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Posts from Player rating for tournaments across all installments of Heroes.Auto-translated
And this provides a justification for using this non-linear function. That is, thanks to this formula, we increase the gap between the leaders and the outsiders in terms of quantitative metrics. This is exactly what was missing!
Let N be the number of players participating in the tournament. Each i-th player has an average rating Ri in this part of the game. Then, we define the tournament category as:
K = (∑(Ri + a)) / 1000, where a is a constant that determines the contribution to the tournament category of a participant who does not have a rating.
A player's base rating consists of two parts: 1. For days: Rd = 50(M/m), where m is the number of days the participant played, and M is the number of days the winner played. 2. For placement: Rp = 50((N - n + 1 + b) / (N + b))^2, where n is the participant's placement, and b is a regulating constant. Rb = Rd + Rp - base rating
The final rating is equal to the product of the base rating and the tournament category: R = K * Rb
Now, about the two constants, it is obvious that they can be discussed. The constant a should not be too large. A new player is a priori not considered strong. My opinion is: a = 20. The constant b may be necessary to smooth out the rather sharp (but this is a matter of taste) behavior of the dependence when N < 10. For large N, it does not play a role at all. Personally, the behavior of the function seemed most correct to me when b = 3
Let's start in order... The constant **a**, which you introduced, represents the average rating of a beginner and is equal to 20, and in principle, it can take values in the range [0; 100]. Did I correctly understand its physical meaning? Now, let's try to justify why you shifted this value towards the zero end of the scale? After all, it is possible that it should be closer to 50. I suggest simply referring to the tournament statistics...
Just look at the average rating values (using the old system) to understand that 50 is too high for a beginner (the probability that this is a strong player who came from another portal, for example, is extremely low; most likely, this is indeed a beginner, and their result will be average). Setting it to zero is certainly not the right approach, but a number greater than 30, in my opinion, should not even be considered.
And don't forget that for strong players, the difference in average rating will not be significant, and by introducing a large constant component to the contribution, we will significantly diminish the role of strong, consistent players.
Setting it to zero is certainly not a good idea, but a number greater than 30, in my opinion, should not even be considered.
I made this calculation and agree that this parameter can be left at 30, but not less.
Next, I propose to calculate the rating of the players in the last tournament in a trio, where 21 people participated, and the winner completed the map in 10 days, and the outsider - in 69 days. Let's calculate according to your calculations:
Winner - 29194 points Outsider - 865 points
Hence the questions:
1. This does not quite align with the general idea of the calculation, where the winner has 100 points, and the rest are calculated from him. 2. Suppose the winner completed the map in 1 day, and the outsider - in 1000 days. What changes?
Winner - 29194 points - the result did not change Outsider - 858 points - decreased by 1%
Conclusion: the only significant parameter is the number of participants. The rest does not matter.
I don't understand which randomizer was used to obtain these numbers.
In your example:
Rb(winner) = 100 (it's always 100)
Rb(last place for b=3) = 8.64
For the second case:
Rb(winner) = 100
Rb(last place) = 1.44. Everything is normal.
And the final rating cannot be determined based on your data, because the average ratings of all participants before the tournament are unknown. But Rb of each participant is multiplied by the same number, ranging from zero to infinity (although in practice, it's somewhere up to 3).
I don't understand how you got these numbers using that randomizer.
Rp=50(N-n+1+b/N+b)^2 For first place, Rp=50*(21-1+1+3/21+3)^2=29144 (since ^2 means squaring). What do you think I'm doing wrong?
Added 4 minutes later
Fireball;103398
But Rb for each participant is multiplied by the same number, from zero to infinity (although in practice, it's somewhere up to 3).
So, shouldn't we calculate it based on a base of 100 points? Because after multiplying the rating of all participants by some number, we should again bring it to a 100-point system... And this is an inverse operation, and in the end, we will get the original values, i.e., introducing the coefficient K as a multiplier loses its meaning.
On the one hand, we can go beyond 100 points; it's up to us to decide. But on the other hand, we will never have 100 points for the leader... We need to think carefully about how not to overdo it.
Rp=50(N-n+1+b/N+b)^2 For first place, Rp=50*(21-1+1+3/21+3)^2=29144 (since ^2 means squaring). What do I think is wrong?
