What is the way to calculate the possibility when throwing multiple dice?

Question

For example, I am throwing 6 six-side dice. What is the probability of me reaching 1 to 6. Is there any concrete formula if I also want to try other situation (for example I need 3 threes and 3 fives)?

Answer 1

6 six-sided dice means that the outcomes are 6-tuples: $(a,b,c,d,e,f)$, with each element being in $\{1,2,3,4,5,6\}$. There are a total of $6^6$ 6-tuples, so any particular 6-tuple has a $1/6^6$ probability of being thrown.

Now, for the first part, let's think about what 6-tuples you want. $(1,2,3,4,5,6)$ works, as does $(2,1,3,4,5,6)$, or $(1,3,2,4,5,6)$. The problem boils down to finding the number of permutations of the numbers 1 through 6, which is just $6!$. 6 choices for $a$, 5 remaining choices for $b$, 4 remaining choices for $c$, and so on until you get $6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1$. Thus for this case the probability is $\boxed{6!/6^6}$.

The second case: you want to find the number of 6-tuples such that there are 3 threes and 3 fives. How can you choose 3 out of the 6 to be threes? Well, $\binom 63 = \frac{6!}{3!(6-3)!}$. Once you have chosen these 3, the other three are frozen in place (i.e. they are already determined). Hence, the number of 6-tuples is just $\binom 63$, and so the probability is $\boxed{\binom 63/6^6}$.

Answer 2

The probability of getting one each of $1,2,3,4,5,6$ is $\dfrac{6!}{6^6} \approx 0.01542$

The probability of getting three $3$s and three $5$s is $\dfrac{6!}{3! \, 3! \,6^6} \approx 0.00043$

It might be slightly more obvious how to generalise the patterns to others if these are written as $\dfrac{6!}{1!\,1!\,1!\,1!\,1!\,1!\,6^6}$ and $\dfrac{6!}{0!\,0!\,3!\,0!\,3!\,0!\,6^6}$ respectively