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.

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