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Use the positive-support convention for the geometric distribution and put $q=1-p$. Differentiating the convergent geometric series ${\sum }_{k\ge 0}{q}^k=(1-q{)}^{-1}$ givesConsequentlySubtracting the square of the expectation from the second moment yields

For the die, this is the coupon collector problem. When $r$ scores are still missing, let $W_r$ count rolls up to and including the next newly observed score. Every fresh roll has success probability $r/6$, so $W_r$ has a geometric distribution of parameter $r/6$. These six successive waiting times are independent: conditional on the entire past at the start of a stage, its waiting-time distribution depends only on the number $r$, not on the identities of the missing scores. Iterated conditioning therefore factors their joint distribution. The total number of rolls is $N={\sum }_{r=1}^6{W}_r$.

Using linearity of expectation givesUsing variance additivity for independent random variables, or the coupon collector waiting-time variance formula just derived by this decomposition, givesHence the requested standard deviation isThe hint rounds the sum of squares from $53.69$ to $53.7$; using that rounded value gives $\sqrt{39}\approx 6.245$, consistent at the precision of the hint. The exact computation preserves the slightly more precise value above.

Use the positive-support geometric distribution, $\mathbb{P}(X=j)=p(1-p{)}^{j-1}$ for $j\ge 1$, $0<p\le 1$. Differentiating the geometric series suppliesFor the jar, let $W_r$ be the waiting time to remove one red ball when exactly $r$ red balls remain. Every draw has success probability $r/n$, so $W_r$ is geometric with that parameter. The $W_r$ are independent: after each success, fresh independent draws begin a new stage, and its conditional waiting distribution depends only on the deterministic count $r$, not on earlier waiting times. Replacing a green ball leaves that stage's count unchanged.

Thus the coupon collector problem decomposition is $T={\sum }_{r=1}^n{W}_r$, andThis is the coupon collector waiting-time variance. For $n=1$, the formula correctly gives zero variance and a deterministic one-minute completion.