Question about combinatoris of four dice

Question

We have $4$ dice with $4$ different dyes : Green, Yellow, Red and Blue. I need to know how many options do i get so that I'll get the number $3$ at least once. These are 6 sided dice, color does matter. The colors matter only because let's say i get 3 on the green dice and then 6, 6 and 6 on the other 3, its not the same as getting a 3 on the blue one and than 6, 6 and 6 on the other 3.

I have two answers that i can think of, don't know which one is true and why :

(1) Lets say i choose the 3 to be on the blue, so 1 option, on the other 3 it doesn't matter so it's $1 * 6 * 6 * 6$, and we have $4$ dices so it will be $6 * 6 * 6 * 4 = 864$.

(2) All cases are $6 * 6 * 6 * 6$, I'll subtract the options with $3$ inside them, so actually $6 * 6 * 6 * 6 - 5 * 5 * 5 * 5 = 671$. Thank you in advance.

Answer 1

In the first one, you are counting the same possibility multiple times as for example the case of all 4 dice rolling up 3 is counted 4 times, those where 3 appears 3 times are counted 3 times over. For the first case to work, you'd have to use the inclusion-exclusion principle to get the count correct where you'd have to subtract the multiple counts to get the number narrowed down correctly.

The second option is the correct answer as it leaves all the cases of at least 1 roll of a 3.

Answer 2

If your question is simply about getting at least one value of 3 on the throw of four 6-sided dice, the classic approach is to consider the combinations for getting no 3s showing and subtracting that from the total possible combinations.

Total combinations is $6^4$

Combinations with no threes is $5^4$

Combinations with at least one three showing is therefore $6^4-5^4$

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...and I see that was your option (2) - as mentioned by others, the issue with your first option was that you were double-counting some instances when 2 or more dice showed "3".