0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ 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. $\endgroup$
    – lulu
    Oct 25, 2019 at 0:14
  • $\begingroup$ As another hint: the last roll is even iff the first roll is odd. $\endgroup$
    – lulu
    Oct 25, 2019 at 0:16

1 Answer 1

0
$\begingroup$

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.

$\endgroup$
2
  • 1
    $\begingroup$ Small correction: Since you have to throw once to decide whether you then want odd or even, you need to add one to the expected number. $\endgroup$
    – lulu
    Oct 25, 2019 at 0:49
  • $\begingroup$ @lulu Yes, thanks. $\endgroup$ Oct 25, 2019 at 0:51

You must log in to answer this question.