# Expected number of rolls for unfair die until both even and odd have shown

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.

• Hint: the individual values don't matter much. Just compute the probability of getting even or odd. If $p$ is the probability of getting even then you expect it to take $\frac 1p$ turns to get an even. And you expect it to take $\frac 1{1-p}$ turns to get an odd. the rest is straight forward.
– lulu
Oct 25, 2019 at 0:14
• As another hint: the last roll is even iff the first roll is odd.
– lulu
Oct 25, 2019 at 0:16

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.