# 2 Dice Rolls vs. 1 Target Number

#### Mechalus

##### New Member
Hi guys, I hope you don't mind my posting this here. I'm a game designer who needs some help working out probabilities for a dice roll mechanic, and this is the first thing that popped up in Google.

Here's the question:

Let's say die #1 has 6 sides. And die #2 has 6 sides.

I know that both #1 & #2, because they have the same number of sides, each have a 66.7% chance of rolling 3 or better.

Ok, that's simple enough. But here's where the dice mechanic comes in.

What if I say. Roll 2 six-sided dice, and you need to roll 3 or better on at least one of them.

What is the percentage chance of succeeding when I roll both dice, but only need a 3 or better on one of them?

Now... what happens when the dice have different numbers of sides. For example, what if die #1 has six sides, but die #2 has eight sides?

What is the percentage chance of at least one of these dice rolling 3 or better? I know that die #1 (six sides) has a 66.7% chance, and die #2 (eight sides) has a 75% chance, but what percentage of success do we have if they are both rolled, but only one needs to hit 3 or better?

What about 8 sided + 10 sided vs. a target number of 6 or better?

For the life of me I can't figure out the formula. Help!

#### Mean Joe

##### TS Contributor

I think your problem becomes a lot clearer if you look at it the right way; instead of working out all the situations where you succeed, try calculating when you fail.

The idea is P(succeed) = 1 - P(fail)

What if I say. Roll 2 six-sided dice, and you need to roll 3 or better on at least one of them.
Fail Event = both dice roll < 3
P(die #1 roll < 3) = 2/6, P(die #2 roll < 3) = 2/6
So P(fail) = P(die #1 roll < 3) AND P(die #2 roll < 3) = (2/6) * (2/6) = 1/9
So P(succeed) = 8/9 (should be higher than 67% right?)

Now... what happens when the dice have different numbers of sides. For example, what if die #1 has six sides, but die #2 has eight sides?
Fail Event = both dice roll < 3
P(die #1 roll < 3) = 2/6, P(die #2 roll < 3) = 2/8
So P(fail) = (2/6) * (2/8) = 1/12
So P(succeed) = 11/12 (should be higher than 75% right?)

#### BGM

##### TS Contributor
You may use the inclusion-exclusion principle:

[math] P(E \cup F) = P(E) + P(F) - P(E \cap F) [/math]

Or follow the way as what Joe suggested above:

[math] P(E \cup F) = P((E^c \cap F^c)^c) [/math] (De Morgan's Law)

[math] = 1 - P(E^c \cap F^c) [/math] (complementary event)

[math] = 1 - P(E^c)P(F^c) [/math] (by independence)

[math] = 1 - (1 - P(E))(1 - P(F)) [/math] (complementary event)

If you expand it out, it would be the same as above.

#### Mechalus

##### New Member
I think your problem becomes a lot clearer if you look at it the right way; instead of working out all the situations where you succeed, try calculating when you fail.
Doh! You were right, and the formula you provided got me exactly what I was looking for.

Thank you!