Suppose you have a regular die with 6 faces numbered 1 through 6, respectively, and roll the die 4 times. What is the probability that the sum of the 4 rolls is 14?
This problem is equivalent to asking the number of ways to partition the number 14 into 4 subsets using 6 different numbers. Hence we can consider 'multichoosing' 4 numbers with repetition out of 6 different numbers such that $14 = x_1 + x_2 + x_3 + x_4$. I suspect that this can be done using either using multinomial coefficients in some way or multisets but I am not entirely sure. One issue is that this does not give a way for removing those cases where the four number chosen $\ne$ 14.
The total number of outcomes from rolling a die 4 times is $6^4$, so our probability will be some number divided by this.
Is this likely a problem where I should just enumerate all possible options to find my numerator?
Best Answer
This can be done using inclusion-exclusion; see Balls In Bins With Limited Capacity. Summing over the number $t$ of dice exceeding their capacity $6$ and assigning $7$ to each of them before distributing the rest of the sum, we count
$$ \sum_{t=0}^2(-1)^t\binom4t\binom{4+14-7t-1}{4-1}=\binom{17}3-4\binom{10}3+6\binom33=206 $$
different rolls that result in this sum, so the desired probability is
$$ \frac{206}{6^4}=\frac{103}{648}\approx16\%\;. $$