How does the probability really work on orbs?

"

Maxmus wrote:

So theorically i could get 6l with just one fuse?

Not just theoretically, in practice, each time a 6L appears, it was rolled with just one fuse. It doesn't matter how many fuses you used in previous failed attempts to roll a 6L, the only fuse that counts is the one that actually linked all six sockets.

Fusings have no memory, none of them ever remember how many previous tries you made. When you try a fusing, you have a 1 in x probability to get a 6L, every single time you try. Some people succeed in 5 tries, some fail after 20,000 tries, there is no set number that will ever guarantee you get what you want. That's the nature of probability.

Took me 16 fuses on my first 6link.

Friend did one with 5.

Second friend used 3k fuses and didnt get it.

The way you do it is by calculating the probaility of fail.

For exemple, lets say that the chances to get a 6L are 1/1250. How many fusings would you need to have a 50/50 chance to link it (the right question is, how many times can you try and get a 50% chance to fail) ?

You do it like this :

First take the probability of the event to happen : (1/1250) = P1

Now you want to calcultate how many times you want to try and get a 50% chance to fail for it to happen.

1-P1 is the chance it will not happen.

For exemple, is the base probability is 1/2, then 1-(1/2) = 0.5 (= 50% for it not to happen).

If you try one time, you have 50% chance to fail (chance to fail is P2), if you try two times in a row you have 0.5*0.5 = 0.25 (25%) chance to fail, three times 0.5*0.5*0.5 = 0.125 chance to fail, etc...

So the formula is (P1 = chance to succed, P2 = chance to fail)

(1-P1)^x = P2

If you replace now with the fusings odds for 6L to want to know how many you need to have a 50% to fail you get :

P1 = 1/1250

P2 = 0.5

So you get

(1-(1/1250))^x = 0.5

You want to find x, x beeing the number of try (the number of fusings)

Lets say (1-P1) = A and that 0.5 = Z

See next post, would show the rest, don't know why

Then (1-(1/1250))^x = 0.5 -> (1-P1)^x = P2 -> A^x = Z

x can be found like this : a^x = z and x = LOGa (z) and LOGa (z) = LN(Z)/LN(a)


Then x = LN(Z)/LN(a) -> x = LN(P2)/LN(1-P1) -> x = LN(0.5)/LN(1-1/1250) = 866


So (1-(1/1250))^866 = 0.5


You will need 866 fusing to have a 50% chance to fail (and to succed) your 6link.


If you want 33% chances to fail (67 chances to succeed) you will need :

x = LN(0.33)/LN(1-1/1250) = 1385 fusings


Using those maths, you can calculate that with 10 fusings you have 1/125 chances for it to happen, its very low but it can happen. With 3000 fusings you have 90.9% chance to happen, and its still a 9.1% chances to fail, it is very high, this is why many people complain about spending thousands of fusings.

I hope it helped to have grasp on what you can expect to spend on a 6L and why some people manage to get it with less than 10 fuse and why some still don't get it with 3000 fusings.

Yeah thanks for the replies. This forum is awesome ;P