If something has an 8% drop rate

[Evelynn-khadgar]

@Aythina-illidan

Something the people crying about 150 kills in on LK and not seeing the mount have not grasped yet.

There is actually ~34% chance of that happening!

[Deos-thrall]

@Nsk-draktharon

That is semantics

The precision and technical use of language actually matters? Wow, who knew.

@Orfieus-area-52

The odds of flipping a coin is 50%. The odds of flipping two heads in a row is 25% but if you flip a coin every morning your odds of it being heads/tails will ALWAYS be 50%

So you have a 8% drop rate no matter how many times you run, that 8% will never increase

But that wasn’t the question. The question was given an 8% chance, how many trials are needed to see at least one success?

The figures above reflect that e.g. 90% and 99% of the population will have achieved success within the given number of trials. You can always be in that 10% or 1%, but statistically speaking the more trials you have performed, the greater your total incidence of at least one success.

You miss 100% of the shots you don’t take.

[Trexx-gorefiend]

I got the Mount second try

[Zemus-hyjal]

@Rgonewildd-sargeras

Theoretically how many times would someone have to kill it to statistically say they would have it?

Anywhere between 1 kill and infinity.

[Deos-thrall]

There is a ludicrously small chance we don’t actually exist too, but it’s small enough to not consider in regular argument.

It would be interesting to set up a population of machines, each representing WoW’s population, and have it run the trial of how many attempts are needed to halt. I’m willing to bet the program halts, the only question is time.

[Shadina-gorefiend (Shadina)]

Divide 8 into 100. 8x12=96. Which means in theory, you should get it within 13 kills because the 8% drop rate would push over into 100% on kill 13.

[Sunzsu-alexstrasza]

In theory

But the probability approaches 1 as you get to 13 kills. Never meets 1 though. So theoretically it could never drop, ever, after infinite kills.

[Shalashaska-stormrage]

0.92^8 = 0.51; so it has a 49% chance of dropping by the 8th kill
0.92^9 = 0.47; so it has a 53% chance of dropping by the 9th kill

The median person will get it on the 9th kill.

[Mute-silvermoon]

This thread makes my head hurt.

[Orfieus-area-52]

@Deos-thrall

@Nsk-draktharon

That is semantics

The precision and technical use of language actually matters? Wow, who knew.

@Orfieus-area-52

The odds of flipping a coin is 50%. The odds of flipping two heads in a row is 25% but if you flip a coin every morning your odds of it being heads/tails will ALWAYS be 50%

So you have a 8% drop rate no matter how many times you run, that 8% will never increase

But that wasn’t the question. The question was given an 8% chance, how many trials are needed to see at least one success?

The figures above reflect that e.g. 90% and 99% of the population will have achieved success within the given number of trials. You can always be in that 10% or 1%, but statistically speaking the more trials you have performed, the greater your total incidence of at least one success.

You miss 100% of the shots you don’t take.

I addressed this, you will always only have a 8% chance of getting the item, on your first run and on the 100th run you will only have a 8% chance

As a population something like 99% of the population will have seen the item at least once after 55 runs, but population odds does not apply to the individual

[Stealthius-borean-tundra (Stealthius)]

@Rgonewildd-sargeras

Everything can be quantified

Except for gravity.

[Hummything-winterhoof]

@Orfieus-area-52

if the ball count always stays at 100 and there is only ever 8 red balls how does doing it more often increase your odds?

Because you’ve actually got two probabilities at work - there’s the raw drop rate, which at 8% says that you have a 92/100 chance of NOT getting the drop on a SINGLE attempt. There’s ALSO a cumulative probability that increases with the number of attempts because it becomes increasingly more unlikely that you’ll fail on EVERY attempt you make. Both probabilities are accounted for in the formula:

P = 1 - ((1 - drop rate) ^ number of attempts)

So, at an 8% drop rate over 10 attempts, the math becomes:

P = 1 - ((1 - 0.08) ^ 10)
P = 1 - (0.92 ^ 10)
P = 1 - (0.4344)
P = 0.5656, or about a 57% chance that you’ll get that drop within 10 attempts.

[Wild-velen]

There’s at least a 95% chance you get the item in ceiling(ln(0.05)/ln(0.92)) = 36 kills.

[Ehiztari-blood-furnace]

@Onikaroshi-mankrik

he more times you do something, higher the probability will be that you will succeed.

Absolutely wrong.

“The dice have no memory” is a statistical truism that you’re breaking here.

What happened in the past has absolutely no influence on the next event unless there is some linkage (like “drawing a red card from a deck” where all your other attempts have been from the same deck and the ratio of red to black changes after every draw).

Vegas would LOVE you.

[Onikaroshi-mankrik]

Absolutely right, that’s how probability works.

It’s not a guarantee that you’ll get something in 1000 tries, but you’ll have a higher probability of having gotten it in those tries.

I feel like probability is far too abstract a concept for some people.

[Wild-velen]

I think he thought you were talking about the probability of the next event and not the collective probability.

[Onikaroshi-mankrik]

Maybe.

Each time is still only 8%.

[Gaahr-whisperwind (Gaahr)]

@Orfieus-area-52

I addressed this, you will always only have a 8% chance of getting the item, on your first run and on the 100th run you will only have a 8% chance

you’re conflating the following two questions:

on any single attempt, what are the chances of a particular item dropping given a drop rate of 8%?

and

how many runs should i expect to do before i see a particular item drop given a drop rate of 8%?

the answer to the first question is easy: it’s 8%. the answer to the second question is covered by something called a binomial probability distribution, the formula for which was linked earlier in the thread.

probabilities aren’t exactly intuitive at first blush, so here’s an example of what i’m talking about. suppose you have two people, both of whom are running stratholme for a shot at the stupid horse. person a runs stratholme once. person b runs stratholme 20000 times. who is more likely to get at least one mount out of their run(s)?

[Evelynn-khadgar]

@Ehiztari-blood-furnace

Vegas would LOVE you.

They do love him because he is using the logic they use to design all their games such that they know they will profit in the long run.

[Yinz-area-52]

It’s funny that not only are people completely ignoring cumulative probability in this thread, they are getting angry about it when someone tries to explain it.