(CLT) Number of rolls of two fair dice to be 90% certain that the percentage of times they show the same face is between 5/36 and 7/36 
How often do you have to roll two fair six-sided dice to be 90% certain that the percentage of times they show the same face is between 5/36 and 7/36?

I was thinking, to apply the central limit theorem, between the two bounds, but I have no idea how to setup it.
First of all thanks to Henry for your answer.

My professor said,
First :
From the statement of the problem one must know, how to apply the
  correction factor to correctly have the limit in which one evaluate
  the normal, and find the correct n.
Second: The distribution is binomial and it will be impossible have a
  greater value (%) for 486 than 487.


I have three question:

1) How Henry obtain (5/36n)^(1/2) for the standard deviation.
2) Where is my mistake evaluating the probability of 486 and 487.
3) How to solve it using the CLT.

Thanks.
 A: You have a binomial distribution with parameter $p$ so the expectation of the number of same-faces is $\frac{n}{6}$ and variance $\frac{5n}{36}$.  
For the proportion you have expectation $\frac{1}{6}$ and variance $\frac{5}{36n}$ so a standard deviation of $\sqrt{\frac{5}{36n}}$.    
If you are going to use the central limit theorem, then since  $\Phi^{-1}(0.95)\approx 1.644854$ the symmetric two-tailed $90\%$ interval is $\pm 1.644854$ standard deviations from the mean.  So you want to find $n$ such that $\frac{1}{36} = 1.644854 \sqrt{\frac{5}{36n}}$, suggesting $n$ should be at least $486.9978$.  That is not quite an integer, so you will need to round up to $487$.
Curiously, while that is a good approximation, $487$ in fact fails, with a probability for the interval of about $0.8996$, while $486$ succeeds with a probability of about $0.90002$.  
Assuming that your "between"  interval is inclusive (though that only matters when $n$ is a multiple of $36$), the exact values to ensure a $90\%$ probability using a binomial distribution seem to be $474, 475,479,480,481,482,484,485,486,489,494,495,496,499,500,$ and any larger numbers, but not $487$, taken from the R code:
n <- 1:600
intprob <- pbinom(7/36 * n, n, 1/6) - 
           pbinom(5/36 * n - 10^(-12), n, 1/6)
which(intprob >= 0.9)

