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.
Best Answer
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.