The issue of player-count when

[[wcip]Angel]

If I join, say, a Baal-run and while the others are knee-deep in Baal-underlings, I'm putting the final screws on the Mephster, do the amount of players factor in on the loot Mephisto will drop?

1. Does it have any effect?
2. Do I need to be in a party for it to take effect?
3. Will it drop more items or better items?

- [wcip]Angel

edit: Same applies for Pindle. Does it make a difference when he's in the same act as the rest of the players?

[Thrugg]

1. Yes.
2. No, but if you are in party they will count more. An unpartied player only counts as half.
3. It will drop "no-drop" less often. When a monster doesn't drop, or a boss drops less than 6 items, it is actually dropping "no-drops".
It has no effect on the quality or type of the items it does drop.

It won't affect Pindle at all, not because of the act, but because Pindle never drops "no-drops" so he can't drop them less often. Pindle always drops 2 items and 4 potions.
You will get more drops from Pindle's minion though.

[Crystalion]

It has no effect on the quality or type of the items it does drop.

Indirectly it can, according to the posts of those more expert than I, due to TC tree traversal (if I understand correctly) for drops of critters like the Countess, when the "type" of drop you wish for is, say, runes.

Admittedly the "type" is not actually changing (she'll drop runes with players 1 or with players 8) but apparently the odds change significantly, so this does seem on-topic.

-- Crys

p.s. I still think I've long ago read posts discussing the drop changes made to nerf v1.10 cow and pindle runs that refer in some way to aspects of no-drop or MFing (but I wouldn't want to be responsible for any wild rumors... next thing you know, people might start behaving strangely, like using "of thieves" and forgoing use of "of harmony" in older versions of D1 ;) )

p.p.s. I don't recall seeing a good explanation of no-drop vis-a-vis Find Item (assuming there is anything to explain)

p.p.p.s. here's a "wild rumor": the drop vis-a-vis players N occurs at monster death, and not according to the HPs/xp of the monster set at monster (players N) creation time... true or false?!? :D

[Obi2Kenobi]

Indirectly it can, according to the posts of those more expert than I, due to TC tree traversal (if I understand correctly) for drops of critters like the Countess, when the "type" of drop you wish for is, say, runes.

Admittedly the "type" is not actually changing (she'll drop runes with players 1 or with players 8) but apparently the odds change significantly, so this does seem on-topic.

Her Rune Drop is used when she does her No Drop, so it actually lowers your chances of getting runes if you kill her in a multiplayer environment.

p.p.s. I don't recall seeing a good explanation of no-drop vis-a-vis Find Item (assuming there is anything to explain)

I think (and don't quote me on this, unless you are proving me right), that it's just the %, and thus no drop doesn't factor in.

p.p.p.s. here's a "wild rumor": the drop vis-a-vis players N occurs at monster death, and not according to the HPs/xp of the monster set at monster (players N) creation time... true or false?!?

From what I've heard, true.

[Jarulf]

>Indirectly it can, according to the posts of those more
>expert than I, due to TC tree traversal (if I understand
>correctly) for drops of critters like the Countess, when
>the "type" of drop you wish for is, say, runes.

I think she is the only case with such a special drop type.


>p.p.s. I don't recall seeing a good explanation of no-
>drop vis-a-vis Find Item (assuming there is anything to
>explain)

If I recall correctly, Find Item sets the NoDrop to 0, but I can remember wrong and mix it up with something else.


>p.p.p.s. here's a "wild rumor": the drop vis-a-vis
>players N occurs at monster death, and not according
>to the HPs/xp of the monster set at monster (players
>N) creation time... true or false?!? :D

Yes, for NoDrop, the player count is done at the time the item is created (that is at death time). AT least it hs been so in previous versions since there is no playercount variable on monsters (I have a vague memory of seeing such a stat added though but I wonder if it would be used for NoDrop. I have no printout of game code any more though so can't check it out (at least not easilly).

[Brista]

If I recall correctly, Find Item sets the NoDrop to 0, but I can remember wrong and mix it up with something else.

I'm sure this is correct. (The skill can of course fail in which case you get nothing)

[Brista]

Her Rune Drop is used when she does her No Drop, so it actually lowers your chances of getting runes if you kill her in a multiplayer environment.

