RGun said:

That's a nice variant idea. I agree that a minimum number of plays prevents someone from posting one lucky first attempt and stopping. I'll think about this more, and maybe in February I'll run one tournament without the variant and one tournament with the variant, seeking feedback as we go. I may also try and do a poll before the February tournaments to elicit feedback ahead of time.

Oh, and I forgot to say another reason why I like your idea:

3) Losses have impact in this system. Losses matter - in life and in games, both.

juicebox said:

The Rabbit decks might take some skill in building the deck, but not in piloting it. If you make a deck that sees less than 10% of the cards IN the deck, and a similar number for the encounter deck? Where is the skill? Where is the decision making .. there is none.. you are just slamming cards and powering though. You are completely relaying on luck, luck in your starting hand, luck in your encounter draw and luck in your card draw.

Is it a legit tactic? Sure. I think the "risk" vs "reward" factor in the rules and scoring system is pretty cool. Like you can build and run such risky decks and get awesome scores. That doesn't mean the Rabbit deck is any less luck based and THAT is the crux of it.. luck has nothing to do with skill.

The great thing about the Rabbit decks is that they are very risky, and it is a big tactical decision to run one in a comp, as you know that MOST times you play it you may loose to a single treachery card or drawing a big monster. So do I think it is a legit deck.. sure I do. Do i think it takes less skill to run one? Yes. Do I think the score system should reflect the losses so Rabbit run decks are not the ONLY legit deck to run? Absolutely

RGun said:

Dude, we are talking about a game that children play. My 9 year old son can not do this math. It is confusing to even adults. I mean look at your instructions!! You are going to introduce a new problem, input error and do not say your not. You have fractions, % calculations, rounding, and averages. .. . . what a mess. You also do not say what you do with your remainders... as it is not 1.67 it is in fact 166.6666666666667, so witch one will people use and even if you did say that you have placed a "rounding" factor on top of it...

Still the REAL problem with it, apart form it being so unwieldy, is that your score system encourages conceding. Like why would I play to the end of a match when I can just concede for a score of 0 that has nearly no effect on the final tally. At least not the same effect as limping over the finish with a high score. The ENTIRE point of trying to do this score system is to minimise people grinding at a quest and just playing it a billion times until they get a good score. Your system doesn't really address that, when you can just call concede get a score of zero and have nearly no effect on your final score. In fact your score system encourages conceding, witch was what we are trying to eliminate in the first place.

We got to just keep this system SIMPLE.

On the plus side, if we are all playing the same number of games (say 5) then the formula simplifies to

"Add up all your winning scores, divide by how many games you won then divide by how many games you won again."

That seems pretty simple.

booored said:

We are talking about a game where you have to be a lawyer to understand all the rules and you complain about some simple maths? Using a win/loss ratio without having periods is almost impossible, as a ratio is per definitionem a quotient. If one can make sure that the denomintaor is always a 5 or a 10, this problem could be solved though.

I know what you want to say about the problem about conceding, but I think that's part of the general scoring system in which an epic fight against a series of mean encounters gives you a lesser score than rushing through some easy encounters. However, as some folks already said: the more you loose, the bigger the impact. Even a good deck and a good player can loose from time to time, if they draw some bad encounters, so I don't consider it a big problem if a single loss has only little influence on the overall result, as it allows you to take at least some risks instead of paying it safe all the time.

the point though is that you can concede at anytime.. like say yo play and you are going to win but you do not like your score.. you can just concede and get a better "meta" score by calculating zero.

Example....

Using the values in the 1st post...
((75 + 100 + 125) / 3) * (5 / 3) = 166.6666666666667 (1.67)

BUT.. what if I did the same.. but decided to concede two of them at the last minuet... this calculation becomes...

((0 + 100 + 0) / 3) * (5 / 3) = 55.55555555555556 (55.6)

so basically what this score system dose is say... just concede ANYTHING that is not a good score. In fact playing a ton of games at 0 is better than just getting a great score 1st turn. This system encourages conceding even more than the old one.

So yeah.. this is useless even worse than the original, good idea .. but needs work to iron out these problems

oh wait.. hang on.. I think I have this all backwards.. the calculation would be...

((0 + 0 + 100 + 0 + 0) / 1) * (5 / 1) = 500

Hmm.. ok .. I see how this works now... still complex.. but your right it dose work a lot better than I thought.

Yeah, I'm astonished, too! I thought I could trick the formula somehow. But that's not possible. At least as long as the score you want to take out of calculation is less than the double of the average of your other scores. Basically, this is because we're dividing 2 times through the number of the games won. (The exact value is not 2 but 2.25).

Example1 : 4 games won with score 100, 1 bad game won with 220.

Without counting the last game as a loss: (100 + 100 + 100 + 100 + 220)/5  * (5/5) = 124. With counting it as a loss: (100 + 100 + 100 + 100 + 0)/4 * (5/4) = 125. So you're better of not counting it as a loss.

But example 2: 4 games won with 100, but the bad game has been won with 230:

With the bad game: (100 + 100 + 100 + 100 + 230)/5 * 5/5 = 126. Without the bad game: (100 + 100 + 100 + 100 + 0)/4 * (5/4) = 125. So here it would be better to willingly get all heroes killed and lose the game.

