# Probability of Obtaining A Particular Sum from Successive Dice Rolls

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?

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Have a look at the [this question][1] [1]: math.stackexchange.com/questions/68045/… – Abe Doe Apr 19 '13 at 17:09
Thank you, the generating function approach makes sense to me. However, I am curious as to how the 'stars-and-bars' approach works. I am familiar with the method but I do not understand how they applied it here. Moreover, are there any obvious modifications of this problem for which the generating function method would fail? – user73041 Apr 19 '13 at 17:22
In the linked question, Quixotic shows how to incorporate the constraint than none of the four can be greater than $6$. His approach would need some updating if you could have more than one part greater than $6$. But your case is the same-you can't have more than one part greater than 6 – Ross Millikan Apr 19 '13 at 17:29
If I am understanding his answer correctly, would the total in this case be $13\choose3$? – user73041 Apr 19 '13 at 17:34
The sum 14 can be represented by 14 stars: **************. To break this into four summands, you put three bars between stars. There are 13 in-between spots, so there are $\binom{13}{3}$ ways to place the bars. The problem is that you might have more than six stars in a row without a bar between them. That's where the trick in Quixotic's answer comes in. To remove solutions where the first summand, $x_1,$ is greater than $6$, let $x'_1=x_1+6,$ so that $x'_1+x_2+x_3+x_4=8$ and use stars-and-bars again to count solutions. Solutions where $x'_1\ge1$ are solutions where $x_1\ge7$. So ... – Will Orrick Apr 21 '13 at 10:04