What is the probability of rolling at least two sixes in 10 rolls of a fair die?

Answer 1

1 - P(0) - P(1)

The old-fashioned way is to calculate P(0) and P(1) using the binomial distribution.

P(0) = C(10,0)(1/6)⁰(5/6)¹⁰ = (5/6)¹⁰ (We can figure out this formula without first using the long-form, but I show this for completeness.)

P(1) = C(10,1)(1/6)¹(5/6)⁹ = 10 (1/6)¹(5/6)⁹ = 2(5/6)(5/6)⁹ = 2(5/6)¹⁰

1 - P(0) - P(1) = 1 - (5/6)¹⁰ - 2(5/6)¹⁰ = 1 - 3(5/6)¹⁰ = 1 - 5^10/2*6^9 = 1 - 9765625/20155392 =

10389767/20155392 = 0.515483251330463

In this day of statistical calculators or Excel, the easier way would be either to use

1 - binom.dist(1,10,1/6,1) = 0.515483251330463 or, equivalently,

binom.dist(8,10,5/6,1) = 0.515483251330463 (between 0 and 8 times, not rolling a 6)