A fair die is rolled repeatedly. Let $X$ be the number of rolls needed to obtain a $5$ and $Y$ be the number of rolls needed to obtain as $6$.

probability probability-distributions conditional-probability

$\mathbb E(X\mid X\geq3)$ is the expectation of the number of rolls needed to obtain a $5$ under the condition that the rolls $1$ and $2$ do not provide a $5$.

We might say that the sequence of relevant rolls actually starts after $2$ rolls that are in vain, so that:$$\mathbb E(X\mid X\geq3)=\mathbb E(2+X)=2+\mathbb EX$$

Here $\mathbb EX=6$ since $X$ has geometric distribution (number of trials needed) with parameter $p=\frac16$.

This together explains that: $$\mathbb E(X\mid X\geq3)=2+6=8$$

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