Die Roll Probability (2d6)

[TheLe]

If I roll 2d6, and want a result of 9 or better, the probability is 27.8%.

Now lets say I get to roll twice. What's the probability that I will get a result of 9 or better on either roll?

Can someone also tell me the formula so I can use it for "10 or better", "11 or better", etc...

Thanks,

-The Le

 

[BeauNiddle]

The way to think about it is to define what would be either the win or the lose condition.

Since you want 9 on EITHER roll then the only way to fail is to get 8 or less on BOTH rolls (that leaves a 9 on either roll and a 9 on both rolls as possible win values)

The chance of 8 or less on a single roll is 1 - 0.278 or 72.2 %
The chance of 8 or less on both rolls is 0.722 * 0.722 or 52.1%

I think

 

[Lord Sessadore]

Yeah, BeauNiddle has it right.

As for a formula, basically all you have to do is find the probability of getting X or higher on one roll, then use his method.

For finding the probability of getting X or higher on one roll of 2d6 .... I think you're going to have to do that manually. If you're going to be doing this extensively, I'd draw up a table that shows all 36 possibilities so you can just count up the possibilities, then divide by 36 to get the probability.

For example, to get 12 on 2d6 there's only one possibility = (6,6). That means it has a probability of 1/36 = 2.78%.
11 has two chances = {(5,6),(6,5)}, so it has probability of 5.56%.
So that means probability of 11 or higher on one roll of 2d6 is 2.78% + 5.56% = 8.34%


.... Does that make any sense?

 

[Spatula]

You could also attack the problem from the other side:
0.278 (chance of 9+ on 1st roll) * 0.722 (chance of 1-8 on 2nd roll) +
0.722 (chance of 1-8 on 1st roll) * 0.278 (chance of 9+ on 2nd roll) +
0.278 (chance of 9+ on 1st roll) * 0.278 (chance of 9+ on 2nd roll)

which will give you the same answer (47.9%). BeauNiddle's method should be a bit faster.