Keep the odds straight (2)

There are $6^5 = 7776$ possible outcomes of rolling five dice. We must count the ones that are Small Straights.

A Small Straight is of one of the forms $\{1,2,3,4,x\};\ \{2,3,4,5,x\};\ \{3,4,5,6,x\}$ . Let's focus on the first of these three.

First, there are $5! = 120$ possible permutations of each of $\{1,2,3,4,x\}$ with $x = 5, 6$ . Moreover, there are $5!/2 = 60$ possible permutations of each of $\{1,2,3,4,x\}$ with $x = 1, 2, 3, 4$ . (The division by two is due to the fact that one value occurs twice.) Thus there are $2\cdot 5! + 4\cdot \frac{5!}2 = 4\cdot 5! = 480$ rolls containing the sequence $1, 2, 3, 4$ .

In the same way we treat the sequences $2, 3, 4, 5$ and $3, 4, 5, 6$ , giving $3 \cdot 480 = 1440$ possibilities. However, we now have counted double all Large Straights, because (say) $\{1, 2, 3, 4, 5\}$ contains both $\{1,2,3,4\}$ and $\{2,3,4,5\}$ . Therefore we subtract the number of Large Straights, which is $2\cdot 5! = 240$ . That leaves $1440 - 240 = 1200$ possible ways of rolling Small Straight. The probability is $P = \frac{1200}{7776} = 0.15432 = \boxed{15.432}\%.$ (Note that this rounded answer is itself a "large straight". Elegance everywhere...)

I solved it by brute force to double check, here is my code: (Python)

def isgood(alist):
    if 1 in alist and 2 in alist and 3 in alist and 4 in alist:
        return True
    elif 5 in alist and 2 in alist and 3 in alist and 4 in alist:
        return True
    elif 5 in alist and 6 in alist and 3 in alist and 4 in alist:
        return True
    else:
        return False

yes = 0
total = 0
for a in range(1,7):
    for b in range(1,7):
        for c in range(1,7):
            for d in range(1,7):
                for e in range(1,7):
                    alist = [a,b,c,d,e]
                    if isgood(alist):
                        yes+=1
                        total+=1
                    else:
                        total+=1
print (yes/total)