Hallow's End 2019 Now Live

[VekniDarkIron]

But that's exactly how math works. While each individual attempt is 1% chance your overall chance increases with each try. It's easier to score a bullseye with 10 darts than with one.

No, your overall chance remains the same. This is grade school level math. The odds of scoring a bullseye do not change with each throw. If I turned away from the bullseye and kept tossing darts, am I going to have a better chance at getting a bullseye with 10 darts than one? Further, you're talking about a factor that can be influenced, as demonstrated by how I can reduce my odds of scoring a bullseye to zero by turning around. Nothing can change the drop rate of that item.

[Alaenore]

No, your overall chance remains the same. This is grade school level math. The odds of scoring a bullseye do not change with each throw. If I turned away from the bullseye and kept tossing darts, am I going to have a better chance at getting a bullseye with 10 darts than one? Further, you're talking about a factor that can be influenced, as demonstrated by how I can reduce my odds of scoring a bullseye to zero by turning around. Nothing can change the drop rate of that item.

First, sorry, english isn't my main language and I don't speak about maths in other languages very often.
The individual chance doesn't change, but the odds of seeing it drop after X run increase. This is also basic maths.

When you flip a coin, you have 50% chances of getting Heads or Tails.
Flip it twice, you can have either (Heads + Heads), (Heads + Tails), (Tails + Heads), (Tails + Tails). So the probability of seeing *at least* Heads once would be 75% (HT, HH, TH, only TT has no Heads).

And yet, these events are independant. The odds of getting tail or head doesn't change, the odds of seeing it at least once does.
Which fits the first equation posted, 1 - ((1-X)^Y) : 1 - ((1 - 0.5)^2) = 0.75


Apply it to our current problematic, the loot!

Do it once, you have 1% of seeing it.
Twice you have 1 - ((1 - 0.01)^2) = 1.99%
10 times, 1 - ((1 - 0.01)^10) = 9.56%
100 times, 1 - ((1 - 0.01)^100) = 63.34%
200 times, 1 - ((1 - 0.01)^200) = 86.60%
Etc.

[VekniDarkIron]

Still wrong. Why haven't some people seen it after a thousand runs, by your terrible math?

One in a hundred twice is two in two hundred which is.... One percent. One in a hundred twice does not just double the part of the equation that you want to double.

[imtrick]

Still wrong. Why haven't some people seen it after a thousand runs, by your terrible math?

One in a hundred twice is two in two hundred which is.... One percent. One in a hundred twice does not just double the part of the equation that you want to double.

So, you're honestly trying to say that, no matter how many times you flip a coin, there's a 50/50 chance you'll ever get heads? Do you realize how ridiculous your argument is?

The math is right. It's statistics. Really basic statistics. Of course your chance of success increases with the number of attempts. That doesn't even take statistics to figure out; even the most basic grasp of logic will tell you it's true.

[imtrick]

Come on, people. It doesn't take much more than a basic grasp of logic to realize that if you flip a coin twice, you have a better chance of one of them turning up heads if you flip it twice. I get people not having a grasp on statistics, since that's fairly advanced, complicated stuff. What I don't quite understand is how people can say with a straight face that the odds of success are the same whether you try once or 100 times. If that was true, there'd never be any use to attempting anything a second time.