Expected number of 1's before a 6 shows up

Question

I have the following problem. How do I compute the expected number of 1s that will be seen by throwing a fair dice until a 6 turns up?

Answer 1

Results of the die-rolling that are neither $1$ nor $6$ are irrelevant. So the answer is the same as the expected number of heads until a tail shows up, where the coin is fair.

You are probably familiar with the expected number of coin tosses up to and including the first tail.

Remark: Instead of using previous knowledge of the geometric distribution, we can attack the problem directly. Let $e$ be the expected number of $1$'s.

If the first toss is a $6$, then the expected number is $0$.

If the first number is one of $2,3,4,5$, then the expected number is $e$.

Finally, if the first toss is $1$, the expected number is $1+e$.

It follows that $$e=0\cdot \frac{1}{6}+e\cdot\frac{4}{6}+(1+e)\cdot\frac{1}{6}.$$ This is a linear equation in $e$. Solve. We get $e=1$.

Edit: While I was typing the Remark, exactly the same argument, using proper conditional probability notation, was posted in a comment by Ian.