Let X = the number of 1's that are rolled. Find E[X] and Var(X).
I can't seem to calculate Var(X).
I've calculated E[X] = 1.
I let R = the number of non-6 rolls, and I let Y = the number of rolls . R + 1 = Y. E[R + 1] = E[Y]. E[R] = E[Y] – 1. E[Y] = 6 because Y ~ Geom(1/6), so E[R] = 5.
R = $R_1 + R_2 + R_3 + R_4 + R_5$, where $R_i$ = the number of times face i has been rolled. All the $R_i$'s are identically distributed, so E[Y] = E[$R_1$] = 1/5E[R] = 1.
Best Answer
For roll $i$, let $X_i=1$ if roll $i$ of the die is 1 and 0 otherwise. Hence, $$R_1 = \sum\limits_{i=1}^R X_i$$
The dice cannot roll 6 in the first $R$ rolls, so $X_i=1$ with probability 1/5 and 0 with probability 4/5. Hence, $$E[R_1|R] = \sum\limits_{i=1}^R E[X_i|R] = \frac{1}{5}R$$
$$Var[R_1|R] = \sum\limits_{i=1}^R Var[X_i|R] = \frac{1}{5}\frac{4}{5}R$$
By the law of total variance, \begin{align*} Var[R_1] &= Var[E[R_1|R]] + E[Var[R_1|R]]\\ &= Var[\frac{R}{5}] + E[\frac{4}{25}R]\\ &= \frac{1}{25}Var[R] + \frac{4}{25}E[R]\\ &= \frac{1}{25}\frac{1-1/6}{1/36} + \frac{4}{25}5\\ &= 2 \end{align*}