God you are not that bright. Let me try again to make it even more simple.

You = per run
Him = over multiple runs

How do you not understand this lol.

I understand that fine.
It's still not the correct way to assess the data.
He is getting a drop rate, that's not correct.

it's 0.03% drop.
Period, there's no way to argue it, if you do, enjoy your slam, i'll gladly hand them out.

This post was edited by iLuxxy on Feb 16 2013 02:16pm

Still ISO how i'm wrong like you say.
5k fg to the winner.

This post was edited by iLuxxy on Feb 16 2013 02:19pm

Lol i thought he was clueless at first as well, but then it's starting to make sense. I'm leaning more towards iLuxxy's claim at this point.

The events that they're describing is mutually exclusive where the events of one outcome does not impact of the outcome of another. 1 character having a .03% chance to get the mount does not increase or influence another character's odds of getting the mount.

by increasing the number of trials (in this case, the number of characters you do it on), you're only proving that the likelihood of getting the mount is .03%...

with the example of tossing a coin, you have a 50% chance to get heads...if you toss that coin 3 times, you'll have a 66.67% chance of getting either heads or tails...increasing the number of trials only supports the claim that the probability of getting heads is 50%...

For example: you toss the coin 1000 times, you are now more likely to get heads 50% of the time than if you were to toss it 3 times. (you get closer to the theoretical probability by increasing the number of trials, not actually impacting the probability itself)

This post was edited by Cyba on Feb 16 2013 02:28pm

Actually the equation only guarantees a drop if you do it an infinite number of times.

The limit as x approaches infinity of 1-(1-0.0003)^x = 1
If you do it an infinite number of times the chance of success will be 100%, for any finite number of times it will be less than 100%.

The probability of getting all failures (Pf) + probability of not getting all failures (Ps) = 100%

Pf + Ps = 1

Pf = (1- 0.0003)^n = (0.9997)^n

Plugging that in to the equation above:
(0.9997)^n + Ps = 1

solve for Ps:
Ps = 1 - (0.9997)^n

probability of not getting all failures = probability of one or more successes = Ps = 1 - (0.9997)^n




Yes, the probability of 'getting lucky' on any individual run is 0.0003 and that is what the equation models.

Think about it in a simpler case. Flipping a coin. You have a 50% chance to get heads on any given flip. Lets consider the probability of getting 1 or more heads in 10 flips.

Pf + Ps = 1
Pf = (1 - .5)^10
(1 - .5)^10 + Ps = 1

Ps = 1 - (1 - .5)^10

Ps ≈ 0.999 ≈ 99.9%

While the chance of getting a heads on any given flip is still only 50%, the chance of getting 1 or more heads in 10 flips is almost 100%. It will never be 100% unless you do an infinite number of flips.

This is not the gamblers fallacy. The gamblers fallacy would be to wait until you had done 5 of the 10 flips (and not having received a heads yet), then using the formula above with n = 10 for the last remaining 5 flips. The correct method would be to use n=5.

So in that situation (after 5 failed flips) someone using the gamblers fallacy would say you have 99.9% chance to get heads in the next 5 flips. Calculating it correctly it would be 96.9% chance.

This post was edited by Azrad on Feb 16 2013 03:14pm

dafuq did i just read

Yes, correct.
I'm trying to talk straight statistic, not theoretical though.
The only statistic is the original drop chance, the point is the ACTUAL chance is not increasing is my point. "theoretically" it is because of more attempts.



Arguing over the laws of statistics for the pure sake of boredom and epoints I guess.

This post was edited by iLuxxy on Feb 16 2013 03:22pm

This thread is littered with mathematically incorrect statement by you. Here is a few gems:

In response to a correctly calculated value:



other gems:





Furthermore your continuous statements that the drop rate on an individual run does not change is a straw man, since no one (as far as I can see) implied it was anything else. The correct equation was posted on post 14 which clearly shows the drop rate is constant, but that your overall chance of getting the success increases with more trials.

This post was edited by Azrad on Feb 16 2013 03:39pm

Thank you for the feedback and well thought out response.

ILuxxy arguements are mute. The original comment was regarding probability it will drop this year... no idea why he keeps trying to answer something THAT WASN'T THE QUESTION.

It's like someone saying "The Sky is Blue" and ILuxxy states "The grass is green". Your statement you keep pounding into the ground has NOTHING TO DO with the probability of the mount dropping in 750x attempts for this season.
Instead, this guy will continue on some tangent unrelated to the original question in a quest to "Slam me".




Mikol, help this kid out. What degree are you going for and what level are u currently studying

This post was edited by Stok3d on Feb 16 2013 03:44pm

It's just a theoretical statistic, and is not correct.
The chance is still 0.03% per try.

If you wan't to believe thats the true percent, the math is not wrong, but it's not a correct chance to obtain it.

Ty men.



Not reading a bunch of boxes.
I'm not incorrect, 5k if you prove i am.
If you can't prove me wrong, i'll take this unsatisfying victory as well.

OH, i realize all you did, was change a bunch of my quotes like a little kid, because you got shown you're retarded, and have the reading comprehension of a monkey.

This post was edited by iLuxxy on Feb 16 2013 03:47pm