Probability that First 'A' before Second 'S' in the word "ASSASSIN"

Question

Consider all the permutations of the word "ASSASSIN". Find the probability that the First 'A' occurs before Second 'S'.

By considering the sequence of letters in reverse, I deduced that the question is equivalent to probability of Second 'A' occurs after Third 'S'. (I found this little easier to attempt).

I split this into two further cases: (I). First 'A' occurs after Third 'S' (II). First 'A' occurs after before Third 'S' and Second 'A' occurs after Third

For (I). Depending upon where the 4th 'S' goes, (i.e. SSSSAA/SSSASA/SSSAAS type) there are $7*8=56$ numbers for each case(considering permutation of letters I,N between their gaps). Hence $56*3=168$ numbers this case.

For (II). We have (ASSSA/SASSA/SSASA) initial possibilities. Now the 4th S has 3 ways to go in the first, 2 ways to go in the second(as one possibility is counted in the previous case), but only one ways in the third type(as in one way it is same as the previous case again, and in another one way it would become case (I) again). The total number of words is $(3+2+1)*(7*8)=336$

Hence the total number of ways is $168+336=504$ ways.

And the required probability is $\frac{504}{(8!/(4!*2!))}=\frac{3}{5}$.

Although this is the correct answer, I am eager to know of any other way to solve this question,preferably with not so many case work and one whose method could be applied to other longer words conveniently as well.

Answer 1

Note that you can ignore the I and the N, as they are irrelevant to the ordering of the A's and S's.

Here are three different ways to proceed.

Solution 1: Direct enumeration--there are only ${6\choose 2}=15$ ways to arrange AASSSS, so this is quite easy to do by hand. Then count how many cases are of interest. Divide this count by $15$.

Solution 2: Find the complementary probability: That is, find the probability that S occurs in both the first and second position: $P(SS...)=\frac{4}{6}\cdot\frac{3}{5}=\frac{12}{30}=\frac{2}{5}$. So the desired probability is $1-\frac25=\frac35$.

Solution 3: Use inclusion-exclusion. Let $E_1$ be the event that the first letter among the A's and S's is A, and let $E_2$ be the event that the second letter among the A's and S's is A. We want $P(E_1 \cup E_2)=P(E_1)+P(E_2)-P(E_1\cap E_2)=\frac26+\frac26 -\frac{1}{15}=\frac35$.