Average multiplier as a function of win rate

[Teeth]

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.

[Panoply]

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.

You can't really calculate this, you can only record it over time. However, you should be able to give a ballpark estimate, which in turn gives you a reasonable estimate of your average multiplier.

Are you sure with that math?

Yes. Like I said before, I ran a simulation of 100,000 rounds at each win rate percentile (eg. 0.01, 0.02, 0.03... 0.99, 1.00), and the simulated data matches up perfectly to the theoretical predictions of my equation.

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.

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.

[Kafein]

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

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.

The formula you are using here doesn't take into account that people don't quit when they have a multiplier. Therefore the only case of difference in multi average compared to the infinite playtime is for people that played a total of less than 5 rounds.

Furthermore, you could easily tweak your simulation program to show that this is true : loop the simulation a small (10) number of times then wait for a multi loss and calculate the average multi. Repeat this simulation a big number of times, make an average and I'm pretty sure you'll find the same average as in the infinite case. It seems very simple to do so I'm coding it right now (EDIT : actually, forget this, I got to reinstall all the crap and I'm lazy :x )

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.

[Kafein]

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.

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%.

[Cosmos_Shielder]

It's hollyday so i'm not gonna do maths
But info is not a problem !

Code: [Select]

> 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 2
So correct formula seems to be 

Code: [Select]

y= sum(x^n with n from 1 to +inf) and don't bother calculating it there is a forumula which is

Code: [Select]

(1-x^n)/(1-x)so (1-0)/(1-0.5)= 1/0.5=2
Damn i said i will not do maths....
For people interested here are the matrix

(click to show/hide)

[[1,1,1,0,1,1,0,1,0,1,1,1,1,0,0,1,0,1,1,1,0,0,1,0,1,1,0,1,1,1,1,1,1,0,0,0,0,1,1,1,0,0,0,1,1,1,0,1,1,1,0,0,1,1,0,0,0,1,0,0,0,0,1,0,0,1,1,1,1,1,0,0,0,1,0,0,0,

1,0,1,1,1,0,0,1,0,1,1,0,1,0,0,0,0,1,1,1,0,0,1,1,1,0,1,1,1,0,1,1,1,0,0,1,0,1,0,0,0,1,0,1,0,0,0,0,0,1,0,0,0,1,1,0,0,0,1,1,0,1,1,0,1,0,0,0,1,0,0,1,1,1,0,1,0,1

,1,1,0,1,0,0,0,1,1,1,1,1,0,0,1,0,1,0,0,0,1,1,1,1,1,0,0,0,0,1,1,0,0,1,0,0,0,0,1,0,0,0,1,1,1,0,1,1,1,1,1,0,1,1,1,0,1,1,0,0,0,1,0,1,0,0,0,0,1,1,0,1,0,1,0,0,0,

1,1,0,1,0,1,0,0,1,0,0,0,0,1,1,1,0,0,1,1,0,0,0,1,0,0,1,0,0,1,1,0,0,1,1,1,0,1,1,0,0,0,1,1,1,0,0,1,0,1,1,1,0,0,1,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,0,0,0,1,1,1,0,0

,0,0,0,1,0,1,1,1,0,0,1,0,1,1,0,0,1,0,0,0,0,0,0,0,1,0,0,1,1,1,1,0,0,0,1,1,0,0,0,1,1,0,1,1,1,0,0,1,0,0,0,0,0,0,1,1,1,0,0,1,1,1,1,1,1,1,1,1,0,0,0,1,0,1,1,1,0,

0,1,0,0,0,1,1,1,1,1,1,0,1,0,0,1,0,0,1,1,1,0,1,1,0,1,0,0,1,1,1,1,1,1,1,0,0,1,1,0,0,1,0,0,1,1,1,0,1,0,0,0,1,1,1,0,0,0,1,1,0,1,1,1,1,0,0,1,0,0,0,0,0,1,0,1,0,0

