Legendary Inventors (and Ultimate Contest)

[braven]

In a winner-take-all system there are only 3 calculations instead of the 12 calculations illustrated above (the winner against each other player individually, but the other players never compare ranks with each other). These means that the winner's rank is the only one that matters and it is always better for the winner to be the highest ranked player (if you can't win yourself).

It's only 6 calculations, not 12.
The 1st against 2nd, 3rd and 4th,
the 2nd against 3rd and 4th, and
the 3rd against the 4th.

That is not correct. If 3rd only went against 4th, than they would always gain points (because, of course, they always beat the player in 4th). Same with 2nd only going against 3rd and 4th.

[dudi2]

In a winner-take-all system there are only 3 calculations instead of the 12 calculations illustrated above (the winner against each other player individually, but the other players never compare ranks with each other). These means that the winner's rank is the only one that matters and it is always better for the winner to be the highest ranked player (if you can't win yourself).

It's only 6 calculations, not 12.
The 1st against 2nd, 3rd and 4th,
the 2nd against 3rd and 4th, and
the 3rd against the 4th.

That is not correct. If 3rd only went against 4th, than they would always gain points (because, of course, they always beat the player in 4th). Same with 2nd only going against 3rd and 4th.

Calculations are done vice versa. If you count them twice, you have to count 1st vs 2nd, 1st vs 3rd and 1st vs 4th also twice, giving 6 calculation versus 12 calculations. So its 3:6 calculations or 6:12 calculations. Either way, its only twice as many calculations as only taking into account the winner.

[braven]

It's only 6 calculations, not 12.
The 1st against 2nd, 3rd and 4th,
the 2nd against 3rd and 4th, and
the 3rd against the 4th.

That is not correct. If 3rd only went against 4th, than they would always gain points (because, of course, they always beat the player in 4th). Same with 2nd only going against 3rd and 4th.

Calculations are done vice versa. If you count them twice, you have to count 1st vs 2nd, 1st vs 3rd and 1st vs 4th also twice, giving 6 calculation versus 12 calculations. So its 3:6 calculations or 6:12 calculations. Either way, its only twice as many calculations as only taking into account the winner.

Oh, I see what you mean. I was counting some match-ups twice. You are right, it is only 6 total. Nice catch.

All of my other conclusions should be accurate still and it is still 3 matchup per player.

[demiurgsage]

These means that the winner's rank is the only one that matters and it is always better for the winner to be the highest ranked player (if you can't win yourself).

I have already heard this argumentation and as I said in another topic it may be easily fixed by give other players a lose from average opponent, not the winner. (under average opponent I mean weighted sum)

Also there was another suggestion:
after the calculations for 1st player are made we can exclude 1st player and give some points to 2nd player on the same basis but divided by some factor (4 or 5 for example).

[sprockitz]

These means that the winner's rank is the only one that matters and it is always better for the winner to be the highest ranked player (if you can't win yourself).

I have already heard this argumentation and as I said in another topic it may be easily fixed by give other players a lose from average opponent, not the winner. (under average opponent I mean weighted sum)

Also there was another suggestion:
after the calculations for 1st player are made we can exclude 1st player and give some points to 2nd player on the same basis but divided by some factor (4 or 5 for example).

Yes, I think the best compromise is "Winner Takes Most" instead of winner takes all. I would say even a divisor of 2 would have the desired impact.

In winner take all (or even most) you want to use average of points won/lost not average of ratings. So with a divisor of 2 Winner Takes Most and All would look like this...

Winner takes most:
1st earns Points x 1 against 2, 3, 4
2nd loses points x 1 against 1, gains points x 0.5 against 3, 4
3rd loses points x 0.75 against 1 & 2, gains points x 0.5 against 4
4th loses points x 0.667 against 1, 2, & 3

Winner Takes all:
1st earns Points x 1 against 2, 3, 4
2nd loses points x 1 against 1
3rd loses points x 0.5 against 1, 2
4th loses points x 0.333 against 1, 2, 3

[demiurgsage]

One of the strongest objection against the current system that last player gets more than one defeat during one game. I don't think that 2 defeats for last player in game with 4 players is acceptable. 1.5 probably is ok.

[Liallan]

A winner-take-all system would have it's own sets of problems. If you fall way behind, you might as well help out another player with a lower ELO ranking win so you lose less points from the loss.

You lost me there. How do you lose less ELO from a lower ELO person winning? Isn't it just about who "beats" you and not their actual place setting? That is, if I'm 4th place, I lost to 3 people, and I'm not aware that the place settings of those 3 make a different to my ELO change. Of course, it makes a difference to the other players' ELO changes.

The amounts of ELO points gained or lost depends on the size of the difference between each winners' and each losers' ELO rank. The bigger the differences, the more points won/lost. A low ranked player beating a high ranked player would result in a large amount of points exchanged and a high ranked player beating a low ranked player results in very few points exchanged. Equal ranked players exchange an average number of points. I did make a mistake in the quote above: You would want the highest ranked player to win in order to lose the least amount of points in a winner-takes-all format, not the lowest ranked player.

Well, that makes total sense, yes. I didn't get it cause you said it backwards. I understand the basic concept of how it works, which is similar to the concept of Trueskill used on Yucata. I don't know how the exact math works, but I get the basic concept. I know darn well if I can't win, I want the highest level person to win, not the lowest.

[ThierryC]

Hello,

I have a suggestion:

I think it would be great to add the year of the invention in the related text-strings, as it is displayed on the cards (but barely visible on BGA).
From my understanding, the developer of Legendary Inventors intended to create a game which both entertains and "teaches" (e.g., there is a booklet with information about the inventors).

What do you think?