In the first 200 votes since I started keeping track (all post cull) I saw 308 unique items.
When I finish my second set of 200 later today, I will calculate the number of unique items for that set, and the total number of unique items over all from the 400 votes... although it should be noted that my sets of 200 are actually being done 100 votes at a time at different times of the day (as part of an experiment to see if time of day does really affect voting as far as items seen and number of repeats)... so the tallies might be different then if somebody did the same experiment but voted straight through on 200 votes for each set (assuming that my other experiment shows that there is a difference in voting by time of day).
my first two hundred is comprised of an early morning set and an early afternoon set... today I am working on a mid morning set (about finished) and a later afternoon set (all based on pacific time).
Eric said that he hit 600 votes the day after the cull.The cull happened somewhere between seeing my 500th unique item and seeing my 600th unique item.
After the cull, I've seen at least 534 unique items. I may have seen as many as 634, depending upon when exactly the cull occurred in the middle of my voting.
Assuming there isn't a second cull between now and then, I should get a chance to put gbonehead's 676 item estimate to the test before the end of voting.
I should have written down the formula; it's on my whiteboard at work. If I remember correctly, it all boils down to the following approximation:
a^(k-1) + aU + (U-1) = 0
Where
k = Total number of non-unique items seen.
U = Number of unique items seen.
a = (1 - 1/T) where T = total number of items.
So if you've seen k items, U of which are unique, you can pretty quickly narrow down a value for T with a little guesswork :)
I started deriving the actual equation today; it was a wee bit more involved than that one.
I wasn't. Now that I'm looking at my whiteboard, it was:
a^(k-1) - aU + (U-1) = 0
:)
(P.S. current estimate after 366 votes (732 items, 448 unique) is 683 items currently in the pool. I've had a recent spurt of unique items, which has bumped it up a bit.)
(P.P.S. and that's not how I'm estimating anyways; I'm making a guess at T, then evaluating a = (1 - 1/T), and then evaluating:
U = (1 - a^(k-1)) / (1 - a)
... until I find a closest value to the observed U; e.g. for U=448 and k=732, a value of T=683 is closest :)
Hrrgh. Math bad.
After nearly 1,000 votes, why is it that...
...the items I've loved the most, the ones with a real spark, I've only seen one or two times each?
...whereas lots and lots of other items have appeared over a dozen times?
Does the logic of the voting system naturally show middle-tier items far, far more often than "top" items?
Yes. I believe that is how it is working out. I think we are mostly just shuffling the mediocre items around a bit.
I think something should be done. I'm not sure another culling is the right thing to do as this supports attempts to vote up a voter's own item as it will likely come up more frequently. What if they just save a copy of the first results and then reset all the scores? At the end the scores from the two sets can be averaged if there is a large discrepancy.
Next year, two weeks voting, max, with a culling after one week.
After nearly 1,000 votes, why is it that...
...the items I've loved the most, the ones with a real spark, I've only seen one or two times each?
...whereas lots and lots of other items have appeared over a dozen times?
Does the logic of the voting system naturally show middle-tier items far, far more often than "top" items?
I am guessing others are seeing at the same time based on couple voting posting together at the same time.
I believe it is weighting the item against similar weighted items and occasionally testing it against higher rated items.
Not all items will be up voted by all people who have different criteria so there will be some breakout. With 20 votes some will have 12 better and 8 not better. Some will be equally weighted 10 better and 10 not better and so on. I believe this gives an item a weighted mathematical average since items seem to float up and down.
IMO doubtful the top 10 have 100% better votes based on the audience size.
Say there are 1,000 items or more IMO Paizo is probably looking for a primary initial judge's pool of 20% +/- 5% to be safe this year. Start at the top computer ranking group and work their way down probably with a Golden Ticket or Two for each judge.
There is some room for an item to climb based on posts in the first few days if it just had bad luck initially at the voting table.
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IMO doubtful the top 10 have 100% better votes based on the audience size.
Say there are 1,000 items or more IMO Paizo is probably looking for a primary initial judge's pool of 20% +/- 5% to be safe this year. Start at the top computer ranking group and work their way down probably with a Golden Ticket or Two for each judge.
There is some room for an item to climb based on posts in the first few days if it just had bad luck initially at the voting table.
I have to agree with this 100%. As the pool of votes grows, each item is more likely to move to its proper place in the rankings. If every competitor voted their item the best, better items may start off lower in the rankings. But the general voting population will not have this bias and vote based on the merits of the item, moving the item to its proper place and the competitor's item will move lower.
I think it's completely random. Since there are fewer top-tier items, it's only natural that we'll see medium-tier items more often, and end up with the illusion that we're not seeing the "really good stuff".
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I think it's completely random. Since there are fewer top-tier items, it's only natural that we'll see medium-tier items more often, and end up with the illusion that we're not seeing the "really good stuff".
I mostly agree with this except there are a few items I just don't see very often at all and others I see constantly...often almost back to back. So I'm trying to not think too hard about the algorithm and probabilities (I'm unlikely to understand them anyway) and trust the system.
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After nearly 1,000 votes, why is it that...
...the items I've loved the most, the ones with a real spark, I've only seen one or two times each?
...whereas lots and lots of other items have appeared over a dozen times?
Does the logic of the voting system naturally show middle-tier items far, far more often than "top" items?
I blame the neither button. The better items always get voted on by everyone. However I'm sure there's a quite a few people who regularly use the neither button when two lesser items come up. As these lesser items get voted on less, the probability mechanism prioritizes these lesser items to get their vote numbers up to the better items. Combine this with the greater number of mediocre items to brilliant items and you have your reason why you see some items a lot more than others and the better items less so than the mediocre ones.
Best Regards
Herremann the Wise
That looks much closer to what I was expecting. I hit a wall while attempting to derive an equation similar to yours, but I was playing around in the same general ballpark.
I started by trying to generalize or use a Fourier Transform or something, my eyes glazed over very quickly, and I then took a different tack; I ignored the fact that we see two items at once, and treated it as a singular series.
Then, for the (k+1)th non-unique item, the expected number of unique items is:
U(k+1) = U(k) + (T-U(k))/T or more simply:
U(k+1) = 1 + (1 - 1/T) * U(k)
If you define a = 1 + 1/T, you can turn that into a series pretty easily, getting:
U(k) = 1 + a + a^2 + ... + a^(k-1), thus
U(k) = (1 - a^(k-1)) / (1 - a)
It's grossly simplified, because it's not a completely independent series. I started to do the math for the actual series, but it's pretty damn ugly, with U(k+2) in terms of U(k); you have to instead go by voting pairs instead of single items and I suspect that it pretty much degenerates into what's above anyways.
So I went with that :)
Thought this would die with the whole gate tags.
It's grossly simplified, because it's not a completely independent series.Yeah, that's where I hit a wall. Thankfully, you seem to have the independent series approximation covered.
(Correct me if I'm wrong, but an independent series approximation is going to slightly overestimate the number of unique items viewed and underestimate the population size, yes?)
Exactly. that (1 - 1/T) term gets a lot more complicated. As the total number of items viewed increases, the two estimates will get closer, and it's a good rule of thumb number, but I'm not sure how far it's off.
Figuring that out would be about equivalent in difficulty to actually figuring out the expression. I came close, but it's pretty hairy and I haven't yet double-checked my equations. Maybe Monday.