,0,0,0,1,0,1,1,0,1,0,0,0,0,1,1,1,0,1,0,1,1,1,0,0,0,1,1,0,0,0,1,0,1,1,0,1,1,0,1,0,0,0,0,0,0,0,0,1,1,1,1,1,1,1,0,1,0,0,0,0,1,0,0,0,1,0,0,1,1,0,1,0,0,0,0,0,1,

0,0,1,1,0,0,0,0,1,0,1,1,0,1,0,0,0,1,0,1,0,1,1,1,1,0,1,1,1,0,0,1,0,1,1,1,1,0,0,0,0,1,1,1,1,0,1,0,0,1,0,0,1,0,0,0,0,0,0,0,0,0,1,0,1,0,0,0,0,1,0,1,0,1,1,1,1,0

,0,0,0,1,0,1,0,1,1,0,0,1,0,1,1,0,0,0,1,1,1,1,0,0,0,1,1,1,1,1,1,1,0,0,0,1,0,0,1,0,1,1,1,1,0,1,1,0,0,1,1,1,1,0,0,1,1,0,0,0,1,1,0,1,1,0,1,0,1,1,1,1,0,1,0,1,1,

0,0,1,0,0,1,1,0,1,1,0,1,0,0,1,1,0,0,0,1,1,1,1,1,0,0,0,0,1,0,1,0,1,1,1,0,0,0,0,0,1,0,1,1,0,1,0,1,0,1,1,0,0,1,1,1,1,0,0,1,1,1,1,1,1,0,0,1,1,0,1,1,1,0,1,1,0,0

,1,0,0,1,1,1,0,0,1,1,0,1,1,1,1,1,0,1,0,1,0,1,1,0,1,0,0,0,1,1,1,0,1,0,1,0,1,0,1,1,0,1,1,0,1,0,1,0,0,1,1,1,1,0,0,0,1,1,0,1,0,0,1,0,0,1,1,1,0,1,1,1,1,1,1,0,0,

1,1,0,0,1,1,1,0,1,0,0,0,0,0,0,0,0,1,0,1,0,0,1,0,0,0,0,0,1,1,0,0,1,1,0,1,1,1,0,1,1,1,0,1,1,0,1,1,1,1,1,1,1,1,0,1,1,0,1,1,0,0,0,0,0,0,1,1,0,1,1,1,0,1,1,0,1,1

,0,0,1,0,0,1,1,0,1,1,0,0,0,0,0,1,1,0,1,1,0,0,1,1,1,0,0,0,0,0,1,1,1,1,0,1,0,1,0,1,0,1,0,1,1,1,1,1,1,1,0,0,0,0,1,0,1,0,1,1,1,1,0,1,0,0,1,0,1,1]]

(click to show/hide)

[[1,2,3,4,1,2,3,1,2,1,2,3,4,5,1,1,2,1,2,3,4,1,1,2,1,2,3,1,2,3,4,5,5,5,1,1,1,1,2,3,4,1,1,1,2,3,4,1,2,3,4,1,1,2,3,1,1,1,2,1,1,1,1,2,1,1,2,3,4,5,5,1,1,1,2,1,1,

1,2,1,2,3,4,1,1,2,1,2,3,1,2,1,1,1,1,2,3,4,1,1,2,3,4,1,2,3,4,1,2,3,4,1,1,2,1,2,1,1,1,2,1,2,1,1,1,1,1,2,1,1,1,2,3,1,1,1,2,3,1,2,3,1,2,1,1,1,2,1,1,2,3,4,1,2,1

,2,3,4,1,2,1,1,1,2,3,4,5,5,1,1,2,1,2,1,1,1,2,3,4,5,5,1,1,1,1,2,3,1,1,2,1,1,1,1,2,1,1,1,2,3,4,1,2,3,4,5,5,1,2,3,4,1,2,3,1,1,1,2,1,2,1,1,1,1,2,3,1,2,1,2,1,1,

1,2,3,1,2,1,2,1,1,2,1,1,1,1,2,3,4,1,1,2,3,1,1,1,2,1,1,2,1,1,2,3,1,1,2,3,4,1,2,3,1,1,1,2,3,4,1,1,2,1,2,3,4,1,1,2,3,4,1,1,1,2,3,4,5,1,1,1,2,3,4,1,1,1,2,3,4,1

