A fair 6-sided die is thrown 10 times. What is the probability of rolling a six three times in a row, and the other rolls not being a 6?

Question

In 10 rolls, there are 8 positions where the chain of three can start, so there are 8 permutations since dice rolls are interchangeable (their order doesn't matter).

Therefore, I believe that the answer should be $8\times \left(\frac 16\right)^3\times \left(\frac 56\right)^7$. However, I am not $100\%$ sure. Any confirmation is appreciated.

Answer 1

Your answer is correct, your logic is sound. It can even be experimentally verified (in Python):

import itertools
from fractions import Fraction

def good_roll(roll):
    sixes = [i for i, r in enumerate(roll) if r == 6]
    return len(sixes) == 3 and sixes[0] == sixes[1] - 1 == sixes[2] - 2

num_good_rolls = sum(good_roll(r) for r in itertools.product(range(1, 7), repeat=10))

print(Fraction(num_good_rolls, 6**10))
print(8 * Fraction(1, 6)**3 * Fraction(5, 6)**7)

Both statements print 78125/7558272.