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If I want to create a function $p_{roll}(x_1, x_2, x_3, x_4, x_5)$ that calculates the probability of getting the given dice (order is not important) in a single roll with 5 dice then how would that look? All the questions I've read so far deals with probabilities of full house in Yahtzee etc. which spans across several possible face values for the dice. This function would be the probability of specifically rolling the given dice values. For example:

$p_{roll}(1,1, 3, 4, 5)$ = ?

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There are $6^5$ possible die rolls. Divide this by $5!/2!$ to account for the irrelevance of order. The 2! accounts for the fact that you have 2 repeated terms in your sequence (the two 1's). So the final answer is $5!/(2!6^5)=0.77$%.

Generally, if you roll $n$ 6-sided die then the probability of rolling a sequence $x_1,...,x_n$ is

$$ p=\frac{n!}{6^n\Pi_{i=1}^6 m_i!} $$

where $m_i$ is the number of occurrences of the number $i$ in your sequence. So in the example you gave $m_1=2,m_2=0,m_3=1,m_4=1,m_5=1$.

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Let $a_i$ be the number of values $i$ in $(x_1,x_2,x_3,x_4,x_5)$ for $1\leq i\leq 6$, so in your example $a_1=2$, $a_3=a_4=a_5=1$ and the others are $0$. Now, the total number of possible outcomes is $6^5$. The number of outcomes with the right number of $1$, $2$, etc. is: $$ \binom{5}{a_1,a_2,\dots,a_5}=\frac{5!}{a_1!\cdots a_6!} $$ Now, the probability of getting a good outcome is $$ \frac{\frac{5!}{a_1!\cdots a_6!}}{6^5} $$

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