What is the probability there are a higher total of red “5’s and 6’s” than white “5’s and 6’s”? You have 3 red 6-sided dice and 3 white 6-sided dice.
When you throw the dice, what is the probability there are a higher total of red “5’s and 6’s” than white “5’s and 6’s”?
Note: When any 6-sided die is thrown three times the probability of three “5s and 6s” will be 1/27, two “5s and 6s” will be 2/9, one “5 and 6s” will be 4/9, and zero “5 and 6s” will be 8/27.
What is actually being asked in this problem?
 A: Change the wording slightly. There are two players, Red and White. Red rolls three red dice, White throws three white dice. Call a $5$ or a $6$ good. Each player counts the number of good numbers  that she got. If Red's count is bigger than White's, then Red wins. The question asks for the probability that Red wins.
Note that it is perfectly possible for a tie to occur. In that case Red does not win.
What is the probability that Red gets a total of exactly $k$ good numbers?  With one die, the probability of getting a good number is $1/3$. So the probability that Red gets $0$ good numbers is $(2/3)^3$. The probability Red gets exactly $1$ good number is $\binom{3}{1}(1/3)^1(2/3)^2$, and so on. 
I hope that this explanation will be enough to let you solve the problem. 
A: By symmetry, the chance that white beats red is the same as the chance that red beats white.  One approach is to compute the chance that the two totals are the same, subtract from $1$ and divide the result by $2$  So compute the chance of each total of $5$'s and $6$'s on three dice:  $$ \begin {array} {c c} \text{total}&\text{chance}  \\0& (\frac 23)^3=\frac8{27}\\5&3\cdot \frac 16\cdot (\frac 23)^2=\frac 2{27}\\11&6\cdot \frac 23(\frac 16)^2=\frac 1{18} \end {array}$$  Other totals are possible, and are for you to calculate.  Sum the squares of the chances to find the chance of a tie.
