I am trying to solve the following problem, from Grimmett and Welsh's book Probability.
The random variable $N$ takes non-negative integer values. Show that $$ \mathbb{E}(N) = \sum_{k=0}^\infty \mathbb{P}(N > k) $$ provided that the series on the right-hand side converges. A fair die having two faces coloured blue, two red and two green, is thrown repeatedly. Find the probability that not all colours occur in the first $k$ throws. Deduce that, if $N$ is the random variable which takes the value $n$ if all three colours occur in the first $n$ throws but only two of the colours in the first $n − 1$ throws, then the expected value of $N$ is $\frac{11}{3}$ . (Oxford 1979M)
The first proof does not seem too hard, indeed: $$ \begin{aligned} \sum_{k=0}^\infty \mathbb{P}(N>k) &= \sum_{k=0}^\infty \sum_{j=k+1}^\infty \mathbb{P}(N=j) = \mathbb{P}(N=1) + 2\mathbb{P}(N=2) + 3\mathbb{P}(N=3) + \dots \\ &= \sum_{k=1}^\infty k\mathbb{P}(N=k) =\sum_{k=0}^\infty k\mathbb{P}(N=k) = \mathbb{E}(N). \end{aligned} $$
The probability that not all colours occur in the first $k$ throws can be found by the inclusion-exclusion rule: $$ \mathbb{P}(\text{not all colours in }k\text{ throws}) = 3 \left(\frac23\right)^k - 3\left(\frac13\right)^k = 3^{1-k}(2^k-1). $$
For the last question, following the suggestion by the authors, one may consider the random variable $N$ defined as above. To compute the average I could use the expression seen in the problem text. This means: $$ \mathbb{E}(N) = 3\sum_{k=0}^\infty \left(\frac23\right)^k - 3\sum_{k=0}^\infty \left(\frac13\right)^k = 3\frac{1}{1-2/3} - 3\frac{1}{1-1/3} = \frac92, $$ which is wrong.
Edit: after the correction by @mike-earnest, we have: $$ \mathbb{E}(N) = 3\sum_{k=0}^\infty \left(\frac23\right)^k - 3\sum_{k=0}^\infty \left(\frac13\right)^k + \left(\frac03\right)^k= 3\frac{1}{1-2/3} - 3\frac{1}{1-1/3} + 1= \frac{11}{2}, $$ which is the correct result.
Your equation for $P(\text{not all colors in }k\text{ throws})$ is incorrect in the case $k=0$. Your formula gives $0$, the correct probability is $1$. Making this change leads to the correct answer.
To make your formula correct for all $k$, you need to add back in the term corresponding to the triple intersection: $$ 3(2/3)^k-3(1/3)^k+(0/3)^k $$ With the convention that $0^0=1$, the above is correct for all $k\ge 0$.