This seems like a very simple problem, but I've been stuck on it for a good while now. Say you have a list of $n$ digits $1, 2, 3, ..., n$ and assume any digit chosen is chosen uniformly at random. After picking your first digit, what is the probability that both of the next 2 digits you choose are greater than the first (assuming no replacement)?
The answer seems to vary greatly depending on the value of $n$ and really I'm stumped on how to proceed. Any insights?
Let be $A$ the event of getting two greater digits than first selected, and let be $A_i$ the event of chosing $i$th digit. Then by Total Probability: $$P(A)=\sum_{i=1}^n P(A|A_i)P(A_i)$$ It is clear that $P(A|A_1)=1$, $P(A|A_{n-1})=P(A|A_n)=0$ and $P(A_i)=1/n$.
For $P(A|A_i)$ with $1<i<n-1$ we have that after chosing first digit, only $n-1$ are left, out of wich there are only $(n-i)(n-i-1)$ greater than $i$, then $P(A|A_i)=\frac{(n-i)(n-i-1)}{(n-1)(n-2)}$, therefore $$P(A)=\sum_{i=1}^{n-2}\frac{(n-i)(n-i-1)}{n(n-1)(n-2)}$$