Probability of a sequence of random letters Suppose an the letters of  a  random sequence  are chosen independently and uniformly  from the set of 26 English alphabet .If the sequence contains n letters what is the probability that it contains  the word "Love" as a  subsequence?Thanks a lot for nay help in advance
The way of dealing with such questions is to think clearly about what is involved. Suppose I am looking for the word "the". Let the probablity that it appears in the first $n$ letters be $p_n$.

Now if "the" appears, either it appears in the first $n-1$ letters, or it appears for the first time at the $n^{th}$ letter. In this case the last three letters chosen are "the" with probability $\frac 1{26^3}$, and the first $n-3$ letters do not contain the word "the".

I am left with $$p_n=p_{n-1}+\frac 1{26^3}(1-p_{n-3})$$ and $p_0=p_1=p_2=0$. This is now a recurrence which can be explicitly solved.



If I were looking for the single letter "a" the same approach would give me $p_n=p_{n-1}+\frac 1{26}(1-p_{n-1})$ or $p_n = \frac {25}{26}p_{n-1}+\frac 1{26}$

The solution to this is $p_n=1+A\left(\frac{25}{26}\right)^n$ and $p_0=0$ gives $A=-1$, which checks with simpler ways of computing, which are available for a single letter.
The total number of combinations is $n!$

The total number of combinations including the word 'love' is $(n-3)!$
This is because we can treat the sequence of the four letters 'love' as one object, so there are only n-3 to arrange.

so $P($String contains 'love'$) = (n-3)!/(n)! = 1/(n*(n-1)*(n-2))$