Expected value of throwing a coloured die - problem from Grimmett's "Probability" book

Question

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.

Answer 1

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

1. The number of arrangements that contain two colors in unequal numbers
2. The number of arrangements that contain two colors in equal numbers
3. 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.

Answer 2

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 $$ \mathbb EN=1+\sum_{k=1}^{\infty}\frac{2^k-1}{3^{k-1}}=1+2\cdot \frac{1}{1-\frac{2}{3}}-\frac{1}{1-\frac{1}{3}}=1+6-\frac{3}{2}=\frac{11}{2}. $$