,1,1,1,1,2,1,2,3,4,1,1,2,1,2,3,1,1,2,1,1,1,1,1,1,1,2,1,1,2,3,4,5,1,1,1,2,3,1,1,1,2,3,1,2,3,4,1,1,2,1,1,1,1,1,1,2,3,4,1,1,2,3,4,5,5,5,5,5,5,1,1,1,2,1,2,3,4,

1,1,2,1,1,1,2,3,4,5,5,5,1,2,1,1,2,1,1,2,3,4,1,2,3,1,2,1,1,2,3,4,5,5,5,5,1,1,2,3,1,1,2,1,1,2,3,4,1,2,1,1,1,2,3,4,1,1,1,2,3,1,2,3,4,5,1,1,2,1,1,1,1,1,2,1,2,1

,1,1,1,1,2,1,2,3,1,2,1,1,1,1,2,3,4,1,2,1,2,3,4,1,1,1,2,3,1,1,1,2,1,2,3,1,2,3,1,2,1,1,1,1,1,1,1,1,2,3,4,5,5,5,5,1,2,1,1,1,1,2,1,1,1,2,1,1,2,3,1,2,1,1,1,1,1,

2,1,1,2,3,1,1,1,1,2,1,2,3,1,2,1,1,1,2,1,2,1,2,3,4,5,1,2,3,4,1,1,2,1,2,3,4,5,1,1,1,1,2,3,4,5,1,2,1,1,2,1,1,2,1,1,1,1,1,1,1,1,1,2,1,2,1,1,1,1,2,1,2,1,2,3,4,5

,1,1,1,1,2,1,2,1,2,3,1,1,2,1,2,3,1,1,1,2,3,4,5,1,1,1,2,3,4,5,5,5,5,1,1,1,2,1,1,2,1,2,3,4,5,1,2,3,1,1,2,3,4,5,1,1,2,3,1,1,1,2,3,1,2,3,1,2,1,2,3,4,5,1,2,1,2,

3,1,1,2,1,1,2,3,1,2,3,1,2,1,1,2,3,1,1,1,2,3,4,5,5,1,1,1,1,2,1,2,1,2,3,4,1,1,1,1,1,2,1,2,3,1,2,1,2,1,2,3,1,1,2,3,4,5,1,1,2,3,4,5,5,5,1,1,2,3,1,2,3,4,1,2,3,1

,1,2,1,1,2,3,4,1,1,2,3,1,2,3,4,5,5,1,2,1,2,1,2,3,1,2,1,1,1,2,3,4,1,2,1,2,1,2,1,2,3,1,2,3,1,2,1,2,1,1,2,3,4,5,1,1,1,2,3,1,2,1,1,2,1,1,2,3,4,1,2,3,4,5,5,5,1,

1,2,3,1,1,2,3,4,1,2,1,1,1,1,1,1,1,1,2,1,2,1,1,2,1,1,1,1,1,2,3,1,1,2,3,1,2,3,4,1,2,3,4,1,2,3,1,2,3,4,5,5,5,5,5,1,2,3,1,2,3,1,1,1,1,1,1,2,3,1,2,3,4,1,2,3,1,2

,3,1,1,2,1,1,2,3,1,2,3,1,1,1,1,1,2,3,1,2,3,1,1,2,3,4,1,1,1,1,1,2,3,4,5,1,2,1,2,1,2,1,2,1,2,3,4,5,5,5,5,1,1,1,1,2,1,2,1,2,3,4,5,1,2,1,1,2,1,2]]

[Cosmos_Shielder]

It seems that 1000 is not enough.
I done it with 5000 :
Average of 2515/5000
and multiplier average 1.9436

I'm trying with 100 000

[edit]
Tried with 100 000 . It got my processor a bit warm but it worked :
Average win as expected of 0.50135
Average multiplier of 1.94047

So previous maths was wrong i'm keeping this multiplier as the correct average multiplier
[edit2] rerunned twice again the 100 000
average win of 0.50003 (got lucky) and second time of 0.50007
and average multiplier of 1.93774 and second time 1.93925

[Panoply]

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.

