No, for exactly three of them show a "1" so the probability is:

4c3 (1/6)^3 = 1/56

and for four of then to show 1 the probability is:

4c4 (1/6)^4=(1/6)^4

So the probability of three or more 1's is:

4c3 (1/6)^3 + (1/6)^4

No, we gave the probability that three or more show a specified face in 2 above, but3) I want at least 3 dice with the same values (any value as long as it's the same)

Ans: 4c3 x 1/6 = 2/3

there are 6 possible specifications, so this is 6 times the answer to 2.

RonL