6 six sided dice (each die is numbered 1 - 6, each die has 6 different face colors), with no color / number duplication i.e. one red 1, one orange 1, one yellow 1, one green 1, one blue 1, one violet 1, etc.

In a single roll of all 6 dice
1. 6 of ANY same number OR color
2. 5 (or more) of ANY same number OR color
3. 4 (or more) of ANY same number OR color
4. 3 (or more) of ANY same number OR color
5. 3 (or more) of ANY same number ONLY
6. A numerical total (from 6 dice) of 21
7.7 or 11 (with 2 die) of ANY same color [if this one proves too difficult don't attempt]

If possible I need them in the format of the following examples;
Any 5 (or more) Same Color:
for 5 of a specified color;
p=(6!/5!1!)(1/6) 5 (5/6)=30/46,656
since there are 6 colors this can occur 6 times
add to this the probability that any 6 of the same color can occur 6 (30 / 46,656) + 6
/ 46,656 = 186 / 46,656
probability against = 46,470 / 46,656
odds against = 46,470 : 186
= 249.8 : 1

7 or 11 of a (specified) color
for 7 (2 die) of a specified color;
for 11 (2 die) of a specified color;
p=(1/6)=1/36
overlap=1/64=1/1,296
probability against = (1/12+1/36)-1/1,296
= 143/1,296 odds against
odds against
= 1,153 : 143
= 8.06:1
4 (or more) of a (specified) color:
for 4 (exactly) of a specified color;
p=(6//4!2!)(1/6)4(5/6)2=375/46,656
add to this the probability that 5 or 6 of the specified color could occur
(computed in #2 and #3)
(375 + 30 + 1) / 46,656 = 406 / 46,656

probability against = 46,250 / 46,656

odds against
= 46,250 : 406
= 113.9 : 1

6 six sided dice (each die is numbered 1 - 6, each die has 6 differ...