TITLE: Expected number of rolls for unfair die until both even and odd have shown
QUESTION [0 upvotes]: An unfair die is rolled until both an even number and odd number have appeared on top. The probability that the roll shows $n$ is proportional to $n$, where $n$ = 1,2,3,4,5,6. 
What is the expected number of rolls, and what is the probability that the last roll shows even?
I believe the answer to the latter is $12/21$, but I am not sure. Additionally, I am unsure of how to compute the expectation value for the number of rolls until both even and odd have shown.

REPLY [0 votes]: You should get the following pmf for the number obtained in a given roll, $X$, $$\begin{array} {|r|r|}\hline x & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline \mathbb{P}(X=x) & \frac1{21} & \frac2{21} & \frac3{21} & \frac4{21} & \frac5{21} & \frac6{21} \\ \hline  \end{array}$$
Thus we have that
$$\mathbb{P}(\text{even number})=\frac2{21}+\frac4{21}+\frac6{21}=\frac47$$
$$\mathbb{P}(\text{odd number})=\frac1{21}+\frac3{21}+\frac5{21}=\frac37$$
Then, applying linearity of expectation, we get
$$\begin{align}
\mathbb{E}(\text{number of rolls})
&=1+\mathbb{P}(\text{even number})\cdot\mathbb{E}(\text{number of rolls to obtain an odd number})\\
&+\mathbb{P}(\text{odd number})\cdot\mathbb{E}(\text{number of rolls to obtain an even number})\\
&=1+\frac47\cdot\frac1{\left(\frac37\right)}+\frac37\cdot\frac1{\left(\frac47\right)}\\
&=\frac{37}{12}\\
\end{align}$$
As @lulu stated in the comments, the probability that the last roll is even is just $\mathbb{P}(\text{odd number})=\frac37$ because we would otherwise throw until we obtain an odd number.
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