So if you have a normal die with 6 sides each value having a 1/6 chance of occurring, what is the probability of it landing on each value twice?
Also how would you compare the probability of a results such as (1:3, 2:1, 3:0, 4:0, 5:5, 6: 3) [1:0 means the value "1" happened "0" times) ?
The probability of getting exactly two 1's on 12 rolls is ${12\choose 2} (\frac 16)^2 (\frac 56)^{10}$
The probability of getting exactly two 1's and two 2's on 12 rolls is ${12\choose 2,2} (\frac 16)^2 (\frac 16)^2 (\frac 46)^8$
can you see where this is going?