You have 3 fair dice each numbered 1 to 6 what is the chance all three dice show the same number?

For this set of data use the formula nCr(p^r)[q^(n-r)]
p = P(a number) = 1/6
q = P(not that number) = 5/6
n = 3 trials
r = 3 successes
n - r = 0 failures
P(the same number in 3 trials) = nCr(p^r)[q^(n-r)]
Substitute what you know into the formula:
3C3[(1/6)^3][(5/6)^0] = 1 x 1/216 x 1 = 1/216 = 0.0046 approximately.

Or shortly we can say :
Since each number has the same probability, 1/6, to appear in this experiment, then the probability of the three dices to show the same number is 1/6 x 1/6 x 1/6 = 1/ 216

The chances are 1 in 36. It doesn't matter what the first one is. There is a 1 in 6 chance of the second one being the same, and same with the third one. 1/6 x 1/6 = 1/36. I have no idea what the rubbish above is all about, but its complete &^%*