How often does it happen that you have a bad game that has a score more than 2.25 compared to your other scores? I guess it could be a few times, because of Dead Marshes (consider a game where you have to skim the encounter deck twice for Gollum...)

We could even improve the formula if we divided 3 times through the number of the games won. So the new formula would be:

(sum of all scores) * (g * g) / (w * w * w)

where w is the number of the games won, and g is the number of all games (won or lost or conceded). The factor needed to take advantage of willingly losing a game is (3g² -3g + 1)/(g² -2g +1). For 5 games, this is 3.8125. Examples like above:

Example 1: 4 games won with 100, 1 bad game with 380.

With the bad game: (100 + 100 + 100 + 100 + 380) * 5 * 5 / (5 * 5 * 5) = 156. Without the bad game: (100 + 100 + 100 + 100 + 0) * 5 * 5 /(4 * 4 * 4) = 156.25

Example 2: 4 games won with 100, 1 bad game with 390:

With the bad game: (100 + 100 + 100 + 100 + 390) * 5 * 5 / (5 * 5 * 5) = 158. Without: (100 + 100 + 100 + 100 + 0) * 5 * 5 /(4 * 4 * 4) = 156.25

How many times do you think it happens that someone loses a game with more than a factor of 3.8 of his points than usual?

And what do you think of the new formula in general?

 So, today I played 5 games of the tournament for this weekend, just to double check how hard it was and if it was winnable.  I went 2-3.  That means the weighted factor is 2.5 times the average score, ouch!  So the quest is winnable, but scores could be high if you lose more games than you win!

Pete... go back to your physics thesis!

so using the old values form before...  5 games, 3 wins, 2 losses.... so

(sum of all scores) * (g * g) / (w * w * w)

(0 + 0 + 75 + 100 + 125) * (5* 5) / ( 3 * 3 * 3) = 300 * 25 / 27 = 277.7777777777778

?

booored said:

hehe :-)

Yes, rounded I also get 278 in that case

Interesting analysis.  One thing to consider is how much influence win % should have on the final score versus the actual quest scores.  There is no right answer, more a matter of preference, but if you increase the win % rating factor too much than the win % becomes essentially what determines the overall winning score, and the average quest score becomes more of a tie breaker when players have the same win %.  When I came up with the original formula I was looking for something that would be simple and intuitive to the average player (i.e. multiply average score times win %) while factoring win % in a meaningful way into the score calculation.  Your analysis raises an interesting fact in that at some point it becomes better to concede then use the final score, which I intuitively don't like yet I wonder how often someone's scores will vary so significantly that this becomes an issue.  Still, I think it warrants some more thought on if there is a tweak that can be made which fixes this issue, while not giving win % too much weight. 

Hmmmmm... very interesting indeed.

Personally, my bias is to keep things as "simple" as possible.

However, if I understand this right, the tension lies between the possibility of there being a certain threshold where a player would decide it is more advantageous to forfeit a game (and take a loss) than see it through to its epic 20+ round conclusion vs the possibility of the system aimed at holding that possibility in check too heavily valuing Win Ratio as compared to Score when it comes to factoring for a final weighted result. Am I following that right?

Yes, correct.  I'm not sure if it is a huge issue if someone is doing so badly on a specific run through compared to their other rounds that they just decide to pack it in rather than finish and count it as a loss.  With current simple approach the win ratio is still fairly heavily weighted so someone with several losses will probably not get a top score anyways.  As well, you have to be doing quite a bit worse than your other scores to even make it worthwile to do this so I don't think it is something that would happen a lot.

RGun said:

That's absolutely right. If we overdo it (e.g. we would divide 4 times through w, the actual scores will really be only a tie-breaker. We can use the "put up or shut up Rhosgobel" thread and Zjb12's current 48-hrs-competition to test our formulas if they "feel right" in determining the winner, when there are enough games posted. Say a week or so. And after that, we could do the same with with the Dead Marshes. I think in all the other quests, you usually don't score with 225 instead of 100 (or 381 instead of 100 for my formula). But in the Marshes, this could happen. (We should ban Denethor and Shadows of the Past because players will especially include them to manipulate the encounter deck. And we have to find players who go to the bitter end if Gollum escapes and search the encounter deck for him...)

For a quick-fix: Good idea, why not count losses and conceded games as fixed 50 pts for solo, 100 pts for 2 player games an so on, in addition to the formula by you, RGun and leptokurt? I calculated it, if A is the average score of the normal games (100 in my previous examples) and B is the score of the bad game, you can take advantage if B > 2.25 * A + 78.125   (that's for 5 games again).

Initial example:

(75 + 100 + 125 + 50 + 50)/3² * 5 = 222.22... (I assumed a solo game and put the 2 last=lost games to 50).

Example so you see the B>2.25A + 78.125 formula works: The limit where it is advantageous to concede a game if the avg score of "normal" games is 100 would be 225+78.125 = 303.125:

without conceding: (80 + 120 + 80 +120 + 303/5²*5 = 140.6      With conceding: (80 + 120 + 80 + 120 + 50)/4²*5 = 140.625     So if someone wants to take advantage of willingly lose a game and not scoring, this game is "worth" a won game of 303 points. Quite ok, I guess. (If he concedes 2 games, then number is even bigger.)

EDIT: I just noticed the Living Tournament. So we have 3 competitions going on for getting data, great :-D