
Originally Posted by
'Frosty',index.php?page=Thread&postID=76408#post76 408

Originally Posted by
'-silencer-',index.php?page=Thread&postID=76404#post76404

Originally Posted by
'daviddoran',index.php?page=Thread&postID=76111#po st76111
A boss that only drops an specific item 20% of the time will need to be run 10 times to have a 90% total chance of dropping that item. That means that even if you run a boss 10 times, there's a 10% chance of that item not dropping, and that's for an item with a 20% drop rate. It gets much worse for those 5% or 1% drop items!
I'm not sure it works out quite that way... every time you run the instance..you have the same 20% chance (assuming it is in fact 20% chance) to drop the item. It doesn't matter if you run it 100 times..you still only have a 20% chance each time. While odds should say that you should get it by then..there is still a chance you won't....always.

Yes you're right, that's the rule of independent probability. However, the formula I posted is correct. Over a group of runs, the *overall* chance of getting your desired item at least once is higher than a single run. Otherwise, no one would ever play the lotto. Here's the simple way to confirm the formula posted:
An item has a 50% chance of dropping. Each *single* run will always have a 50% chance of dropping, but we're adding up all the runs. Compare this to flipping a coin. A 50% chance of having tails on any given flip, but if we're given two opportunies, we have a 75% chance of getting tails on at least one of those flips.
(1-.5)^2=1-Y
.5^2=1-Y
.25=1-Y
Y=.75
75% chance of getting at least one tails with two coin flips. This same fomula works for drops in WoW, given that there are no problems with the random number generator seeding.
Visually, here's the possilbe results of two coin flips:
HH
HT
TH
TT
Clearly, 75% of those combinations contain tails, and since there's an equal chance of heads or tails at 50%, 75% is the final probability of having at least one tails on two coin flips.
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