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Joystiq

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## Reader Comments (Page 1 of 1)

9-02-2010 @ 4:20PM## Artificial said...

"The drop rate for the ashes is thought to be around 1 percent, and Tempest Keep is still on a weekly raid lockout, so you'll probably be farming Kael for a long time before you see one."

Right. Also, if you're going in a group of ten, that's 0.1% chance per run of getting one. At 52 attempts per year, your odds of getting one in a year of attempts is 5%.

Realistically speaking, the odds are against you ever getting one. They'll be turning off the WoW servers long before your odds rise above 50/50.

Reply

9-02-2010 @ 4:36PM## Hih said...

Which is why you make a pug reserving the mount for yourself but offering 10k gold to the winner of the roll if it does drop.

9-02-2010 @ 5:25PM## Kunikenwad! said...

Not how statistics works, really. Your chance is still 1% no matter what, but in layman's terms, yeah, it's hard to get :D

9-02-2010 @ 9:45PM## Dazaras said...

Yes, actually, that is how statistics work. In 52 tries with a 0.1% chance of success each time you have a 5.07% chance of success overall. You are referring to the fact that each individual attempt has a 0.1% chance of success regardless of the number of previous failed attempts, which was not disputed. Please actually learn statistics before you argue with people about statistics, people say that a lot and it really annoys me.

In 693 tries you have a 50% chance of getting the mount.

9-03-2010 @ 3:20AM## elphie said...

Actually, if the chance of it dropping is 1% for arguements sake, then if you run it five times or one hundred times or one million times it is still only ever a 1% chance that the mount will ever drop. Following your logic, if one was to run it for over twenty years you would be guaranteed to see it, which you are not. Maybe you need to learn about statistics too ;)

9-03-2010 @ 4:42AM## Matt said...

What elphie wrote is incorrect, as would be obvious to anyone who has taken a serious course in statistics. Artificial and Dazarus are correct, but they haven't supplied their math, which may be the cause for dispute.

First off, the assumptions: 52 raid lockouts per year, .1% chance you'll get the drop on any given run. This .1% is converted to a .001 for use in calculations, as something with a probability of 1 is statistically a guaranteed, or 100%, outcome, and .001 is 1% of 1.

Next, the method. Often in statistics, it is convenient to determine the likelihood of some event occurring within some number of attempts by first determining the likelihood that said event does NOT occur during the attempts, then subtracting that number from 1 (or 100%). Since there are only two possible outcomes here (either the Ashes drop or they don't), we can deduce the probability of an UNsuccessful attempt to be .999 (1 - .001 = .999). From there, we must determine the probability of having 52 successive unsuccessful attempts. This seems to be the area most people have trouble with, so I'll start with only two successive attempts to illustrate the logic behind the math. For each individual attempt, there is a 99.9% chance you will not get the mount. However, over two attempts, there is a .999 * .999 = .998 chance you will not get the mount, or a 1 - .999 * .999 = .002 chance you will. What confuses people is that this does NOT mean that, having failed to get the mount in the first lockout, your odds of getting the mount in your second lockout increase to .002. What it DOES mean is that if you go for two lockouts, you have a .2% chance of getting the mount in that period of time. So next, we'll do the math for 52 lockouts: .999^52 = .9493, and 1 - .9493 = .0507, which translates to 5.07%, exactly as Artificial and Lazaro said. (For those interested in how this works, look up Bernoulli Trials)

For those out there who still doubt the math, I will also demonstrate how this behaves at the extremes for the possible number of trials, zero and infinity.

Chance of getting a drop after zero attempts: 1 - .999^0 = 1 - 1 = 0

Chance of getting a drop after an infinite number of trials: A little more complicated than an equation, as writing some number to the power of infinity is bad mathematics. What this boils down to is a converging series in the form of r^n, where r is a constant less than 1 (say, .999) and n is the number of trials. As n approaches infinity, r^n gets closer and closer to 0. (You can see this for yourself on a calculator, just pick a number between 1 and 0, multiply it with itself, then multiply this solution by the original number, then multiply THAT solution with the original number, etc. it will eventually decay to near zero). Therefore, the probability here roughly equates to 1 - .999^infinity = 1 - 0 = 1.

The fact that the numbers perform so beautifully at the extremes lends credence to the methodology as well, as formulas that don't hold up in extreme cases are usually dismissed as incorrect in mathematics. (Yes, the numbers say doing this raid an infinite number of times is the only way to guarantee yourself a phoenix, but this is clearly is an impossible feat. Them's the breaks.)

9-04-2010 @ 5:27AM## elphie said...

I understand that you are saying that by running it more then you have increased your chances oif seeing it drop but I am saying that the actual drop rate of the mount never changes, so if it was run three hundred times with the mount being a one per cent drop rate, then the chance of seeing it drop out of all runs is three per cent, but the actual drop rate is still one per cent. I obviously didn't explain what I meant very clearly but you could run a million times and the mount will only drop at a one per cent rate each time. The simplified form of 3 out of 300 being one out of 100. And I did not do a serious course in Maths, but the maths teacher sitting next to me did severeal years ago :). So to summarise, yes if it was a one per cent drop and you ran it 500 times you would have increased your chance of seeing it, but it is always going to drop at the same rate every time....one per cent.