Difficult Problem!

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  

This problem appears in a Mathematical Site on the Internet. It is classified under "Probability problem #113." Both the problem and the solution are posted there. Here is the main site:

http://www.cut-the-knot.org/content.shtml

Note: The small error in CPhill's counting of EVEN sixes neglected to take "0" into account as follows: C(10,0)*(5/6)^10*(1/6)^0 =0.1615056......., which would give a total of: 0.50867....etc., so that both odd sixes and even sixes must ADD up to 1.