I think this is not quite correct

What I believe happens is the game first checks her regular drop and then her special rune drop for each pick. If unsuccessful with both it then assigns a no drop for that pick

With more players each pick is more likely to succeed in getting a regular drop so it reduces the rune drop chance simply because for that pick it didn't get so far as to check the rune drop

[Brista]

Simple answer:

Yes it does effect Meph's drop. He has roughly 20% nodrop. In a full game with 7 running Baal and you running Meph you'd cut that to about 1% at the cost of Meph being tougher

Complex answer:

More players reduce the nodrop

D2data.net provides nodrop information: here's Meph's Treasure Class chart:

http://www.d2data.net/tc/Mephistoq_tch.html

Now from Jarulf's mf guide for 1.09 here is the explanation of how nodrop is calculated:

1. Calculate the fraction of NoDrop, that is, NoDrop/Sum_Prob 2. Calculate "number of people". Here each character in the party (and in same area) of the killer count as 1 and those not in the party (but in the game) as a half. This value IS truncated to an integer. 3. Raise the value calculated in step 1 to the power of the value from step 2. Assume, the example above, we had 75/104=0.7212. Assume there are 2 people in the party and same area, and 3 other people, so the value from step 2 is 2+1.5=3 (truncate). So we calculate 0.712^3 = 0.375. 4 Adjust the NoDrop value so that it is a value that makes it the fraction calculated in step 3. That is multiply the Sum_prob minus the old NoDrop value (plus the value to be calculated) with the value from step 3. In our example it would be 29*0.375/(1-0.375) = 17. This is the new No Drop value.

The full article is here.
http://phrozenkeep.it-point.com/forum/kb.p...de=article&k=58

The main change of 1.10 was to the numbers not the mechanics* so if you get your head around Jarulf's explanation and then apply it to the D2data.net numbers you will have pretty accurate numbers to work with

(*except in one regard which does not effect Mephisto:
http://www.theamazonbasin.com/d2/forums/in...showtopic=33345 )

[Jarulf]

I think this is not quite correct

What I believe happens is the game first checks her regular drop and then her special rune drop for each pick. If unsuccessful with both it then assigns a no drop for that pick

With more players each pick is more likely to succeed in getting a regular drop so it reduces the rune drop chance simply because for that pick it didn't get so far as to check the rune drop

The countess drop works the following way. It is made up by two parts, done after each other. The first part is in turn made up of a "normal" drop, that is, it is a number of random items, each of which can be a NoDrop as well. The rune drop is a drop made up by droping random runes.

All this works OK except for the fact that the game has a hard cap on a maximum of 6 items on any particular kill/drop. Now, the countess Treasure Class and its drop is set up so that the first part, if no NoDrops are generated, plus the number of runes will be higher than 6 items. In that case the game aborts the drops immediately and the full run drop will not take place.

Since more players in a game means a smaller chance for a NoDrop, it means that the more players there is in a game, the higher is the chance that the drop for the Countess will be aborted due to the cap at six items and it will be the special run drop that will not be fully done.

[gta-maloy]]

I think this is not quite correct

What I believe happens is the game first checks her regular drop and then her special rune drop for each pick. If unsuccessful with both it then assigns a no drop for that pick

With more players each pick is more likely to succeed in getting a regular drop so it reduces the rune drop chance simply because for that pick it didn't get so far as to check the rune drop

The countess drop works the following way. It is made up by two parts, done after each other. The first part is in turn made up of a "normal" drop, that is, it is a number of random items, each of which can be a NoDrop as well. The rune drop is a drop made up by droping random runes.

All this works OK except for the fact that the game has a hard cap on a maximum of 6 items on any particular kill/drop. Now, the countess Treasure Class and its drop is set up so that the first part, if no NoDrops are generated, plus the number of runes will be higher than 6 items. In that case the game aborts the drops immediately and the full run drop will not take place.

Since more players in a game means a smaller chance for a NoDrop, it means that the more players there is in a game, the higher is the chance that the drop for the Countess will be aborted due to the cap at six items and it will be the special run drop that will not be fully done.

To elaborate, I did research on the countess rune drop:

The first decision on rolling is that of normal items as stated by Jarulf. The second, known as the special rune drop is divided in three seperate chances at attaining a successfull drop. The less chances the countess has for "no drop" causes the special rune drop to have better drop odds.

The only thing this really effects is her ability to drop runes above Ist in hell difficulty. When in a full game she has better odds of dropping three successfull special drops which cannot be higher than ist runes. I say this because no one really runs the countess for items.

Nice to see you back Jarulf.

[Thrugg]

I suggest you reread Jarulf's reply.

A full game will result in more drops from the item drop, in fact, all 5 possible items will drop because a full game will reduce the Countess' no-drop to 0.
The game has a hard cap of at most 6 items dropping any time you kill any monster.
Which means that, even though you are more likely to get a successful drop from each of the three rune drops in a full game, there is only room under the 6 cap to drop one rune. Even if that one rune drops the first try, and you still have two tries left, too bad, 6 items have dropped, and the drop function ends.

It ends up being a combinatorial problem - is it better to have more chances with worse odds or less chances with better odds?
If you crunch the numbers, it turns out it is weighted towards the former. In a solo game you expect the Countess to drop around 1.85 runes from her rune drop each kill. Whereas in a full game you expect precisely 1 every time. If you are seeking runes, the Countess should be run solo.

[CelticHound]

Angel,Aug 23 2004, 02:34 PM]2. Do I need to be in a party for it to take effect?

2. No, but if you are in party they will count more.  An unpartied player only counts as half.

I thought that a partied player in a different map/act also counted for half. (As noted in Brista's post quoting Jarulf.)

-- CH