Maximize Invincible's Reins farming in ICC 25 Heroic!

[Owning]

1,672 runs = 32 years.

Then again, this is useful. +rep

Divide that by 112 characters.
Seeing how i have 4 accs all linked/have tons of 80 +.. Im bound to see one soon if i run on all

[rath999]

I know this is old but it confused me a bit.

I done what you said with sindragosa on my druid alt on 25 man. I then extended the lock on my alt and my friend when inside, but he kicked me instead of letting me leave. Then after a kill lich king it said i had 0/12.

Also at first i couldn't get inside it said 'another member of the group has killed a boss' bla bla, so i changed it to 25 HC while i was inside ICC and then left the instance, could this have caused me to remove my lock that i had at lich king?

[ketrish]

soon ppl will write "guides" - "how to proper log in"
....
&
....
old since i guess cata & mop ?

[csluyuan]

Thanks for sharing this!

[Razor33]

You certainly took a good amount of time to write this for people who are farming and desire this wonderful mount +rep from me

[Taftvalue]

this doesnt work, if the party kills the Lich king your id will be 12/12 aswell.
contacted a gm about it some time ago and he confirmed it.

[TommyT]

1,672 runs = 32 years.

Then again, this is useful. +rep

for 1 character
the whole point of this is to run multiple times

[4L3X]

On every thread like this people always find the opportunity to argue about it, but quite frankly the more often you do it, indeed the probability INCREASES, although you can technically go on for years and never see the mount. And true, if the drop rate's 1%, each kill is independent; you might even see b2b2b2b invinciblkes, ebcause each drop is independent. But yeah, this works with anything farm-able, and i know a lot of people on my server do this for Mimiron's head (whcih I'm not a great fan of, but each to their own)

[meleevirus]

Assume that the drop rate for Invincible is 1%.
The probability of it dropping at least once in 100 times is 1 - (0.99)^100 = 0.63396765872 ~ 63.4%. In other words, the probability of you seeing it at least once is ~ 63.4%
Reasoning: P(not dropping in a single run) is 0.99 or 99%. P(not dropping at all in 100 runs) is (0.99)^100 = 0.36603234127 ~ 36.6%. So P(dropping at least once) = 1 - P(not dropping at all in 100 runs) = 1 - 0.366 = 0.634 = 63.4%

Other fun facts:
P(dropping only once in 100 runs) = 0.01*(0.99)^99 * 100C1 = 0.36972963765 ~ 37.0%
P(dropping twice in 100 runs) = (0.01)^2*(0.99)^98 * 100C2 = 0.18486481882 ~ 18.5%
P(dropping three times in 100 runs) = (0.01)^3*(0.99)^97 * 100C3 = 0.0609991658 ~ 6.1%
P(dropping four times in 100 runs) = (0.01)^4*(0.99)^96 * 100C4 = 0.01494171485 ~ 1.5%
P(dropping five times in 100 runs) = (0.01)^5*(0.99)^95 * 100C5 = 0.00289778712 ~ 0.3%
... and so on

Double check: 37%+18.5%+6.1%+1.5%+0.3% = 63.4% (tick)

[rdruid69]

Edit: delete this