.
Think of it this way: if I said "on a scale of 1-10, how attractive is this girl?" to one group and they averaged a 5, and I did the same thing with another group/*another girl* and they averaged a 7, we COULD NOT combine those two scales because they measure different things, even though they're both technically 1-10 scales.
We might be able to ask "if person P rated girl X with a 6, how *might* person P rate girl Y, who averages a 7?" That's a nontrivial statistics problem and requires domain knowledge about the problem.
Basically, we need more info.
So, if I gave them a 10 and an 8.5, respectively, is there any way I can convert those scores to his scale? Because I am dumb and I didn't really understand what you said.
If you say somebody is a 7.5 for example (half way between 5 and 10), to get to his scale you could just call it 8.5 (half way between 7 and 10). Although, I'd have to say his scale is... odd at best. He's allowing for a very specific description of ugly people, and a very narrow description of attractive people. Who the hell does something like that? Either that, or he's somehow trying to account for the fact that he has low standards, so the run-of-the-mill woman is considered rather attractive, which leads to a whole different argument against his measurement methodology that I'll save for later.
Damnit, studying for finals makes me overly analytical.
edit: A more formal description would be to make an odd linear interpolation. For, something like an 8 on a 5 being average scale could be solved thusly:
(10-5)/(10-7) = (8-5)/(x-7) where solving for x would yield an 8 adjusted for the "7 is average" scale.
So...wait...you're using algebra to figure out the comparative hotness of the Jessicas? Seriously, that's like comparing a fine steak to tiramisu. Either way, you win.
My advice...smack your friend in the head, tell him to start using a scale that makes sense to the rest of the world. Then find some alone time to do more research, if you get what I'm saying.
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so this is actually a tricky problem and it requires a lot of simplifying assumptions about how hotness scores are distributed and how we can infer parameters on those distributions.
here's the easiest way to do it: come up with anchor points. get him to plot out his scale for various girls, and then use that scale as a reference point for your own numbers.
let's say I have some girl susie that I want to convert to his scale. I pick one girl A slightly hotter than susie, and one girl B slightly uglier than susie. Your buddy ranks A and B, and susie should be approximately at the midpoint of the two numbers.
if you want a permanent conversion factor, have him go through each number and name a girl who would qualify for each score. do the same with your own scale. then you can make a mapping between the two based on the girls. e.g. "jessica alba is an 8 on your scale but a 9 on mine, so when I say a girl is a 9, she's really an 8 for you"
if that seems annoying, think about the problem this way: if you did something stupid like subtract 2 from his scores (to move his average of 7 to your average of 5), then 10's on his scale become 8's on yours. He could never get above an 8, and that's obviously wrong. so it's a tricky problem, but the solution I gave should work.
Not if you use nonparametric tests :winky:
Lets pretend that you have an average of 5, with a standard deviation of 2, and your friend has an average of 7 with a standard deviation of 1. So in this case, your 5 would equal his 7, your 7 would equal is 8, your 9 would equal is 9, etc.
GT: Tanky the Tank
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Start using a percentile score, that is, the score a person receives is the percentage of people they score higher than. Now you are each using the same numbers (0-99), and a percentile score of 50 is always right in the middle.
The problem you were having before comes from the fact that you were making a 5 the 50th percentile and your friend was making a 7 the 50th percentile score. Hope this helps.
Quick apology: I'm studying for a stats final so this is tmi, but I figured it's interesting enough to think about/mention.
Your strategy assumes a single normal distribution on hotness scores even though the data is bimodal with a normal "true" parameter for the girl's attractiveness, a *second* normal parameter for the individual scale bias, and a *third* zero-mean parameter for noise. All three have separate standard deviations. You would then need to learn all parameters (the expectation maximization algorithm will do the trick) from parallel ratings, and then take a girl's true attractiveness and add it to the scale bias parameter for both of you to get two numbers that represent her score on both scales. It's exactly problem 2 from this: http://www.stanford.edu/class/cs229/materials/ps4.pdf
But even if you did it your way, you need enough unbiased data points to see where everyone falls. If I just pick hot girls and they fall in the 8-10 slot, you're going to get biased mean/std numbers, which is probably why your friend's scale has a mean of 7. No one really argues about whether a girl is a 1 or a 2, so the scale is only realistically used to measure girls who are maybe 6-7+ and that throws everything off.
That's why the anchor points work-- it enforces a uniform distribution and you should be able to mentally interpolate between girls to decide how the density falls.
Not that this is a very serious topic in the end, but no, that's not simple it's wrong. You couldn't even attain a value of 10 on the 5-scale that way, the highest would be 7ish for a "perfect ten."
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I gotta say, I had no idea it would get this complicated. Maybe the real solution is to just call him an idiot for not seeing that Biel is definitely the hotter Jessica?
You seem to be considering "average" to signify "mean;" i.e., if we consider all women, their "average" hotness should be the middle of the scale. Your friend may be using "average" to signify "median," or possibly "mode."
my unofficial autobio will be accompanied with tips on how to smile
cause I've found that when they don't see you frown, they never know that you're a threat
and they don't sweat you when you came around
well if you want to take the massive skew into account
from his, for values above 7: (value-7)/(10-7) * (10-5) + 5
for below: (value-7) / (7-0) * (5-0) + 5
But... Average does mean "mean". That is exactly what it means.
Call him an idiot.
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As i wrote above, simplied
from his, for values above 7: (value-7)/3*5 + 5
for below 7: 5 - (7-value)/7*5
This'll give you what you want. Anything more complex is silly for a hotness scale
not always
my unofficial autobio will be accompanied with tips on how to smile
cause I've found that when they don't see you frown, they never know that you're a threat
and they don't sweat you when you came around
For example, even lightly equipped starships in the Star Trek universe can travel Warp 7, but only the most powerful and well equipped can travel warp 9 and beyond: and it takes some malevolent force to bring you past warp 9.75.
So in this way, you could equivocate women to starships. A lot of them can make it to warp 7, but how many can reach warp 9.75? Not many, that's what.
And yes, for those who must be picky, that does mean that Janeane Garofalo is the hotness equivalent of Transwarp. Does that mean that you would occupy all points in space simultaneously if you were to meet her in person?
Perhaps.
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