Dice mechanic formula

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kjkage

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Hello!
I am creating a game and would like to know the formula to determine the odds of rolling a particular result:
if rolling a 12 sided dice(1-12) with a +4 to the roll, and a 6 sided dice (that has a different symbol on each of the six sides). What are the odds of rolling a 10 or greater AND rolling on the 6 sided dice a specific symbol that is only on 1 of the 6 sides?

I am hoping to get a formula so I can adjust any of the numbers to determine other odds that use the same mechanics.

something tells me this isn't that hard of a formula to figure out but I am just pretty poor at math. :)

thank you.
 
First, my standard gripe: you mean a six sided die and a twelve sided die. "Dice" is the plural of the singular "die".
Now: there are 12*6= 72 possible results when you roll a twelve sided die and a six sided die. Two of the faces of the twelve sided die are "greater than 10" (11 and 12). The "specific symbol that is only on 1 of the 6 sides" is, of course, 1 of the six sides so that 2(1)= 2 out of the 72 ways the two dice are "successes". The probability is \(\displaystyle \frac{2}{72}= \frac{1}{36}\). This can also be calculated as \(\displaystyle \left(\frac{2}{12}\right)\left(\frac{1}{6}\right)= \frac{1}{36}\)
 
I assume that this is one of those games where you are rolling to determin "hit strength" or "defense" and that "with a +4 to the roll" means that you add 4 to whatever the roll gives.

Of course, that is irrelevant to the question "What are the odds of rolling a 10 or greater AND rolling on the 6 sided dice a specific symbol that is only on 1 of the 6 sides?" since it only asks about rolling the dice.
 
Unless I have completely missed it, in this setup "ten or more" on a twelve sided die means ten, eleven, or twelve.
In that case the probability of that event is \(\displaystyle \frac{3}{12}\).
 
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