I have a probability question. The problem goes: John (J) and Max (M) are rolling a die (one die together). John wants to see an odd number on the die and Max wants to see a 6.
X ... number of rolls until they both get what they want
Y ... number of rolls until at least one of them gets what they want
What can we say about X and Y? So I was wondering what E(X) and E(Y) would be and also what are the distributions of X and Y? I know that X and Y are geometric. I have also done some simulations in R which give me the following answers: E(X) = 1.5 and E(Y) = 6.5.
Edit: Now that I know what the expected values are, I was wondering how we can get Var(Y). I know that $Var(X)=\frac{1-2/3}{4/9}=0.75$. Then we can say get Y form X and characteristic functions: $Y = X+\mathbb{1}_{M<J} J +\mathbb{1}_{J<M} M$.
\begin{align*} E(Y^2) &= E((X+\mathbb{1}_{M<J} J +\mathbb{1}_{J<M} M)^2) \\ &= E(X^2 + \mathbb{1}_{M<J}^2 J^2 + \mathbb{1}_{J<M}^2 M^2 + 2 X \mathbb{1}_{M<J} J + 2 X \mathbb{1}_{J<M} M + 2 \mathbb{1}_{J<M} \mathbb{1}_{M<J}) \\ &= E(X^2) + E(\mathbb{1}_{M<J}^2 J^2) + E(\mathbb{1}_{J<M}^2 M^2) + E(2 X \mathbb{1}_{M<J} J) + E(2 X \mathbb{1}_{J<M} M) + E(2 \mathbb{1}_{J<M}\mathbb{1}_{M<J}) \\ &= E(X^2) + E(\mathbb{1}_{M<J}^2)E(J^2) + E(\mathbb{1}_{J<M}^2)E(M^2) + 2E(X)E(\mathbb{1}_{M<J} J) + 2E(X)E(\mathbb{1}_{J<M} M) + 0 \\ &= 3 + \frac{1}{4} \cdot 6 + \frac{3}{4} \cdot 66 + 2 \cdot \frac{3}{2} \cdot \frac{3}{4} \cdot 6 + 2 \cdot \frac{3}{2} \cdot \frac{1}{4} \cdot 2 \\ &= 69 \end{align*} Is this correct?
And then also a follow up question: What is the probability that John gets what he wants (an odd number) after Max gets what he wants (a 6) if we know that we needed $n$ rolls for both of them to get what they wanted (y=N)? Here I would write $P(J<M|Y=n) = \frac{P(A)}{P(A) + P(B)}$, where A is n-1 even numbers and at least one 6 and n-th roll is an odd number. B is n-1 numbers which are not 6 and at least one odd number. n-th number in B's case is a 6. But then I don't know how to get P(A) and P(B).
Thank you in advance for your help!
Obviously $X>Y$ in every case since to have had both things happen (a six rolled and an odd rolled) one of them must have happened first before the other.
$E(Y) = \frac{6}{4}$ as $Y$ is simply a geometric distribution with probability $\frac{4}{6}$ each roll of having reached one of the desired outcomes.
Once we have reached one of said outcomes, we must wait an additional period of time depending on which outcome it was. With probability $\frac{3}{4}$ it was us having hit an odd number first at which point we need to wait an additional $6$ turns on average for our next six to roll. With probability $\frac{1}{4}$ it was us having hit a six first instead.
This gives $E(X) = \frac{6}{4}+\frac{3}{4}\cdot\frac{6}{1}+\frac{1}{4}\cdot\frac{6}{3}$