Apparently, this was my mistake. I meant (N-n+1+b)/(N+b), and all of that squared. From a formal mathematical notation standpoint, you were absolutely right, but I couldn't have imagined that the notation would be interpreted differently. That's why the misunderstanding arose. Now everything has been corrected. It's good that this question came up right away.
That is, we won't be calculating from a base of 100 points? Because after multiplying the rating of all participants by some number, we should again bring it to a hundred-point system...
The 100-point system is the base rating, but the final rating awarded for the tournament can be higher or lower. This is the simplest and most accurate way to show the difference between a serious tournament and a warm-up. Awarding 100 points to both the winner of the Grand Tournament and the winner of a small, informal competition is not right.
...but I couldn't have imagined that someone would interpret the entry differently...
Well, I guess I'm the only one who thought that way... But it's not rocket science; it's your invention, and who knows the meaning better than the author? Our task is to read the words and understand their essence, not to guess what was meant... Apparently, that's really the case... :) Let's continue.
1. Tournament difficulty coefficient.
K=(∑(Ri+a))/1000 - where a is a certain constant that determines the contribution to the tournament category of a participant who does not have a rating.
I think we'll have to adjust this as well. Why is the constant a under the summation sign? The number of players without a rating is not equal to the number of players in the tournament... And in your case, it turns out that this coefficient will be taken into account exactly N times in the final sum... Something is not right... Maybe it was meant to be ∑(Ri)+a*n(a), where n(a) is the number of players without a rating?
2. What does the divisor 1000 represent? Why exactly that value? And what range will this coefficient be in, in your opinion? For now, it seems to me that it depends exclusively on the number of participants, and not on their average rating. Why not divide the sum by N? If you compare a tournament with 5 players (all of whom are aces) and a tournament with 50 players (all of whom are beginners), the difficulty of the second will be almost an order of magnitude higher! That doesn't make sense...
It seems we are moving further and further away from the main goal of the rating system: to clearly and simply show the player their strength in comparison to others. Instead, it's becoming overly complicated. No one will understand the final formula and will genuinely vote for the system where their rating is higher. I think it would be better to simply decide on the principle for calculating the rating. And then adjust the existing formula accordingly. My bonus shouldn't be too large, so as not to create an imbalance between the actual strength of the players. Indeed, it currently assumes that a newcomer has a strength of 0. This is easily fixed by making their strength 1/3 of the number of participants. That is, if there are 10 players in the tournament, the newcomer will be considered not tenth, but seventh. This is not difficult.
2AlexeyPank As I mentioned before, your bonus is too small to differentiate tournaments by strength.
2AmberSoler
I think adjustments will be needed here as well. Why is the constant 'a' under the summation sign? The number of players without a rating is not equal to the number of players in the tournament... And in your case, it turns out that this coefficient will be taken into account exactly N times in the final sum... Something is not right... Maybe it was meant to be ∑(Ri)+a*n(a), where n(a) is the number of players without a rating?
No, now everything is written exactly as intended. When I wrote the formula, I meant that the constant 'a' represents not so much some assumed rating for a beginner, but rather a contribution "for participation." And only players with a rating have a contribution "for strength." An alternative to this approach is, of course, to assign a certain rating to a beginner and then sum ONLY the ratings. But I stayed with the first option based on the following reasoning: Let players A and B have the same skill level in some part of the game, but one of them participated in tournaments, and the other did not. Due to some nuances of the tournament, their results will differ in favor of the experienced player, but this difference will disappear after 1-2 tournaments. Next, let the average rating reflect the strength of the participant, and we assign a certain rating value to the beginner - X, then the beginner's contribution is (X-Y) - in the first tournament, he will not realize his full potential!!!! The contribution of an experienced player remains equal to his rating - R. Now, we simply shift the scale to the right by Y, the beginner's contribution is -X, the experienced player's contribution is R+Y. For now, it is assumed that X=Y=a. In this case, we can take a=30. If we write it as you suggest, it will probably be clearer.
∑(Ri)+a*n(a),
THEN the physical meaning will remain the same, but a will be no more than 20.
