y = x^4 + x^3 + x^2 + x + 1
wat
I believe that is what could be considered as 'l33t speak.' I understand it to have been popular with internet users a couple of years ago, though it seems to have fallen with style as people have grown older.
Average multiplier depends on the average amount of rounds played in succession but goes rather quickly(after 4 rounds) towards to aforementioned 1.9375 (50% win chance). So just playing 1, 2 or 3 rounds and then quitting is bad for the average multiplier but I guess that's common sense.
So more interesting is a formular that contains the average amount of rounds(n) played in succession as a parameter instead of just assuming the case n->∞.
I'm impressed by both the OP and Paul's reasoning. But judging from the OP, for me to calculate my average multiplier, I would have to know my win rate. So if I required to keep track of my win rate, why won't I just keep track my average multiplier instead, which isn't that much harder.
Playing battle you have a 50% winrate when you play an infinite number of rounds.
Playing siege is a whole different matter though. I rarely get past x3 on a siege server UNLESS there's a clan steamrolling the server and I'm lucky enough to be in their team for some rounds.
Some clans can hold a an x5 for hours on end by playing well together.
Overall, I make much less gold/exp on siege compared to battle.
The internet went down at my work today, so I decided to see if I couldn't come up with the equation that describes the relationship between average multiplier and the probability of winning a round. I've been working on a tool for players to calculate upkeep and equipment, and I wanted to be able to determine average multiplier if given a certain win rate. In my previous encounters with this problem, I've either programmed a simulation, or I've used simple cases, such as a win rate of 50%.
Today, however, I decided to generalize it. Worse yet, I would do it by hand, without the aid of WaltF4's previous posts, which when I look at now, would have given me some solid hints on how to proceed. The algebra was a bitch by hand, but it was all worth it, when all the terms canceled out to yield this gem:
y = x^4 + x^3 + x^2 + x + 1
y is your average multiplier
x is your win rate, eg. if you win 50% of the time, x = 0.5; at 87.22%, x = 0.8722, etc.
So at 50% win rate, your average multiplier (not including valour) would be 1.9375. This is well established and simple to do by hand, since x = 1 - x.
I ran a simulation over 100,000 rounds at each percentile, and the data matches up perfectly with the above equation, so I can say it is correct with confidence.
Most of you are likely indifferent to this information, and some of you will hopefully have a passing curiosity, but I thought I'd share it anyway. Personally, I'm pretty proud, as probability often makes my head spin.
Playing siege is a whole different matter though. I rarely get past x3 on a siege server UNLESS there's a clan steamrolling the server and I'm lucky enough to be in their team for some rounds.
Some clans can hold a an x5 for hours on end by playing well together.
Overall, I make much less gold/exp on siege compared to battle.
really nice! but... there is a but. for instance if i know i'll have a free hour to play, i'll play for an hour and i'll probably leave only when my multiplier goes back to x1 (as most of us do). so, sometimes, i hop in, win 5 rounds and when i get x1 again i'll leave instead trying to regain multi (stuff to do outside the games).
and i still think people with really good teamwork skills have a higher average multiplier. even with a unlucky unbalanced random team, can flip the tide of battle. or for instance, if a full clan will only jump in a server when they can bannerstack will probably have a higher multiplier than the average player.
so.. i really don't know.. it's not just a matter of probability. only devs can tell if it's right or wrong by looking at the stats to see who is in the top for round won and how much is the average multy.
The internet went down at my work today, so I decided to see if I couldn't come up with the equation that describes the relationship between average multiplier and the probability of winning a round. I've been working on a tool for players to calculate upkeep and equipment, and I wanted to be able to determine average multiplier if given a certain win rate. In my previous encounters with this problem, I've either programmed a simulation, or I've used simple cases, such as a win rate of 50%.
Today, however, I decided to generalize it. Worse yet, I would do it by hand, without the aid of WaltF4's previous posts, which when I look at now, would have given me some solid hints on how to proceed. The algebra was a bitch by hand, but it was all worth it, when all the terms canceled out to yield this gem:
y = x^4 + x^3 + x^2 + x + 1
y is your average multiplier
x is your win rate, eg. if you win 50% of the time, x = 0.5; at 87.22%, x = 0.8722, etc.
So at 50% win rate, your average multiplier (not including valour) would be 1.9375. This is well established and simple to do by hand, since x = 1 - x.