This was exactly my initial intuition. Urist's posts made me question that intuition, and hence my second batch of formulas. I should have done it to begin with, but I ran a simulation similar to the one you described where you only quit at x1 and after playing at least 5 rounds in one sitting, and you are exactly right. As long as those conditions are fulfilled, you get the theoretical average multiplier predicted by my first equation. It seems I must revise my posts again such that the earlier post, with its recommendation of playing at least five games a sitting and only quitting on x1, is correct, while the later one following Urist's model.

I've updated the original post accordingly to make clear the practical implications that can be drawn from this.

Actually, I think it changes everything (but the trick to increase gold-upkeep profit related to this doesn't really change).

Argh! Such an amateur mistake. I did say earlier I wasn't very good with probabilities. You're absolutely right again. 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. With this latest patch and its upkeep increase, they claimed that break chance was reduced by 1% but repair cost was increased by 2% to yield a 5% increase in upkeep. So, comparing 0.04*0.05 = 0.002, and 0.03*0.07 = 0.0021, and 0.0021/0.002 = 1.05, representing the 5% increase in upkeep. Do you have reason to doubt this? It wouldn't be the first time the upkeep rates were not as advertised.

Also, I'm curious as to this "trick" you keep mentioning.

It's hollyday so i'm not gonna do maths

Haha, thanks for your input. However, I already ran a simulation with 100 000 and 1 000 000 rounds. As you can see if you plug your win rates into the equation in my OP, the average multipliers are very close.

[Torp]

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%.

this is exactly why i said it was important, i was just too lazy to come with a long explanation.

You can also explain it with gold amounts; say you have total equipment worth 100k.
If the break percentages adds up, you would, in a 5 minute round with 4 % break chance have a total break chance of 20%, on average giving you 20k worth of broken items. If you paid 5% equipment cost, that would cost you an average of 1k per round.

If they dont add up, you would in the first minute have 4% break chance, giving you, on average, broken equipment worth of 4k. That means 200 gold in upkeep for that minute (on average with a 5% upkeep cost)
That means you will only have 99,8k of equipment that can break at the next tick, lowering the upkeep cost for that minute by 0,2%, and it would keep lowering by 0,2% per tick for the rest of the round, giving you a lower total upkeep cost.

I don't know the exact numbers for break chance and upkeep cost after the patch, and i don't have a calculator atm so i'm not gonna do the big math and give upkeep costs as a function of your equipment if it don't add up, and i won't come with a final comparison of a 5 min round with and without adding up; the important thing here is that there is a difference.

EDIT: didnt read previous post which says pretty much the same (lazy and jst had a drink), sorry for that.

Also; if you only quit on x1, the equations based on the number of rounds you play will start from the very first round of warband you played (since december patch) instead of the number of games in your playing session since you would start with x1 the next round no matter what. That means that since people have played very high number of rounds since then, the average multi will be very close to the multi you would have if you played an infinite number of rounds.

[Teeth]

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! 

[WaltF4]

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.

The data collected for my more recent upkeep study (unfortunately, done before the latest patch) suggests that the breakage checks are calculated as:

chance to break per round = chance to break per tick * ticks per round

without additional complications from a decaying chance to break per tick. From the upkeep post:

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

At the time, the cost per repair was 5% of the buying price and all items had the same chance to break, so:

breakage chance = rate of maintenance per tick / fraction of equipment cost spent per repair
breakage chance = 0.0020 / 0.05 = 0.04

which is in agreement with the 4% breakage chance per tick that the development team had stated. This can be taken further since it is know that the data was collected over 913 rounds. With the assumption that all equipped items broke the same number times, we can combined all of the equipped items into a single entity with a value equal to the sum of the component item values, which in this case was 9686 gold and such an item would have a repair cost per breakage of 0.05 * 9686 gold = 484.3 gold. With a known total gold lost to maintenance of 58816 gold, the total number of breakages for the equipped item set can be calculated as:

number of breakages = total gold lost to maintenance / cost per breakage
number of breakages = 58816 gold / 484.3 gold = 121.4

