What is the probability that when 5 dice are rolled, the sum is a prime number?

Question

What is the probability that when five dice are rolled, the sum is a prime number?

This problem is giving me a tough time. I know that there are $6^5=7776$ different possibilities, and the maximum sum is $6+6+6+6+6=30$. However, if I take, for example, $29$, how do I find the number of ways in which the dice rolls can sum to $29$ or any other prime number?

True there are $6^5$ different die rolls, but there are only $25$ different sums, and not too many of those are prime. — Matthew Leingang, Aug 14 '19 at 18:55
The key word is generating function. There are many posts here on MSE with this content. Here is the link to the results: — callculus42, Aug 14 '19 at 18:55
You want positive solutions for $x_1+x_2+\dotsb+x_5=p$, where $p$ is prime number and $x_i \geq 1$ for all $i$. Are you familiar with stars and bars idea (https://en.wikipedia.org/wiki/Stars_and_bars_%28combinatorics%29)? — Anurag A, Aug 14 '19 at 18:57
@AnuragA Note that $x_i \le 6$ for all $i$, so it's a little more complicated than the standard stars and bars. — eyeballfrog, Aug 14 '19 at 19:03
@eyeballfrog Yes I agree. But I wasn't sure if OP is familiar with the basic idea or not. — Anurag A, Aug 14 '19 at 19:05
Just count in an orderly fashion. 29 is easy, since you know that the only way to obtain 29 is 5 + 6 + 6 + 6 + 6, and that order can be permuted in 5 ways. Numbers in the middle have more combos. For instance, take 17. You can get 17 with 6 + 6 + 3 + 1 + 1, which permutes in (5 C 2) * 3 = 30 ways. You can also do 6 + 6 + 2 + 2 + 1, which also has 30 ways. Then count 6 + 5 + 4 + 1 + 1, which permutes in 5 * 4 * 3 = 60 ways. And so on. — K. Jiang, Aug 14 '19 at 19:05
@Math There's been several comments regarding using generating functions. If you're not particularly familiar with them & would like to learn more, I suggest you read How can I learn about generating functions?. — John Omielan, Aug 14 '19 at 20:24

score 5 · Answer 1 · answered Aug 14 '19 at 19:06

You can use the stars and bars method.

Valid sums are: 5,7,11,13,17,19,23,29 (2,3 are not valid since the minimum on five dice is 5)

So, you want: $$x_1+x_2+x_3+x_4+x_5 = p, \forall i, 1\le x_i \le 6$$

For each of these equations, it corresponds to a similar equation:

$$y_1+y_2+y_3+y_4+y_5 = p-5, \forall i, 0\le y_i \le 5$$

Now, $p=5$ and $p=7$ are easy. There is 1 way to get a sum of 5 and $$\dbinom{2+5-1}{5-1} = \dbinom{6}{4} = 15$$ ways to get 7.

For 11 and above, we need to use Inclusion/Exclusion. We take the total number of solutions of nonnegative integers and subtract the total number of solutions that violate the upper bound for one of the variables.

For $p=11$ we have $$\dbinom{6+5-1}{5-1} - \dbinom{5}{1}\dbinom{0+5-1}{5-1} = \dbinom{10}{4}-5 = 205$$

For $p=13$, we have $$\dbinom{8+5-1}{5-1} - \dbinom{5}{1}\dbinom{2+5-1}{5-1} = 420$$

For $p=17, p=19$, we can have two upper bounds violated.

$p=17$: $$\dbinom{12+5-1}{5-1} - \dbinom{5}{1}\dbinom{6+5-1}{5-1} + \dbinom{5}{2}\dbinom{0+5-1}{5-1} = 780$$

$p=19$: $$\dbinom{14+5-1}{5-1} - \dbinom{5}{1}\dbinom{8+5-1}{5-1} + \dbinom{5}{2}\dbinom{2+5-1}{5-1} = 735$$

For $p=23$, we can violate 3 of the upper bounds: $$\dbinom{18+5-1}{5-1} - \dbinom{5}{1}\dbinom{12+5-1}{5-1} + \dbinom{5}{2}\dbinom{6+5-1}{5-1} - \dbinom{5}{3}\dbinom{0+5-1}{5-1} = 305$$

For $p=29$, we can violate 4 of the upper bounds: $$\dbinom{24+5-1}{5-1} - \dbinom{5}{1}\dbinom{18+5-1}{5-1} + \dbinom{5}{2}\dbinom{12+5-1}{5-1} - \dbinom{5}{3}\dbinom{6+5-1}{5-1} + \dbinom{5}{4}\dbinom{0+5-1}{5-1} = 5$$

Total number of ways to yield a prime: $$1+15+205+420+780+735+305+5 = 2466$$

Total probability:

$$\dfrac{2466}{6^5} = \dfrac{137}{432}$$

Using a generating function confirms the numbers above. – JMoravitz Aug 14 '19 at 19:19 — JMoravitz, Aug 14 '19 at 19:19

score 0 · Answer 2 · answered Aug 21 '19 at 14:01

I wrote a Python program that gets the count for all sums. The program just adds and subtracts
(c.f. inclusion/exclusion principle).

Python Program

roll_vec = ['dummy',0,1,1,1,1,4]
sum_buckets = [0] * 31

def Increment_Dice():
    if roll_vec[1] < 6:
        roll_vec[1] = roll_vec[1] + 1
        roll_vec[6] = roll_vec[6] + 1
        return True
    else:
        for i in range(1,6):
            if roll_vec[i] < 6:
                roll_vec[i] = roll_vec[i] + 1
                roll_vec[6] = roll_vec[6] + 1
                return True
            else:
                roll_vec[6] = roll_vec[6] - 5
                roll_vec[i] = 1
        return False

while Increment_Dice():
    sum_buckets[roll_vec[6]] = sum_buckets[roll_vec[6]] + 1

for i in range(5,31):
    print(i, sum_buckets[i])

OUTPUT

What is the probability that when 5 dice are rolled, the sum is a prime number?

2 Answers2