# /players command and noDrop

#### javcek

##### Diabloii.Net Member
/players command and noDrop

So, I was checking the drop calculator at Link to competing forum deleted, sorry - AlterEgo and according to that page the extra players that come from the /players command only count as 0.5 regarding chances of no drop. Anyone knows if that's right?

#### jjscud

##### Diabloii.Net Member
I'm not exactly sure about the no drop but I know that each additional player (or player setting) adds .5 of the original monsters HP and .5 of the original experience. It makes sense that the no drop chances are affected in a similar manner.

Edit: Thanks Uzziah, i was trying to figure out how to correct my response without making stuff up, when I thought about it a bit more I knew it wasn't linear, especially with bosses but I didn't know what it was.

#### Uzziah

##### Diabloii.Net Member
The chance of no drop is exponential compared to the /players. IE the more players the slower the no drop decays, you gain the most increase in drops with the first increase in players.

Item drop percentage increase greatly with the decrease in no drop. I wish I had the knowhow, as thrugg explained it to me long ago. but basically.

When the drop occurs you have xx chances at equipment drop, xx chances at a crap(gold scrolls etc) drop, xx chance at a gold drop, xx chance at a no drop, . . . etc. the total chance at each is a set number except for the xx chance for no drop (which happens to be controlled by number of players in game).

so as an example:
10 chances for equipment
25 chances for crap
45 chances for gold
70 chances for no drop

Thats 10+25+45+70 = 150 total chances now thats 10/150 for equipment 25/150 for crap etc.

Now when the players increase that 70 for no drop decreases exponentially I have to wait on thrugg to get the formula again. but say it drops to 35 when you add a second player.

you now have 10/115 for equipment 25/115 for crap etc.

now say a third player is added the no drop goes down to say 10

you now have 10/90 for equipment 25/90 for crap etc.

with the bosses this no drop disappears extremely quickly and you are left with 6 drops normally with 3 players in game.

I hope my half confusing explaination helps you understand that is not a 0.5 increase/decrease or whatever. That would be linear where as the item drop is exponential.

#### javcek

##### Diabloii.Net Member
Let me see if I can explain it a bit better...
Apparently, the chances of noDrop depends on a variable named "playerbonus" that varies according to the number of players in the game. If the extra players are in your party, they count as 1; otherwise, as 0.5. So i guess what I'm asking is: does the extra players from the /players X command are in your party?

PS: AE, sorry for the link; won't happen again.

#### Thrugg

##### Diabloii.Net Member
Actually javcek is asking a pretty deep question here.
I had indeed always believed that the players set by the /players command in SP count in full for the nodrop exponent, but an off hand remark by Ruvanal (who almost always deserves to be listened to), at another forum which would also be deleted, made it sound that it is in fact only 0.5 per additional player. I quickly replied asking him to confirm but he never responded.

To make this more clear:

When you play a multiplayer game, you have the total number of people in the game, say N.

All monsters get HP and exp multiplied by (N+1)/2. This is well known and also doesn't matter whether the players are partied, close by, whatever.

If you are playing off-realm you can also use /players X to raise the number of players to X > N. Then the HP and exp will scale up further to (X+1)/2 as you would expect.

Drops, however, are not the same. When you kill a monster in a multiplayer game, the game calculates a number to use as the nodrop exponent. This is not simply N.
It counts 1 for you, the killing player.
It counts 1 more for each player that is (a) partied with you and (b) within two screens of you.
It counts 0.5 for each other player, either unpartied or far away.
It rounds the final total down.

So, if you sneak into a full public game on the realms and don't party up but go MFing on your own, you actually only get a nodrop exponent of 4 (1 for you, 3 for the 7 unpartied players rounded down), while the other partied guys are getting an exponent of 7 (7 people in their party, and you rounded down to 0). Even though all of you are killing p8 monsters with their 4.5x HP. It is one of Blizzard's many measures to encourage party play.

javcek's question is, since when you type /players 8 (say) there is no information as to whether those imaginary extra players are partied or whatever, how does it count the nodrop exponent in that case? If it assumes you are all partied and close by it may be being too generous; if it assumes you are all unpartied it may be being too unfair.

