more dice i dont understand

What would be the formula for figuring the chance to roll at least two 6s with x number of 6 sided dice?
Thank you again for any help.

Binomial distribution, with number of trials x and p=1/6.

You want 1 - P(k=0 successes) - P(k=1 success). That is:

1 - (5/6)^x - x * 1/6 * (5/6)^(x-1)

Wow I was making this way harder than it really is. Although I still really don't understand how you arrived at the formula but, it seems to work.
Can this same formula be adapted to figure the chance of rolling at least a total of 10 on two dice while rolling x number of dice?

Thanks again for the help.

To maybe help understand dwheatley’s formula a little….if you want at least two 6s, this will be the same as not rolling two 6s and subtracting from 1.

So,
1 – probability of rolling zero 6s – probability of rolling one 6

The probability of not rolling a six on a single die is 5/6.
The probability of not rolling a six on two dice is (5/6)(5/6) = (5/6)².
The general formula is then….(5/6)^x

Now, the probability of rolling one 6 on a single die is 1/6.
The probability of rolling a 6 and some other number when rolling two dice is then 2*(1/6)(5/6).
You multiply by 2 since there are two chances to roll the 6.
And for three dice….3*(1/6)(5/6)(5/6)
The general formula is then…x*(1/6)*(5/6)^(x-1)

Put it all together to get what dwheatley wrote.


As for you second question, it is much more difficult. You have to count the number of ways you can get a sum of at least 10 on two dice. Which would be every combination involving the pair 4,6 or 5,6 or 5,5 or 6,6. When you add more and more dice this becomes very tedious to count (at least, I can’t think of a super easy way...maybe someone knows). Here are some probabilities for different number of dice found by brute force:

Number of Dice	Combinations to roll at least 10	Total combinations of x dice (6^x)	Probability
2	6	36	0.1667
3	77	216	0.3565
4	676	1296	0.5216
5	5067	7776	0.6516
6	34958	46656	0.7493
7	229777	279936	0.8208
8	1465512	1679616	0.8725
9	9166679	10077696	0.9096
10	56599330	60466176	0.9360

Quote: PeeMcGee

When you add more and more dice this becomes very tedious to count (at least, I can’t think of a super easy way...maybe someone knows).

p(all combos) - p( all 4 or less) - p( one 5 and the rest 4 or less) - p(one 6 and the rest 3 or less )
6^x - 4^x - x*4^(x-1) - x*3^(x-1)

Quote: miplet

p(all combos) - p( all 4 or less) - p( one 5 and the rest 4 or less) - p(one 6 and the rest 3 or less )
6^x - 4^x - x*4^(x-1) - x*3^(x-1)

Beautiful. Thanks miplet.