I ran a simulation over 100,000 rounds at each percentile, and the data matches up perfectly with the above equation, so I can say it is correct with confidence.
Most of you are likely indifferent to this information, and some of you will hopefully have a passing curiosity, but I thought I'd share it anyway. Personally, I'm pretty proud, as probability often makes my head spin.
I was just thinking about something... Each tick gives you 50x gold and 1000x exp (at gen 1 with x being multi)... and gives you y% chance of each item breaking... But do those break chances add up and give you 4Y% break chance in a 4 minute round, or is it calculated each minute? cause if it's calculated each minute the total break cost would be lower since the same item cant break twice, and if it breaks during the first minute, it cant break again.I believe the breaking chance increases with every minute that passes and your items are subjected to this chance at the end of the round.
The biggest problem is not calculating your gold/exp gain as a function of win rate, the real hard stuff is calculating your win rate since it depends on so many things.
Are you sure with that math?
I was just thinking about something... Each tick gives you 50x gold and 1000x exp (at gen 1 with x being multi)... and gives you y% chance of each item breaking... But do those break chances add up and give you 4Y% break chance in a 4 minute round, or is it calculated each minute? cause if it's calculated each minute the total break cost would be lower since the same item cant break twice, and if it breaks during the first minute, it cant break again.
For n = 3, x = 0.5
y = (x^2 + 2*x + 3) / 3 = (0.5^2 + 2*0.5 + 3) / 3 = 1.41667
For n = 6, x = 0.6
y = (2x^4 + 3x^3 + 4x^2 + 5x + 6) / 6 = (2*0.6^4 + 3*0.6^3 + 4*0.6^2 + 5*0.6 + 6) / 6 = (0.2592 + 0.648 + 1.44 + 3 + 6) / 6 = 1.8912
For n = 1000, x = 0.5
y = (996x^4 + 997x^3 + 998x^2 + 999x + 1000) / 1000 = (62.25 + 124.625 + 249.5 + 499.5 + 1000) / 1000 = 1.935875
Getting damn close to the n->∞ of 1.9375
It shouldn't matter either way. If you look at the round as a whole, it doesn't matter if the break chances are added up and used to determine if an item breaks at the end of a round, or if the break chance simply has a chance to flag an item as broken every minute. The probabilities add up to be the same. In neither system would an item break twice. Break chance for most items is 3%, so in a 5 minute round, overall there's a 15% chance that a particular item will break. Even if the item break flag is checked every minute, there would be 5 chances to break at 3%, which comes out to 15%. This is an average over the longer term.
> roll := rand(0 .. 1);
proc() ... end;
>
1
> A := matrix(1, 1000, [seq(roll(), i = 1 .. 1000)]);
> MULTI := matrix(1, 1000);
> MULTI[1, 1] := 1;
for i to 999 do
if A[1, i] = 0 then MULTI[1, i+1] := 1 end if;
if `and`(A[1, i] = 1, MULTI[1, i] < 5) then MULTI[1, i+1] := MULTI[1, i]+1 end if;
if `and`(A[1, i] = 1, MULTI[1, i] = 5) then MULTI[1, i+1] := 5 end if end
do;
1
> print(MULTI);
MULTI
> Somme := 0; for i to 1000 do Somme := Somme+A[1, i] end do;
0
> Somme1 := 0; for i to 1000 do Somme1 := Somme1+MULTI[1, i] end do;
0
> print(Somme);
513
> print(Somme1);
1991
>
I'm getting a win average of 0.513 (close to 0.5) and a multiplier average of 1.991 which is close to 2y= sum(x^n with n from 1 to +inf) and don't bother calculating it there is a forumula which is (1-x^n)/(1-x)so (1-0)/(1-0.5)= 1/0.5=2 After a little thought, I have come to the conclusion that is it false that longer sessions will get you more multi. As long as the player only quits the game when he looses his multi, his situation is exactly the same as the infinite playing session situation. Imagine two players that loose their multi. One continues the game and starts over from x1. The other one quits. When the second player joins a server again, he will start from x1 just like the guy that didn't quit the game, so their average is the same.
Actually, I think it changes everything (but the trick to increase gold-upkeep profit related to this doesn't really change).
It's hollyday so i'm not gonna do maths
Actually, I think it changes everything (but the trick to increase gold-upkeep profit related to this doesn't really change).