What does the divisor 1000 represent? Why exactly this value?
1000 is a simple normalization so that the multiplier to the base rating is 1 for an average tournament. For example, in a tournament with 5 beginners and 10 participants with an average rating of 60, the multiplier will be (a=20) K=0.9
And within what limits will this coefficient be, in your opinion?
On our portal, it's somewhere between (0.5;3)
For now, it seems to me that it depends exclusively on the number of participants, and not on their average rating. Why don't you divide the sum by N? If you compare a tournament with 5 players (all of whom are aces) and a tournament with 50 players (all of whom are beginners), the difficulty of the second will be almost an order of magnitude higher! This is not logical...
You are wrong. A tournament with 5 aces (R is around 100), the coefficient K=0.6, a tournament with 50 beginners will have K=1. Everything is not so bad even in this completely unrealistic (extreme) example.
Why don't you divide the sum by N?
It's obvious why, otherwise a tournament with 10 aces will receive a higher category than a tournament with 10 aces and 10 almost aces, but to win, you need to beat both of them. More precisely, to win in the second case, you need to beat all those that you need to beat in the first case, and also 10 almost the same players.
No, now everything is written exactly as intended. When I wrote the formula, I meant that 'a' should represent not so much some hypothetical beginner rating, but rather the contribution for "participation."
You consider a beginner to be a player without tournament experience. I consider a beginner to be a new player who doesn't have a tournament rating. However, they may have extensive experience gained on another platform... It's necessary to average this out, which is what was initially proposed. I believe the formula ∑(Ri)+30*n(a) is quite optimal, as already mentioned.
1000 is a simple normalization, so that for an average tournament, the multiplier for the base rating is 1. For example, in a tournament with 5 beginners and 10 participants with an average rating of 60, the multiplier will be (a=20) K=0.9.
The calculation mechanism is unclear. From the formulas, it seems like (10*60+5*20)/1000=700/1000=0.7, or something else? At least provide an explanation of your calculations...
But there's also a nuance here. According to your proposal, the difficulty coefficient for playing against 10 aces (rating 100), for playing against 20 average players (rating 50), and for playing against 50 beginners (R=20) would be the same: k=1. Intuitively, this seems contradictory... This isn't a situation where quantity turns into quality... Because if you combine all the opponents, in your case, the coefficient would be k=3, i.e., we see a linear dependence on the number of players. But it seems to me that by solving the problem of how to beat the top ten strongest opponents, we automatically solve the other two... That is, the coefficient shouldn't increase significantly when these groups are combined, but it also shouldn't decrease. It's necessary to take into account the difficulty due to the increasing number of participants and, consequently, the increasing probability that someone from the relatively weak players will make a leap forward – this is quite acceptable. But this doesn't depend linearly on the number of players...
You are wrong. A tournament with 5 aces (R around 100), coefficient K=0.6, a tournament with 50 beginners will have K=1. It's not that bad, even in this completely unrealistic (extreme) example.
Beating 50 or 100 beginners (if they are truly beginners in the sense you mean) won't be difficult... Beating 5 aces is much harder. I say this from experience in tournament battles, not from pure theory. Therefore, your coefficient isn't tied to reality, which speaks against it.
Obviously, why, otherwise a tournament with 10 aces would receive a higher category than a tournament with 10 aces and 10 almost aces, but to win, you need to beat both. More precisely, to win in the second case, you need to beat all those that you would beat in the first case, and also 10 almost equally strong players.
The number of opponents should increase the coefficient, but not in a linear way...
This is all theory. It's useful for a warm-up.
But in practice, it's not very applicable due to the lack of an intuitively understandable explanation. It doesn't have a physical model as an example. It can't be explained "on the fingers." And that's a minus in our case. I still believe that complicating things is not rational.
You consider a novice to be a player who has no tournament experience. I believe it's a new player who doesn't have a tournament rating. Moreover, they may have extensive experience gained on another platform...
