I want to do coupon collector's for a dice roll, or the expected number of rolls to get all 6 numbers from definition of expected value as opposed to using linearity of expectation.
$\tau = min\{ t \vert X_t = i, \forall k \neq i,\ \exists j $\mathbb{E}[\tau] = \sum_{j = 1}^{\infty} j\mathbb{P}(\tau = j)$ is the definition. I try to compute $\mathbb{P}(\tau = j)$, and for j rolls, you must have 5 numbers in the first $j-1$ rolls, and a 6th number in the $j$th spot, and the 6th number can't be in any of the first $j-1$ spots. So $\mathbb{P}(\tau = j) = 6[\frac{1}{6^6}5! \binom{j-1}{5} (\frac{5}{6})^{j-6}$]. $\frac{1}{6^6}$ is for six numbers being in those spots (say, 1-5 being in the first j-1 spots, 6 being in the last), $5!$ is the ways of arranging the 5 numbers, $\binom{j-1}{5}$ is to choose the ways of putting $1,2,3,4,5$ in $j-1$ spots, and $(\frac{5}{6})^{j-6}$ is so the $j-1$ spots which you didn't place 1 to 5, don't contain a 6. then multiply by 6 since 1 through 5 can also be at the jth position. This doesn't work out to be the right sum. Where is this wrong?
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