Let's say we have only one item and that the breaking state is a flag that can be set each tick with a 3% probabilty. A broken item can't break again (except if taken multiple times which is not the case here).
So after one tick, he has 97% chances of staying ok. After the second tick he has only 97% of the earlier 97% chances of being intact, which equals 94,09%, which is more than the 94% of the simple break chance adding calculation. Let's continue with the third tick. Now only 97% of the 94.09% of "not broken after two ticks" items will not break, which means 0.97*0.9409 = 0.912673, slightly more than 0.91.
The difference can reach 4% for rounds longer than 10 minutes.
Furthermore, I think 3% isn't the base break chance per tick, It seems to me it's closer to 5%. If we consider 10 ticks, we have 0.95^10 = 0.5987 probability of staying intact for the "possible break each tick" calculation and 0.50 probability of staying intact for the "break chance adding each tick" case. The difference is nearly 10% and increases if the round is longer. If we take it to the extreme, for 20 minutes rounds, in one case the break is certain and in the other it is only ~36%.
Also, there ARE a lot of misconceptions about upkeep vs ticks. If the what the dev said about it's calculation is correct, I know at least 2 very serious ways of reducing it a little for the same equipment set. But I prefer keeping this secret :p Actually I prefer you thinking I'm lying here.PM me this info! 8-)
It would seem that the longer the rounds, the more money you can expect to make as the average break chance per tick is less the more ticks their are in a round. As described by this equation:
y = (1- (1-x)^n)/n
Where:
y - average break chance per minute
x - break chance
n - length of the round in minutes
What makes you say that the base break chance is 5% and not 3%? It should be 3% for all items except arrows and low wpf weapons. Previously, upkeep used to be 4% chance to break with a 5% repair cost.
starting experience: 1041390 exp
ending experience: 9853970 exp
starting gold: 1499164 gold
ending gold: 1771701 gold
generation: 12
average multiplier: 2.23
equipment cost (buying price): 9686 gold
My character gained 8812580 experience and 272537 gold over this period. This amount of gold gain has been reduced due to maintenance. The gold gain without maintenance is proportional to experience gain and character generation such that:
gold gain/exp gain = 50 gold/(1000 exp+30 exp*(generation-1))
With a values of generation greater than 16 counting as 16. For my character:
gold gain/exp gain = 50 gold/1330 exp = 0.0376 gold/exp
So without maintenance I would have gained:
8812580 exp * 0.0376 gold/exp = 331353 gold
Therefore, over this period I lost 58816 gold to maintenance. As gold gain, experience gain, and equipment breakage are incremented at regular intervals, or clock ticks, it is easier to look at these tick time units instead of number of rounds. The number of ticks included in this set of games can be determined as:
Number of ticks = (end exp-start exp) /((1000 exp+30 exp*(generation-1))*average multiplier)
Again, values of generation greater than 16 count as 16. For this set of data:
Number of ticks = 8812580 exp/ (1330 exp * 2.23) = 2971.3 ticks
And the amount of gold lost to maintenance per tick is:
58816 gold / 2971.3 ticks = 19.79 gold/tick
To determine the rate of maintenance, that is the gold lost to maintenance each tick per gold worth of equip items, the following equation is used:
rate of maintenance per tick = (gold lost to maintenance per tick)/(equipment cost)
Using my values:
rate of maintenance per tick = 19.79 gold / 9686 gold = 0.0020
(click to show/hide)
It's actually true. If you have the discipline to only quit at x1 then it is the same as n->∞ and we get an average multi of 1.9375 (or 2 if multi wasn't capped at x5).
It's actually true. If you have the discipline to only quit at x1 then it is the same as n->∞ and we get an average multi of 1.9375 (or 2 if multi wasn't capped at x5).
No, you are wrong. When quitting from time to time (disconnect, have to got, etc.) with a multi of >1, the theoretical average multiplier is lower than the 1.9x value.
No and I'm too lazy to check the code.
Even if I would take a look I wouldn't be allowed to tell.
No, you are wrong. When quitting from time to time (disconnect, have to got, etc.) with a multi of >1, the theoretical average multiplier is lower than the 1.9x value.You mean "can be" lower? If I always leave in my first round, I will only have an average of 1x, but if I play for thirty days and nights and then leave on a 2x, I think it will be >1.9x.
(words)
y = x^4 + x^3 + x^2 + x + 1
(another collection of words)
TL:DR Latest Recommendation: Only quit on x1.