I would like to know when they say "a little more thought" reveals their answer for $T_{k-1}-T_k$, what is the thought exactly? If $T_2$ is distributed geometrically with probability $\frac{n-1}{n}$, then surely $T_k$ is distributed geometrically with probability $\frac{n-k-1}{n}$ and $T_{k+1}$ is distributed geometrically with probability $\frac{n-k}{n}$? How is it so that they claim $T_{k+1}-T_k$ ddistributed geometrically with probability $\frac{n-k}{n}$?
2026-02-23 06:35:33.1771828533
Coupon collector's clarification
28 Views Asked by Bumbble Comm https://math.techqa.club/user/bumbble-comm/detail AtRelated Questions in PROBABILITY
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