Let's look at things realistically. In this year's tournaments, newcomers (Maximix doesn't count; he was already known as a leading player in 1v1 matches on another portal) can only showcase themselves in 5v5 matches because the game itself is new. And as the portal develops, the chances for newcomers to immediately stand out will decrease. The standard situation for a novice is: they come, they see, they are amazed (by the winner's result). Only the aforementioned Maximix has jumped in from another portal for a single tournament, and that's it. Therefore, it is reasonable to assume that if a person doesn't have a rating, they also don't have experience. Exceptions will be rare, and note that each such exception can only slightly influence a single tournament. Moreover, the number of newcomers will always be small.
I believe the formula ∑(Ri)+30*n(a) is quite optimal, as already mentioned.
As I already wrote, it's the same thing, just viewed from a different angle. If this makes it clearer, I'm all for it.
The calculation mechanism is unclear. From the formulas, it seems like (10*60+5*20)/1000=700/1000=0.7 or something else? Please provide an explanation of your calculations at least...
Of course, I calculated using the original formula: (10*(60+20)+5*20)/1000=0.9
It won't be difficult to beat 50 or 100 newcomers (if they are truly newcomers in the sense you mean). It's much harder to beat 5 aces. I'm saying this based on my experience in tournament battles, not just pure theory. This means your coefficient is not tied to reality, which speaks against it.
It is very strongly tied to reality, specifically to how things happen on our portal. There are no friendly matches between top players here, and there are no tournaments specifically for newcomers. In a tournament, there are always strong, average, and weak players in the most ordinary proportions, and it's not even possible to gather all the strongest players in a given area. This year, for example, there were 3 tournaments in 3v3, and the compositions were very different. For example, out of the top 5 in the 3rd stage of the VK tournament, only vbn was present at Atamana's tournament. So, your examples are not realistic.
But it seems to me that by solving the problem of how to beat the top ten strongest opponents, we automatically solve the other two...
Again, no. If the places were always distributed according to the rating, then why hold tournaments? The rating is a measure of the probability of a strong performance.
But there is a nuance here. According to your proposal, the difficulty coefficient for playing against 10 aces (rating 100), for playing against 20 average players (rating 50), and for playing against 50 newcomers (R=20) is the same: k=1. Intuitively, this creates a contradiction... This is not a situation where quantity turns into quality... Because if you combine all the opponents together, in your case, the coefficient would be k=3, i.e., we see a linear dependence on the number of players.
As I already wrote, all tournaments have a very uneven composition of participants. And if the proportion between strong, average, and weak players was always the same, then the coefficient would indeed depend only on the number of participants, which is very logical. If the ratio changes, the coefficient changes.
But in practice, it is not very applicable due to the lack of an intuitively understandable explanation. It does not have a physical model as an example. It cannot be explained "on the fingers." And this is a minus in our case.
Here, you are definitely very wrong; there is a physical model, and it is taken from this portal. The model is as follows: -tournaments involve players with very different skill levels -there are certain proportions between the number of players of different levels (conditionally strong, average, and weak), which change from tournament to tournament -the fluctuations of these proportions are not too large (at most, a few times, but not an order of magnitude)
There is a physical model, and it was taken from this very portal. The model is as follows: - Tournaments involve players with very different skill levels. - There are certain proportions between the number of players of different levels (conditionally strong, average, and weak), which change from tournament to tournament. - The fluctuations of these proportions are not too large (at most, a few times, but not by orders of magnitude).
Correct, you provided a physical model for conducting a tournament, but not a physical model explaining your calculation mechanism. That's what was meant. Try to explain your method for calculating the coefficient in simple terms:
(10*(60+20)+5*20)/1000=0.9
After all, we gave a new player 20 rating points, which is quite justified – because their experience is higher than zero (and I think it would be more appropriate to indicate the number 30 here). But it is not clear why you are adding those same 20 to the player's already formed rating again? This is not a gift to newcomers "for free"; they will not receive these 20 points in their total, which you are trying to compensate for by adding the same amount to everyone else. This is an attempt to approximate the total strength of all participants in the tournament to reality. And it is used only to calculate the coefficient...
We have 15 players who have a clearly defined rating. Their sum is the criterion, i.e., 10*60+5*20=700, which is understandable. Next, you adjust the final sum by an amount equal to the number of players with a formed rating, multiplied by the average rating of newcomers... Why? The essence of this adjustment is not intuitively clear...
AmberSoler
