Let's work out the probability of the first 4 throws coming up with a 5 each time: (1/6)^4. The probability that no fives will be thrown in the next 6 throws is (5/6)^6. So combine these: 5^6/6^10. But the 4 successful throws could occur 210 ways=10*9*8*7/(1*2*3*4), in other words, any combination of 4 out of 10. So we need to multiply our combined probability by this factor: 210(5^6)/6^10=0.0543 approx or 5.43%.
Another way is to use binomial expansion: (5/6+1/6)^10=1^10=1. This equation covers every possibility so that the sum of each of them comes to 1, certainty. Expanding into individual terms we get:
(5/6)^10+10*(5/6)^9*(1/6)+45*(5/6)^8(1/6)^2+120*(5/6)^7(1/6)^3+210*(5/6)^6(1/6)^4+...
The separate terms stand for: No fives at all; only one five; only two fives; only three fives; only four fives; ...
[If we were to add together the first five terms we would be finding the probability of up to four fives, or four fives at the most. It doesn't have to be a five; any specific number would have the same probability.]