I'm afraid I cannot be 100% confident of the answer. I am still more sure than not that p8 means nodrop exponent=8. And, Atma currently figures it that way.

AE, I am actually not aware of the calculator javcek referred to. I might be able to find out more about the situation if I had a look at it. Feel free to PM me if that is acceptable.

Addendum: the way the nodrop exponent is actually used...
Uzz explained most of it. You have your nodrop number (eg for Meph, this is 15) and your total of other drops (for Meph this is 65). This means the total ratio of nodrops in solo play is 15/(15+65) = 15/80 = ~19%.

You then apply your nodrop exponent (call it n) like this:
new nodrop rate = (base nodrop rate)^n
new nodrop number = new nodrop rate / (1 - new nodrop rate) * total drops, rounded down.

For example, Meph, nodrop exponent = 1
new nodrop rate = (15/80)^1 = 15/80
new nodrop number = 15/80 / (1 - 15/80) * 65 = 15/65 * 65 = 15 (just showing you that it works even for n=1)

Meph, nodrop exp = 2
new nodrop rate = (15/80)^2 = 0.03515625
new nodrop number = 0.03515625 / (1 - 0.03515625) * 65 = 0.036437 * 65 = 2 after rounding.
So with exponent = 2, Meph is already only missing drops 2/67 of the time.

Meph, nodrop exp = 3
new nodrop rate = (15/80)^3 = 0.00659
new nodrop number = 0.00659 / (1 - 0.00659) * 65 = 0.00664 * 65 = 0 after rounding.
So with exponent = 3, Meph is always dropping in full.

It is unfortunately extremely hard to test this empirically. Enough wierd stuff happens that also seem to hide drops, plus Meph gets 7 drop chances and only 6 items can drop, that it is too hard to tell 2/67 apart from 0. So it could actually be that all this time we've been running Meph at p3 to get full drops when really we are just getting exp=2 drops - you'd need to run him at p5 to get guaranteed full drops if Ruvanal was right.

Intriguing stuff.

 ah, javcek posted again after I started. OK, replace no-drop exponent with "playerbonus" in the above

#### Nightfish

##### Diabloii.Net Member
I got 3 items from meph on players 3 before. Did I hack my game? :teeth:

#### AlterEgo

##### Diabloii.Net Member
Thrugg said:
AE, I am actually not aware of the calculator javcek referred to. I might be able to find out more about the situation if I had a look at it. Feel free to PM me if that is acceptable.

#### bill_n_opus

##### Diabloii.Net Member
My brain hurts.

I'm gonna have to digest Thrugg's dissertation later ... although I think I get the gist of it.

#### javcek

##### Diabloii.Net Member
I'm guessing some field work is necessary in order to clear things up. I was thinking in doing some meph runs in /players 1, write down the average number of items per run, and compare with the average number of items per run on /players 2. There should be no meaningful difference between the two numbers if the chances of no drop are indeed based in the "playerbonus" variable.
According to Urlik's RF Guide (If i got it right) :
Number of players (apparently 1+"playerbonus") : Mean number of Drops per Run
1:5.6875
2:5.9822
3:6.00

(The above should be a table, but that is beyond my current skills...)

So if you play at /players 2 ( effective number of players: 1.5 rounded down to 1) you should get the same average number of items per run as in /players 1; and have to play at /players 5 to reduce the chance of no drop to 0.

: That data is for mephisto, just in case.

#### JicamaEater

##### Diabloii.Net Member
Thrugg said:
So, if you sneak into a full public game on the realms and don't party up but go MFing on your own, you actually only get a nodrop exponent of 4 (1 for you, 3 for the 7 unpartied players rounded down), while the other partied guys are getting an exponent of 7 (7 people in their party, and you rounded down to 0). Even though all of you are killing p8 monsters with their 4.5x HP. It is one of Blizzard's many measures to encourage party play.
Yeah, but they don't TELL you about it, which means nobody knows about it to utilize this feature. Except us, of course...for all the good it does us and all of our 8-player MP games...

