Probability of rolling a die until a 6 comes up using linearity of expectation

Question

Suppose that we roll a die until a 6 comes up.

Using linearity of expectation, define indicator random variable $X_K$ to be 1 if X≥K and 0 otherwise.

Argue that $X=∑_KX_K$ so that $E[X]=∑_KE[X_K]$. Calculate the expectation of each term $E[X_K]=Pr[X≥k]$. Then simplify.

Example where X = 10. $∑_KX_K$ will equal 10 because $X_K$ will be 1 for all of the values less than or equal to 10. More generally, for any value X, $X_K$ will be 1 for all of the values less than or equal to X so the sum of all $X_K$ equals X.

How do I get the probability that X $\ge$ K?

Answer 1

How do I get the probability that $X≥k$?

$\mathsf P(X\geq k)$ is the probability that the first six occurs on or after roll $K$.

Thus that is the probability for $k-1$ consecutive rolls of not-sixes.

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So $Pr(x≥K)$ would be $(5/6)^{k−1}$? $5/6$ is the probability of not getting the six and we do this $k-1$ times.

Exactly.

So because you know that $\mathsf E(X_k)=\Pr(X\geq k)$, therefore $\mathsf E(X)~{=\sum_{k=1}^\infty (5/6)^{k-1}\\ = \sum_{j=0}^\infty (5/6)^j}$

And you should be familiar with that named series.