2011-01-26, 04:45 PM
It seems I've stumped myself with my own math. I think I’m over-thinking the simple. I’ve got all my combinations worked out, but the probabilities are confusing me a bit. I think I’m over thinking it. In Brandish/IS spam; there are 4 things that can happen:
B(1) = 1 orbed Brandish (0.78)
B(2) = 2 orbed Brandish (0.22)
F(1) = 1 orbed FA (0.78) (wrt orb charge, not FA activation)
F(2) = 2 orbed FA (0.22)
Now let’s deal with the scenario of 5, 1orbed Brandishes in a row:
B(1)->B(1)->B(1)->B(1)->B(1) = 5 orbs
At first I assumed it was:
(0.78)^5 = 0.2887174368
There’s no way there’s a 28.87% chance to have 5 Brandishes in a row with 1 orb and no FA activations. Then I figured I’d account for the FA’s by multiplying each term by 0.6 for the fact it didn’t activate. Since there should be 5 unsuccessful FA’s
0.2887 * (0.6)^5 = 0.022449312 = 2.24% <- seems a lot more accurate. This problem solved itself.
So say I take a random outcome from my tree:
B(1)->F(2)->B(1)->B (1 orb brandish -> 2 orb FA -> 1 orb brandish -> brandish)
This is where I ran into a bit of trouble, at this point, it wouldn’t matter whether or not FA activates, so I just threw down a B with no subscript while pondering how to deal with this dilemma. I came up with these symbols.
B(0) = Brandish at 4 orbs (you'd get to 5 orbs regardless of the ACA "Proc" or not)
F(0) = FA activates/wouldn’t charge orbs
F(00) = FA doesn’t activate/wouldn’t charge orbs
B(0) accounts for a Brandish at 4 orbs, since you’re going to get to 5 anyway, the amount of orbs charged doesn’t matter. F(0) and F(00) account for an FA activating from the final brandish
Redoing the scenario I get
B(1)->F(2)->B(1)->B(0)->F(0)
And
B(1)->F(2)->B(1)->B(0)->F(00)
Using the former scenario [B(1)->F(2)->B(1)->B(0)->F(0)]
(0.78 * 0.4) -> (0.22 * X) -> (0.78 * 0.6) -> (?) -> (?)
This is the part that’s confusing me, since the first term has already accounted for the FA activation, can I leave out the X? And what do I do for the “?” so I accurately account for those scenarios (The subscript 0’s)? I haven’t taken probability or stats in at least 3-4 years so I’m going off memory here.
B(1) = 1 orbed Brandish (0.78)
B(2) = 2 orbed Brandish (0.22)
F(1) = 1 orbed FA (0.78) (wrt orb charge, not FA activation)
F(2) = 2 orbed FA (0.22)
Now let’s deal with the scenario of 5, 1orbed Brandishes in a row:
B(1)->B(1)->B(1)->B(1)->B(1) = 5 orbs
At first I assumed it was:
(0.78)^5 = 0.2887174368
There’s no way there’s a 28.87% chance to have 5 Brandishes in a row with 1 orb and no FA activations. Then I figured I’d account for the FA’s by multiplying each term by 0.6 for the fact it didn’t activate. Since there should be 5 unsuccessful FA’s
0.2887 * (0.6)^5 = 0.022449312 = 2.24% <- seems a lot more accurate. This problem solved itself.
So say I take a random outcome from my tree:
B(1)->F(2)->B(1)->B (1 orb brandish -> 2 orb FA -> 1 orb brandish -> brandish)
This is where I ran into a bit of trouble, at this point, it wouldn’t matter whether or not FA activates, so I just threw down a B with no subscript while pondering how to deal with this dilemma. I came up with these symbols.
B(0) = Brandish at 4 orbs (you'd get to 5 orbs regardless of the ACA "Proc" or not)
F(0) = FA activates/wouldn’t charge orbs
F(00) = FA doesn’t activate/wouldn’t charge orbs
B(0) accounts for a Brandish at 4 orbs, since you’re going to get to 5 anyway, the amount of orbs charged doesn’t matter. F(0) and F(00) account for an FA activating from the final brandish
Redoing the scenario I get
B(1)->F(2)->B(1)->B(0)->F(0)
And
B(1)->F(2)->B(1)->B(0)->F(00)
Using the former scenario [B(1)->F(2)->B(1)->B(0)->F(0)]
(0.78 * 0.4) -> (0.22 * X) -> (0.78 * 0.6) -> (?) -> (?)
This is the part that’s confusing me, since the first term has already accounted for the FA activation, can I leave out the X? And what do I do for the “?” so I accurately account for those scenarios (The subscript 0’s)? I haven’t taken probability or stats in at least 3-4 years so I’m going off memory here.