Arnold, I thought about the same new-points-system to get it more balanced - the old system is overrating the good places and underrating the bad places. So solved levels divided by average place will be better - a bad score damage more your overall-performance. Thanks for your support.

My idea for table 5 in July 2013 was to estimate more than first places (shortest solutions) only. A player that consistently scores second places is very good, but can't be found in tables 2 and 4. However I agree that the current computation of points in table 5 is not entirely fair and objective. For instance, in the spirit of utimm and MnLsDad: two players solve 2 levels each and score the same average place 5. The first player scores places 1 and 9, and gets 1.1 points now. The second player scores places 5 and 5, and gets 0.4 points only. So maybe this is a more fair and objective formula: total solved levels divided by average place per level. In the above example both players would get the same amount of points: 2/5=0.4 points each.

admin1, thanks a lot for the extra column in table 5! By the way, utimm is right: Average points per level in table 5 isn't the inversion of Average place, but the outcome of Points in table 5 divided by Completed levels in table 1. In my case for instance: 6064 / 15078 = 0.4. So, the extra column isn't redundant, but offers new information.

Utimm: I am intrigued by your "What about a new list with solo first places?" idea. Not from a "statistic" standpoint, but one that would then link to each puzzle that has a solo 1st place solution. It would be a "king of the hill" list that sends a challenge to everyone to come knock me off the top. Comments anyone? Admin1?

Utimm: You are correct. Since that table was Arnold's baby (See posts: 11 Jul 2013 18:21 GMT, 27 Jul 2013 15:41 GMT) I should let him defend it, but it has some merit. I see Admin has added the requested column. Sorting on place I see no large alteration. JotaCartas stands out. I've seen that he radically changed his play style a couple of years ago. (He may actually be an AI agent being tested by Google - j/k - but if he is, he got a software rev.) I would still question the entire place scheme. If 25 people found a shortest 50 move solution and you do it in 51 moves, you get 2nd place status. Realistically you deserve 26th place. In your example, the table rewards a player with 1st & 99th over one with two 5th place finishes. If you were at a horse (or dog, people, turtle, auto)race, would you bet on a participant with consistent 5th place showings, or one who's won half his competitions?

Here is a last example, that there is something wrong with the point-system in Rating 5. Player A: Places: 5 and 5 - av. Place=5 - av. Points=0,2 -- Player B: Places 1 and 99 - av. Place=50 - av. Points=0,505. Which player had a better performance?

My average place is about 3,8 and that's not the inversion of my average points.

Sorry, but the average place is not the inverted average points: Here are some easy eaxamples - 2 Games, Places 1 and 9 - average place: 5 - average points: 0,555 --- Places 1 and 99 - average place: 50 - average points: 0,505. The av. places difference is 45 and the av. points diff. is 0,05!!!

Arnold: My point - if I have one - is that displaying both would be redundant. Either can be used to compare player's "efficiency" assuming that both are striving to find the shortest solution. I suspect some - perhaps many - are content finding any solution. They play a different game, so comparison is meaningless. Realistically, the one thing shown by all these tables is that if your username is on these lists, you are hopelessly addicted to the pointless pursuit of pushing little, digital boxes around a phantom maze and you probably should get up and go outside more... I think I will... after I finish *one* more level.

MnLsDad, you're right, value of last column is inversion of average place, in my case for example: 1/2.5 = 0.4 (cause 1/0.4 = 2.5). I agree with utimm however that average place 2.5 is a more concrete and comprehensible value than its inversion 0.4.

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