16

Happy holidays everyone! Bricks Stack Exchange is once again participating in the Winter Bash celebration with the rest of the Stack Exchange network!

While this is a "just for fun" activity, there are also a few different prizes available this year.

75253

Any ties will be resolved randomly. You can track your hat progress on the Bricks SE leaderboard. Bricks mods are excluded from this portion of the competition for fairness reasons.

  • 3 winners will be selected randomly from everyone who earns at least 3 hats, excluding the user with the most hats, as they already get a prize. They will each win 10267 Ginger Bread House:

10267

Thanks go to TLG for supporting this activity by providing prizes for us!

All winners will need to provide me with a shipping address within 1 week of the end of the competition, or their prize will be forfeited. I'm willing to be flexible if winners would prefer a different set of equal value if they already have the prize.

This contest is unfortunately not open to residents of Iran, North Korea, Myanmar/Burma, Zimbabwe, Sudan, Syria, and any other U.S. sanctioned country and where prohibited or restricted by law. This is a legal requirement that is outside of both our control and The LEGO Group's control.

The goal here is to facilitate positive engagement on the site both to celebrate the holidays and because this is the time of year when we get a lot of new traffic from parents and others just getting into LEGO. My hope is that providing prizes for this will facilitate meaningful engagement on the site. Please don't make a mess of things just to earn hats. :)

Good luck and have fun!

Update

The winners have been announced here.

  • 1
    I feel the stated goal ("The goal here is to facilitate positive engagement on the site") conflicts with the observed reality. See my question on meta for the details and please share your opinion. – zovits supports GoFundMonica Dec 12 '19 at 12:15
  • 3
    "3 winners will be selected randomly from everyone who earns at least 3 hats" so that includes the user with the most hats, meaning there is a very small chance of getting two sets? – David Dec 12 '19 at 15:04
  • 2
    @David Good point. I've changed this to prevent the top user from winning a second prize. – jncraton Dec 12 '19 at 15:59
  • 3
    I assume this only includes hats earned specifically on this site? – Chris says Reinstate Monica Dec 12 '19 at 17:22
  • 3
    @ChrissaysReinstateMonica Correct. This is the hat count from the Bricks leader board linked above. – jncraton Dec 12 '19 at 18:02
  • I qualified for the Copy Editor gold badge before WB ended, but it looks like it hasn't been awarded yet, and I don't know if it will before WB ends. Does that count for anything? i.imgur.com/RDX6ZoW.png – Alexander O'Mara Jan 1 at 23:40
  • How do I provide you with a shipping address? I'm not very good at figuring this kind of stuff out. – technicguy1 Jan 2 at 17:00
  • @technicguy1 I'll ask the winners to send me info via email after I draw the names. – jncraton Jan 2 at 17:07
  • @jncraton Oh ok – technicguy1 Jan 2 at 17:11
11

I have a suggestion for next year: perhaps we can make the lottery draw for the prizes weighted? So everyone who gets at least three hats gets as many entries as they have hats. This would stimulate further participation, as it is fairly easy for a regular Bricks.se member to get 3 hats, but getting beyond 5-6 hats requires several secret hats, which can typically only be gained through very active engagement. And the more you participate, the higher the chance are that you win a prize, as your name gets entered multiple times.

As a slight variation we could award people extra entries for each hat they have over the required minimum 3 hats, so basically #entries = #hats - 2, with #hats >= 3 to qualify.

  • 1
    I think this is a good suggestion, and I was thinking about this myself. – Alexander O'Mara Dec 28 '19 at 4:42
  • 3
    This sounds like a good change to me. – jncraton Dec 28 '19 at 14:20
  • 1
    I was planning to make this suggestion as well, maybe with an additional restriction on the maximum number of "lottery tickets" one can receive, to discourage the kind of cutthroat competition observed this year (and to a lesser degree yesteryear). – zovits supports GoFundMonica Jan 3 at 9:30
  • 2
    @zovitssupportsGoFundMonica: the change I propose is to counteract the fact that there is an initial surge in participation which then levels off when people have reached the 3 hats, as competing for the top spot requires a whole different level of dedication :). With my proposed change, hopefully some people will continue to interact to gain more hats and increase their chances of a prize. That said, if we make the maximum high enough (getting 10 hats is quite difficult) that could work, or perhaps we say 1 ticket for every 3 hats, with a max of 4 tickets per person? – Phil B. Jan 3 at 11:31
  • @PhilB. "competing for the top spot requires a whole different level of dedication" You're absolutely right, and in my opinion if somebody has this level of dedication then being on the top spot should be it's own reward. - "That said, if we make the maximum high enough (getting 10 hats is quite difficult) that could work, or perhaps we say 1 ticket for every 3 hats, with a max of 4 tickets per person" I'm completely on board with this, 10 seems attainable but only through work and dedication. – zovits supports GoFundMonica Jan 3 at 23:17

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .