
Protect Companion + Called Shot
Re: Protect Companion + Called Shot
Just to preserve my own interwebz dignity, I did know the maths I originally posted were offs. But I confess I didn't know what the correct maths formulez were, and I didn't care too much. 

Rignuth: Barding Wordweaver Wanderer in Southron Loremaster's game.
Amroth Ol'Hir: High Elf Vengeful Kin Slayer in Zedturtle's game.
Jakk O'Malli: Dwarven Orator Treasure-Hunter in Hermes Serpent's game.
Amroth Ol'Hir: High Elf Vengeful Kin Slayer in Zedturtle's game.
Jakk O'Malli: Dwarven Orator Treasure-Hunter in Hermes Serpent's game.
Re: Protect Companion + Called Shot
Absolutely and the main point still stands. I just felt that the 5/6 probability made it sound like it happened nearly all the time whereas in reality it's less than 2 shots out of 3. Personally I feel that makes called shot a little less attractive and so it is relevant.
James Semple, occasional composer of role playing music
Re: Protect Companion + Called Shot
bluejay, just out of curiousity, what's the probability when you have 6 attack dice?
Rignuth: Barding Wordweaver Wanderer in Southron Loremaster's game.
Amroth Ol'Hir: High Elf Vengeful Kin Slayer in Zedturtle's game.
Jakk O'Malli: Dwarven Orator Treasure-Hunter in Hermes Serpent's game.
Amroth Ol'Hir: High Elf Vengeful Kin Slayer in Zedturtle's game.
Jakk O'Malli: Dwarven Orator Treasure-Hunter in Hermes Serpent's game.
Re: Protect Companion + Called Shot
Not Bluejay, but 66.5%Rocmistro wrote:bluejay, just out of curiousity, what's the probability when you have 6 attack dice?
I was recently looking at this math, for a different thing where the chance of getting a 6 on 5 dice would trigger something special, so that's why it's fresh in the mind.
And, your approximation earlier was good enough for a gut feeling thing.
Jacob Rodgers, occasional nitwit.
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Re: Protect Companion + Called Shot

Basically the maths is like this...
You calculate the chance that you don't roll a 6 on every dice and then subtract that probability from 1.
So for each dice there is a 5/6 chance of not rolling 6. For multiple conditions that have to occur you multiply the probability. So for 1 dice the chance of not rolling 6 is 5/6. For 2 dice the chance of not rolling 6 on either dice is 5/6 x 5/6. You could write this as (5/6)^2.
Therefore the chance of rolling at least one six on n dice is
1 - (5/6)^n
For the specific case you asked (6 dice) the probability of rolling at least 1 six is: -
1 - (5/6)^6 which is around 66.5%
Hope that helps! Probability isn't even intuitive for me and my degree was Mathematics!

James Semple, occasional composer of role playing music
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