With 121.3 breakages over 913 rounds, the chance for a breakage per round is 13.3%. The average number of ticks per round is also known to be 2971.3 ticks / 913 rounds = 3.25 ticks per round. Assuming that the chance to break per round is proportional to ticks per game:

chance to break per round = chance to break per tick * ticks per game
chance to break per round = 0.04 breakages per tick * 3.25 ticks per round= 0.130 breakages per round

Assuming that the chance to break per tick decays with increasing numbers of ticks:

chance to break per round = 1 - ( 1 - initial chance to break per tick ) ^ ticks per game
chance to break per round = 1 - ( 1 - 0.04 ) ^ 3.25 = 0.124 breakages per round

The proportional model yields a better fit (2% error vs 6% error) and is also the less complex option.

[Kafein]

(click to show/hide)

The data collected for my more recent upkeep study (unfortunately, done before the latest patch) suggests that the breakage checks are calculated as:

chance to break per round = chance to break per tick * ticks per round

without additional complications from a decaying chance to break per tick. From the upkeep post:

At the time, the cost per repair was 5% of the buying price and all items had the same chance to break, so:

breakage chance = rate of maintenance per tick / fraction of equipment cost spent per repair
breakage chance = 0.0020 / 0.05 = 0.04

which is in agreement with the 4% breakage chance per tick that the development team had stated. This can be taken further since it is know that the data was collected over 913 rounds. With the assumption that all equipped items broke the same number times, we can combined all of the equipped items into a single entity with a value equal to the sum of the component item values, which in this case was 9686 gold and such an item would have a repair cost per breakage of 0.05 * 9686 gold = 484.3 gold. With a known total gold lost to maintenance of 58816 gold, the total number of breakages for the equipped item set can be calculated as:

number of breakages = total gold lost to maintenance / cost per breakage
number of breakages = 58816 gold / 484.3 gold = 121.4

With 121.3 breakages over 913 rounds, the chance for a breakage per round is 13.3%. The average number of ticks per round is also known to be 2971.3 ticks / 913 rounds = 3.25 ticks per round. Assuming that the chance to break per round is proportional to ticks per game:

chance to break per round = chance to break per tick * ticks per game
chance to break per round = 0.04 breakages per tick * 3.25 ticks per round= 0.130 breakages per round

Assuming that the chance to break per tick decays with increasing numbers of ticks:

chance to break per round = 1 - ( 1 - initial chance to break per tick ) ^ ticks per game
chance to break per round = 1 - ( 1 - 0.04 ) ^ 3.25 = 0.124 breakages per round

The proportional model yields a better fit (2% error vs 6% error) and is also the less complex option.

tl;dr : So you mean the system is likely working with breakage chances per item that add up each tick and are tested when the round ends ? Btw : Thanks a lot, collecting all that data must be tiresome.

Also, @Panoply : I think you don't need to play at least 5 rounds per session. If you already played 5 rounds in your entire cRPG gaming history, then you only need to quit when you loose your multi to be in the optimal infinite gaming session case. For example, you can play one round, win it, then play at x2 during one round, loosing it and quitting.

About the "tricks", both are relatively easy to find out if you analyse everything upkeep and gold income related (equipment, round length, multiplier) very precisely. You can imagine a profit function of many variables. If you find it out correctly and don't forget anything, good approximations to maximise it are quite simple to state intuitively.

[Paul]

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).

[Torp]

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).

not true, the number of rounds is just counted from your very first warband round with this system instead of the number of rounds in a play session - you still had to start at x1 in the very beginning

Also, thanks for the info Walt, that pretty much puts an end to that discussion

Edit: i have thought about something for some time; is the upkeep based on the items you are using when the tick hits? Or does it give each item you've had that round the same breakage chance per tick even if some of it had been dropped?
and what about killed horses and broken shields?

SO i think we need Paul to answer this: Which items gets an increased breakage chance per tick?

[Paul]

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.

[Overdriven]

I don't understand half this stuff. But I'm curious...would it be possible to add an ability for the website to record wins/losses/streaks under your character page?