I roll a die repeatedly until I get 6, and then count the number of 3s I got. What's my expected number of 3s?

Consider the following experiment. I roll a die repeatedly until the die returns 6, then I count the number of times 3 appeared in the random variable $X$. What is $E[X]$?

Thoughts: I expect to roll the die 6 times before 6 appears (this part is geometric), and on the preceding 5 rolls each roll has a $1/5$ chance of returning a 3. Treating this as binomial, I therefore expect to count 3 once, so $E[X]=1$.

Problem: Don't know how to model this problem mathematically. Hints would be appreciated.

Solution 1:

We can restrict ourselves to dice throws with outcomes $3$ and $6$. Among these throws, both outcomes are equally likely. This means that the index $Y$ of the first $6$ is geometrically distributed with parameter $\frac12$, hence $\mathbb{E}(Y)=2$. The number of $3$s occuring before the first $6$ equals $Y-1$ and has expected value $1$.

Solution 2:

There are infinite ways to solve this problem, here is another solution I like.

Let $A = \{\text{first roll is }6\}$, $B = \{\text{first roll is }3\}$, $C = \{\text{first roll is neither }3\text{ nor }6\}$. Then $$ E[X] = E[X|A]P(A) + E[X|B] P(B) + E[X|C] P(C) = 0 + (E[X] + 1) \frac16 + E[X]\frac46, $$ whence $E[X] = 1$.