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Proposed Change

I have done 15 runs on the chest during Straight to the Heart and have found that every time, a Spiked Eggnog, Wintergreen Candy Cane, Snowman Summoner, and Fruitcake are produced. Then one random one is thrown in the mix. I do not wish to change the article because I can not confirm what the chest produces during the quest The Strength of Snow.


The Wintersday Chest is a chest which spawns upon completion of the Wintersday 2007 quest Straight to the Heart and The Strength of Snow.

The chest will contain one of each from the following list.

Spiked Eggnog
Wintergreen Candy Cane
Snowman Summoner
Fruitcake

The chest will also contain one random item from the following list.

Spiked Eggnog
Eggnog
Rainbow Candy Cane
Wintergreen Candy Cane
Peppermint Candy Cane
Fruitcake
Snowman Summoner
Yuletide Tonic

A total of 5 items will be produced from the chest.


Can anyone else confirm this? Unindal 19:14, 26 December 2007 (UTC)

Yes, i got over ten chest drops and they all had one of the first four listed above. They all droped on the same spot each time I opened the chest. The fifth drop is one of the items of the second list. I'll change the article. -{[ PUL ]}- 11:59, 4 January 2008 (UTC) edit: i did 'The Strength of Snow'.
100 runs produced a total of 100 snowman summoners. A friend of mine did 200 runs for a similar result, making me think summoners are not part of the random component of the drop. Tonics appear to be rare (~3% of the drops). Everything else appears to be occurring with roughly equal rates. Yamagawa
206 runs, nary a single drop of a snowman summoner in the random slot. I'll be removing it from the main page. Yes, it is part of the guaranteed drop, just you never get a 2nd one from the random drop. Yamagawa 17:21, 25 December 2008 (UTC)

Drop Rates

I've 3 ideas on how to go about getting a drop rate for the miniature.

  1. Count how many minis drop after x runs. -- I deem this invalid as many people would stop after they get a mini, skewing the results.
  2. Count how many runs it takes to produce a mini -- This can probably be worked into a valid scheme, but would likely require hundreds (thousands?) of reports to produce an accurate figure
    I think this would work by taking all the 'xxx runs for a mini' reports. Discard any reports with fewer than 200 runs -- those were the lucky few and the light weights. Discard any reports of no runs. Take the remaining reports, and discard all but the last two digits. So if someone reports a mini after 337 runs, count it as 37. If someone reports a mini after 1243 runs, count it as 43. We are essentially breaking down any report of xxx runs into sets of 100 runs, and ignoring any set of 100 that fails. Take the average of these figures. The lower the average, the higher the drop rate. As an example: If the drop rate is 10%, the average would be near 10. If the drop rate is near 5%, the average would be closer to 20. If the drop rate is 2%, the average approaches 50, but here is where things get a little hinkey. Lets say for this example widget x drops 2% of the time, and we get 10000 reports of xxx opens to get it. The first try, all 10000 people opened the chest and 2% (200) got the widget. The second try, 9800 people opened the chest, and 2% (196) of them got the widget. On the third try, 9604 people opened the chest.... Yes, we know all of them got the mini somewhere in that 100 opens, but more of them got it towards the beginning of their opens than towards the end. Ultimately, the closer the average # opens for the widget is to 50, the flatter the curve that averages to 50, and the more rare the item is.
    I'll need to see if I can come up with proper averages for a 0.2% drop rate, a 0.1% drop rate, and a 0.01% drop rate, to have any idea how workable it is.
  3. Process of elimination: Identify the other drop rates. They are probably fairly round numbers. Once you know how often everything else drops, what is left is all mini.
    This works by starting with an assumption: That a single die roll is determining the outcome from the chest. This may not be true, and if it isn't it may not work at all. Drops appear to be breaking down into 3 categories: very rare (mini), rare(tonic), and common (the other items). Add a second factor: Assume a 1000 sided die is cast (REMEMBER: Programmers are people too. They often like things nice and simple, this drop rate may be no different). Assume that the 6 common items share a common drop rate. Assume a standard drop rate for the tonic: 5%. (Lotta assumptions, yes, house of cards and all that, yes....). 5% drop rate means 50 of the 1000 sides on the 1000 sided die go to the tonic. That leaves 950 sides to split between the 6 common drops and the mini. 950 / 6 ~ 158. This gives us a chart like:
    Wintergreen CC: 158/1000
    Peppermint CC: 158/1000
    Rainbow CC: 158/1000
    Fruitcake: 158/1000
    Spiked Eggnog: 158/1000
    Eggnog: 158/1000
    Tonic: 50/1000
    That gives us a remainder of 2/1000, which goes to the mini.
    Now, that's a house of cards there, but a darn sight more than I've seen elsewhere, and gives people a reasoned guess to look at. Can we prove or deny it? Either showing the tonic drop rate is not 5%, or showing the minis are dropping at a significantly different rate than 1/500 will do.
Yamagawa 17:21, 25 December 2008 (UTC)
Lets see here... for method 2 above, a 10% drop rate would average 9.997, a 1% drop rate would average 42.265, a .2% drop rate averages 48.832, a 0.1% drop rate is 49.666, and my math must be kaput because a 0.01% drop rate averages 50.4.... my understanding is that the average should approach 50%, but will not pass it.... !%#@%^ pentium floating point math! In any rate, to figure the drop rate with any accuracy with this method, I would see us needing 100s of dropsYamagawa 00:40, 26 December 2008 (UTC)
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