The problem is: "Given 6 coupons, how many coupons do you expect you need to draw with replacement before having drawn each coupon at least once?"
The solution is well-known: $$E[X] = E\left[1+\sum_{i=1}^5X_i\right]=1+\sum_{i=1}^5E[X_i]=14.7$$ where $X_i$ is the number of tries between i-th and (i+1)-th new coupon (geometric random variable).
However, the sum of geometric random variables $\sum_{i=1}^5X_i$ can be seen as the negative binomial distribution with parameters $p=1/6$ and $r=5$. But calculating the mean of this distribution according to the known formula for the negative binomial distribution $$\dfrac{pr}{1-p}=\dfrac{1/6\cdot 5}{1-1/6}=1$$ I obtain the wrong resut. What is wrong in my reasoning?
Here, each random variable $X_i$ has different success probability $(\frac{6-i}{6}$ for $X_i)$.
Therefore, you can't model the sum of $X_i$'s as negative binomial distribution where the success probability remains constant.