likelihood in die throw

The probability of getting no fours is: (5/6)^10
The probability of getting exactly one four is (1/6)(5/6)^9(10!/9!/1!)
Of getting exactly two is (1/6)^2(5/6)^8(10!/8!/2!)

And so forth. This makes the probability table look like:
0 16.2%
1 32.3%
2 29.1%
3 15.5%
4 5.4%
5 1.3%
6 0.2%
7 0.0%
8 0.0%
9 0.0%
10 0.0%

The chance of getting at least 2 fours (or any other value) is 51.5%

Thanks for your reply. I am not quite sure I can pull out from your response the general expression, which is what I was really looking for. I tried to keep my question simple, but in fact here is the problem I am dealing with specifically: Consider arrays of 40 blocks assembled from an infinite supply of equal numbers of red, green, white, and black blocks. How many arrays must be developed such that in the ultimate collection of arrays there is a 90% likelihood of at least 2 identical arrays?

Any help would be appreciated. Thanks.