#### Thrugg

##### Diabloii.Net Member
Well I've searched everywhere I can think of, including forum deleted, forum deleted and forum deleted. Oh, and forum deleted too

The only real hits were repeated comments by Ruvanal that it works the way javcek suggests, which is (I'm afraid) not the way Atma handles it. I am convinced Ruvanal believes it to be true, and that's close enough to good enough for me.

I think some empirical tests would still be useful. I'm not sure which would be clearer, running a boss or killing dozens of regular monsters. Boss killing is more controlled but there is less of a difference between players settings. Killing regular monsters should show a clearer distinction between what should and what does happen, but you'd have to do a lot of killing to bring the variance down far enough to make actual claims. Plus not all regular monsters ahve the same no-drop profile...

I have not found any answer for "what happens if you have 7 partied players in a game and type /players 8", however. Open question.

#### javcek

##### Diabloii.Net Member
Well, I'm off to doing some Meph runs then... posting back the results ASAP.

#### KatoZyEl

##### Diabloii.Net Member
Would it surprise you to know that the nodrop rate varies depending on which monsters you kill?

Treasure classes (which include nodrop rates, normal, magic, set, unique and rare drops) are all defined by monster and by game difficulty. Anything that can drop treasure is defined as having a treasure class - including treasure chests, corpses, barrels, etc...

Rather than measuring the drops on the runs you do on a particular map, you should be measuring the drops on a particular monster and on a particular game difficulty. Suffice it to say, killing a Skeleton on Nightmare will yield better drops than killing a Skel on Normal. Of course, you could also check the no drop rates by measuring what you get from treasure chests only or barrels only. Save you the hassle of having to look for the same monsters to kill over and over again.

#### javcek

##### Diabloii.Net Member
Did 60 hell mephisto runs on /players 1:
average number of drops per run: 5.50.

I'm guessing I need a bigger sample to get more accurate results, but still...

Going to try on players 2 and see if I get a big difference. Anyone willing to sacrifice some of their time for the cause are welcomed to contribute

On a sidenote, I found 3 unique amulets while doing that 60 runs (atma scarab, the cat's eye and the mahim-oak curio). What are the odds of that?

#### essojay

##### Diabloii.Net Member
3 amulets and not a single one of them Nokozan? That's pretty lucky. Kinda like getting 3 unique rings without any of them being nagel or manald heal *shudders*.

#### bill_n_opus

##### Diabloii.Net Member
javcek said:
Did 60 hell mephisto runs on /players 1: ....

On a sidenote, I found 3 unique amulets while doing that 60 runs (atma scarab, the cat's eye and the mahim-oak curio). What are the odds of that?
You are damn lucky. There are only two ammys that i've yet to find for some strange reason. Atma's and The Cat's Eye. You found them in 60 runs. I've done maybe over a thousand runs easy. At least, if not over 1500 Meph runs. I'm being conservative too. Haven't found them yet.

#### javcek

##### Diabloii.Net Member
60 more runs on /players 2:
average number of drops per run: 5.41667 (what? lower? :scratch: )
Definitely more testing needed; but that is gonna be tomorrow.

Add another amulet to the others: The Rising Sun.

#### PinkeyandtheBrain

##### Banned
Please explain what exactly the "no drop" is i am quite confused.
thanks

#### javcek

##### Diabloii.Net Member
PinkeyandtheBrain said:
Please explain what exactly the "no drop" is i am quite confused.
thanks
I think Uzziah in an earlier post in this thread explained it quite a bit better that I can, but basically is the chance of a monster not dropping anything everytime it gets a pick from its TC. This chance decreases with the number of players in the game, and what we are trying to find out here is the way the formula works when you user the /players X command.

#### Randall

##### Diabloii.Net Member
Just to get a quick overview

Am I right when I say:

- When this is true, ATMA is not correct for higher player settings. (it is for P1)

- Most likely all chances of items on higher player settings will be lower as chance for no-drop will be higher.

Randall