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In the winterbash leaderboard for Windows Phone.SE, there are currently 7 hatted users, 2 of whom are in 1st place, and 5 of whom are in... third?

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2nd place is just the first loser and we wouldn't want anyone to feel bad, now would we? ;) –  Anna Lear Dec 20 '12 at 20:39
    
The simplest way to explain this: Consider a competition with ten competitors. If nine of these people tie for first place, the last person is not in second place, and he does not get a silver medal. He is tenth, dead last, and he gets nothing. –  meagar Dec 20 '12 at 21:26
    
@meagar - That's a great distillation of what most people use for ranking. I have seen systems where you want to give out as many prizes as possible - but even toddlers quickly get suspicious when everyone gets a blue ribbon for 1st place. I suppose there are "systems" where you juice the results to have more people place highly, but this listing of more than one person in first and third is a well-established interpretation of a placement algorithm IMO. –  bmike Dec 20 '12 at 21:43
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Can't believe people would downvote this. Thanks for the bug report. –  Lance Roberts Dec 31 '12 at 20:28

3 Answers 3

What else would you expect to happen in the event of a tie? Randomly sort them into place? The first two users are tied for 1st place, and the next five users are tied for third place. If either of the first two users gains another hat, one will be in 1st and one in 2nd, so 2nd place has to be "skipped" to keep the leaderboard accurate. Saying the next five users are tied for 2nd place is untrue as without gaining another hat, there is no possible way they could ever end up in 2nd place.

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I get the point (it's not a bug), but I think counting places ordinally makes just as much sense - if the top two (or three or four) are #1, they're in the same place, right? So the next hattish echelon down would be the second place attained by x users. –  Daniel Dec 20 '12 at 20:46
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@Danielδ - if you watch any sport in the world...have you ever seen the sort you're proposing? I always see exactly what we're doing, which is accurate. –  Nick Craver Dec 20 '12 at 20:48
    
@NickCraver Point conceded. I haven't watched ranked sports :) –  Daniel Dec 20 '12 at 20:51
    
@NickCraver It's pretty cool the algorithm handles ties so well. If Jin weren't so overworked, it would be cool to have the icons sit on a three tiered podium Olympics style.... –  bmike Dec 20 '12 at 20:52
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What? So because you have five people in "third" place, you call the next person after that in "ninth" place? That makes no sense at all. The first three positions are positions 1, 2, and 3, not 1, 3, and 9! –  tchrist Dec 20 '12 at 21:05
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@tchrist: It would be 8th place, and it makes perfect sense. You have to think of the leaderboard in a way that there are no ties. If they were all broken, the first two users who end up in places 1 and 2, and the next five would end up in 3, 4, 5, 6, and 7. Since there is no tie-breaking method here, they are all listed at the highest possible rank they could achieve with their current score. –  animuson Dec 20 '12 at 21:09
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@animuson This is the nuttiest thing I ever heard. We never did it that way. If two people tie for the first-place prize, the next person doesn’t get the third-place prize, they get second-place prize. There would be mayhem otherwise. –  tchrist Dec 20 '12 at 21:11
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@Danielδ This is how Foursquare sorts their leaderboard as well. /random info –  Aarthi Dec 20 '12 at 21:16
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@tchrist This isn't nutty, this is how ranking works. If 9 people are ahead of you, and they are tied, you are not in second place. You are in 10th place. The 1st ranked through 9th ranked people are above you in the standings. You may have played differently in whatever amateur made-up rules you used, but this isn't how it works in any organized system. Your claim of "mayhem" is particularly dumb, as this is how it is done everywhere. –  meagar Dec 20 '12 at 21:32
    
I can see it in competitions which award prizes, but this is just a leaderboard with no prizes (only hats). Prize competitions are contractually obligated to award those prizes to its participants if they meet the criteria for the prize (in this case 1st place). If the competition rules don't outline any way to break a tie for 1st, they're contractually obligated to award a 1st place prize to both 1st place participants, which leaves a 2nd place prize unawarded. Many competitions will award the 2nd place prize anyways, because they have it. –  animuson Dec 20 '12 at 21:42
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Consider the Olympics. In event of a first-place tie, they award gold, gold and bronze medals. When there are two people who are better than you, you are not second best. Those two people can however both be tied for being the best. –  meagar Dec 20 '12 at 21:46

This is what happens when there is a tie. Rather than forcing one person into first or second place, you let both be #1 and resume counting the next contestant(s) from #3 since the two people scored higher than them.

A mathematician might feel that both people in slots one and two should have (1+2)/2 place or something nonsensical, but most sports events just put more than one person in a slot and then fill in from below when there is no procedure to break a tie.

If we wanted to be nice, perhaps we would let all the losers get bronze?

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Actually (|1>+|2>)/sqrt(2)'th place would kind of make sense... never give a quantitative scientist ideas ;-) –  David Z Dec 20 '12 at 20:47
    
If there are three times: 2 second, 4 seconds, and 5 seconds, then those are by definition 1st, 2nd, and 3rd. It doesn’t matter how many people fall into each bucket. –  tchrist Dec 20 '12 at 21:06
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@tchrist No, those most definitely are not "by definition" 1st 2nd and 3rd. If you have 5 people clocking in at each time, you have fifteen people, and by definition you have 1st through 15th. All those people who clocked in at 5 seconds are tied for last place, not tied for third place. I'll say it another way: If the top three runners in the world race, and two of them tie for first, the last runner is the third fastest, not the second fastest. –  meagar Dec 20 '12 at 21:39

I am the Real Number 8

Real Number 8

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