Expected value of throwing a coloured die - problem from Grimmett's "Probability" book I am trying to solve problem 2.6.6 from Grimmett and Welsh's "Probability" book. 
"The random variable $N$ takes non-negative integer values. Show that $E(N) = \sum_{k=0}^{\infty} P(N>k)$, provided that the series 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 $\frac{11}{2}$."
I have shown the first part (the series formula for $E(N)$ given above) and for the probability that not all colours occur in first $k$ throws I've got
$$ P = \frac{ 6\sum_{l=1, \quad l\neq \frac{k}{2}}^{\left \lfloor{\frac{k}{2}}\right \rfloor } {k\choose l} + 3{k\choose \frac{k}{2}} + 3 }{3^{k}},$$
where the middle term in the formula (i.e. $3{k\choose \frac{k}{2}}$ ) is only included if $k$ is even.
But I don't know what to do for the last part to obtain $\frac{11}{2}$. I am also not certain if my derived probability formula is correct ( I have verified it for the cases $k=3, 4$.)
I would appreciate any help for this problem.
 A: Let $p_k$ denote the probability that not all colors occur in the first $k$ throws, and observe that $p_k=P(N>k)$ and thus by the first part $\mathbb EN=\sum_{k=0}^{\infty}p_k$.
Without loss of generality, assume the first throw is blue. Then in the remaining $k-1$ throws,
$$
p_k=\mathbb P(\textrm{only blue and green appear})+\mathbb P(\textrm{only blue and red appear})-\mathbb P(\textrm{only blue appears}).
$$
Thus if $k\geq 1$ we have that $$p_k=\left(\frac{2}{3}\right)^{k-1}+\left(\frac{2}{3}\right)^{k-1}-\left(\frac{1}{3}\right)^{k-1}=\frac{2^k-1}{3^{k-1}}.$$
Using the geometric series formula $\sum_{k=0}^{\infty}x^k=\frac{1}{1-x}$ twice, we find that
$$
\mathbb EN=1+\sum_{k=1}^{\infty}\frac{2^k-1}{3^{k-1}}=1+2\cdot \frac{1}{1-\frac{2}{3}}-\frac{1}{1-\frac{1}{3}}=1+6-\frac{3}{2}=\frac{11}{2}.
$$
A: I don't think your formula is correct for $k=3,4$. For instance, for $k=3$ it gives $\frac{27}{27}=1$, but it is definitely possible to not see all three colors in the first three throws. Based on the form of your expression, I suspect you are counting


*

*The number of arrangements that contain two colors in unequal numbers

*The number of arrangements that contain two colors in equal numbers

*The number of arrangements that contain one color


For 1 and 2 you've counted $\sum_{l=0, l \neq k/2}^{\lfloor k/2\rfloor}\binom{k}{l}+3\binom{k}{k/2}$. However, by starting the sum at $l=0$ you are allowing arrangements with only one color, which you have counted separately in 3. So this sum should really start at $l=1$. 
This awkward sum with $k/2$ can also be simplified by noting that $\binom{k}{l}=\binom{k}{k-l}$. So,
$$2\sum_{l=1,l \neq k/2}^{\lfloor k/2\rfloor} +\binom{k}{k/2} = \sum_{l=1}^{n-1}\binom{k}{l}$$
$$\Rightarrow 6\sum_{l=1,l \neq k/2}^{\lfloor k/2\rfloor} +3\binom{k}{k/2} = 3\sum_{l=1}^{k-1}\binom{k}{l}.$$
Finally, if this sum started at $0$ and ended at $n$, you could employ the binomial formula to get a nice closed form. You can achieve this by adding and subtracting $\binom{k}{0}$ and $\binom{k}{k}$. I'll leave you to work that out.
For 3, you've correctly counted 3 arrangements with only one color. 
In the end you should find $P(N>k) = \frac{3\cdot 2^k - 3}{3^k}.$ This simple form suggests there might have been an easier method to count. Indeed, we could instead count by saying there are $\binom{3}{2}$ ways to choose two colors out of the three. For any choice of $2$ colors, there are $2^k$ ways to color the throws with those two colors. However, each of the $3$ outcomes that have just one color has been counted twice (e.g., for the choice of red and green, we counted the all-green coloring, and we also counted it for the choice of green and blue). So we need to subtract $3$ to account for this overcounting.
