Exact 5 II

Let $X$ be the event representing number of times we get a $5$ before hitting our first $6$.

Let’s try to break this down using conditional expectation.
From any given state, every die roll gives us 6 possible outcomes:

• With probability $\frac{1}{6}$, we get a $5$.
• With probability $\frac{1}{6}$, we get a $6$ → game stops.
• With probability $\frac{4}{6}$, we get something else ($1$, $2$, $3$, or $4$).

Let’s write the expected value in terms of these outcomes:

$$E[X]=\frac{1}{6}\cdot E[X\mid \text{5}]+\frac{1}{6}\cdot E[X\mid \text{6}]+\frac{4}{6}\cdot E[X\mid \text{Other}]$$

Now, analyze each term:

• $E[X\mid \text{6}]=0$ → because we stop immediately, no more $5$s possible.
• $E[X\mid \text{5}]=1+E[X]$ → we got one $5$, and then we're back to square one, since the process restarts.
• $E[X\mid \text{Other}]=E[X]$ → getting $1$, $2$, $3$, or $4$ doesn't change the structure. We're still in the exact same state with same probabilities.

Substituting all this in:

$$E[X]=\frac{1}{6}(1+E[X])+\frac{1}{6}(0)+\frac{4}{6}(E[X])$$ $$E[X]=\frac{1}{6}+\frac{1}{6}E[X]+\frac{4}{6}E[X]$$ $$E[X]=\frac{1}{6}+\frac{5}{6}E[X]$$

Bring terms together:

$$E[X]-\frac{5}{6}E[X]=\frac{1}{6}\Rightarrow \frac{1}{6}E[X]=\frac{1}{6}\Rightarrow E[X]=1$$

Final Answer

So the expected number of times Abd rolls a $5$ before his first $6$ is:

$$\boxed{1}$$