2) The probability of no sixes is 5/6 in a single roll. In two rolls, it would be (5/6)*(5/6), since the rolls are independent. In n rolls it is (5/6)^n. How many rolls before the probability is 1/2 or less? Solve for:

(5/6)^n = 1/2

log( (5/6)^n ) = log(1/2)

n*log(5/6) = log(1/2)

n = log(1/2) / log(5/6) ~= 3.8

So, after about 3.8 rolls, there's 50% chance of no sixes. At 4 rolls, the probability is less than 1/2, so the probabiltiy of at least one six is greater than 1/2.