The integers $ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10$ are written on a blackboard. Each day, a teacher chooses one of the integers uniformly at random and decreases it by $ 1$ . Let X be the expected value of the number of days which elapse before there are no longer positive integers on the board. Estimate $ X$

The answer is given to be $ 120.7528$ .

I solved it in the below manner and wonder if it is right.

Let $ X_i$ be a random variable denoting number of trials required to hit 0. The discrete values it can take are $ (1,2,.. i,0)$ where I club all non positive integers on trials to 0.

For $ X_1$ , the sample space is $ {0,1}$ . Actually, it should be represented by hypergeometric distrbution with the teacher striking down each integer by 1. As an approximation as the number of trials can be infinite, binomial distribution is used. Thus the probability of success for $ X_1$ is $ \frac{1}{10}.\frac{1}{2} = \frac{1}{20}$ . The probability of failure is $ 1-p_{success} = \frac{19}{20}$ .

Similary for $ X_2$ , the sample space is $ (0,1,2)$ and the probability of success for $ X_2$ is $ \frac{1}{10}.\frac{2}{3} = \frac{2}{30}$ . and that of the failure is $ = \frac{28}{30}$ .

And thus for $ X_{10}$ , the sample space is $ (0,1,2,3,..,10)$ and the probability of success for $ X_{10}$ is $ \frac{1}{10}.\frac{10}{11} = \frac{10}{110}$ and that of the failure is $ = \frac{100}{110}$ .

Let us define the $ X$ as the total number of trials required for the teacher to strike down all integers to 0. Then $ X = X_1+X_2+\cdots+X_{10}$

$ E(X) = E(X_1+X_2+\cdots+X_{10}) = E(X_1)+E)X_2)+\cdots+E(X_{10})$

$ E(X_i) = \frac{1-p_i}{p_i}$

I did the calculation and I hit a very close answer of $ 119.2897$ which is $ 1$ less the contest answer 120.7528.

Let me know if the approach is right?. There is no solution provided so I do not know what approach they have taken.