1- What is the probability of an odd number of sixes turning up in a random toss of 10 fair dice?
We can either have 1 - 3 - 5 - 7 or 9 6s
We have a (1/6) probability of a six on any roll and a (5/6) probability of "not 6" on any roll...so....
P( one 6) = C(10,1)(5/6)^9 (1/6) = 9765625/30233088 ≈ 32.3%
P(three 6s) = C(10,3) (5/6)^7 ( 1/6)^3 = 390625/2519424 ≈ 15.5%
P ( five 6s) = C(10,5)(5/6)^5 (1/6)^5 = 21875/1679616 ≈ 1.3%
P ( seven 6s) = C(10,7) (5/6)^3 (1/6)^7 = 625/2519424 ≈ .02%
P ( nine 6s) = C(10,9)(5/6) (1/6)^9 = 25/30233088 ≈ 0%
Adding all the fractions, we get 58025/118098 ≈ 49.1%
2-What is the probability of an even number of sixes turning up in a random toss of 10 fair dice?
We can have 2 - 4 - 6 - 8 or 10 6s
P(two 6s) = C(10,2) (5/6)^8 (1/6)^2 = 1953125/6718464 ≈ 29.1%
P(four 6s) = C(10,4) (5/6)^6(1/6)^4 = 546875/10077696 ≈ 5.42%
P( six 6s) = C(10,6) (5/6)^4 (1/6)^6 = 21875/10077696 ≈ .21%
P(eight 6s) = C(10,8) (5/6)^2(1/6)^8 = 125/6718464 ≈ 0%
P (ten 6s) = (1/6)^10 = 1/60466176 ≈ 0%
Adding all these fractions, we get 20991751/60466176 ≈ 34.7%
They are not equal....notice that P( one 6 ) > P (two 6s) and P ( three 6s) > P (four 6s) and P( five 6s) > P (six 6s)...etc.
Thus it appears that P(2n - 1 6s) > P(2n 6s) for n = 1 - 5
