Ok, well me and ovi have been having a discussion about dice,

if you have 1 loaded dice, with a 1/4 chance of getting a 1 and a 1/4 chance of getting a 6

and a normal die with 1/6 chance of getting everything

what is the chance of getting a 7 when rolling 1 loaded die and 1 normal die?

and here is my math

1/2 chance of it laninding on 1 or 6 on the loaded die, right

1/2 * 1/6 = 1/12

that is the chance of it landing on 1 or 6 on the loaded die, or any number on the other die

how many ways can you make seven out of that,

2 ways

so, you can get 6 and 1, or 1 and 6, so thats 2 ways

so that is 2/12 or 1/6

1/6th chance to get 7 with it landing on either the 1 or 6 on the loaded die

then the rest of that die

you have a 1/2 chance to get 2-5 on the loaded die

1/2 * 1/6 = 1/12

as in a 1/12 chance of it not landing on one of the rigged numbers and any other number on the other die

1/12 chance of it being a 2 2 , 2 3, 2 4 etc etc

how many ways can you make seven,

4 ways

so thats 4/12'ths

4/12'ths chance of getting a 7 without it landing on 1 or 6 on the loaded die

so that is 4/12ths, or 1/3

so you have a 1/6ths chance of it being a 7 with it being 1 or 6 on the rigged die

or a 1/3 chacne of it being a 7 without getting a 1 or a 6 on the rigged die

so you have a 1/2rd chance of getting a 7 if you roll 1 rigged die and 1 normal die