I am asking this to understand the coupon collector's problem better.
Say that we are trying to collect $n$ coupons, and every time we obtain a box of cereal, we get a coupon, which has an equal probability of giving us any of the $n$ coupons. I am trying to understand why if we let $X_i$ be a random variable which returns the amount of boxes we bought until we got the $i$th new card, measured from the boxes we bought right after getting the $(i-1)$st new card, that $\mathbb{P}[X_i=j]= (\frac 1 n)^{j-1} \cdot (\frac {n-i +1} 1)$.
My chief difficulty in understanding this is that the sample points we have vary in terms of which coupons their $i$th first element and $(i-1)$st first element are, as well as which boxes it were on which we have found the $i$th first element and $(i-1)$st first element. If the question simply was, what is the probability of finding coupon $a$ on the $b$th box we have opened, and to open $j-1$ boxes after this and still get coupon $a$, and then to on the $j$th box after this to get coupon $b$, I could see that $(\frac 1 n)^{j-1} \cdot (\frac {n-i +1} 1)$ is the probability; but because these things vary on each sample point I am not seeing this.
So far, I have tried thinking that the question is essentially that given we have grabbed $i-1$ different things out of $n$ available choices, what’s the probability that for the next $j-1$ grabbings, we grab the same thing, and for the jth grabbing, grab something different? However, the matter is still rather unclear for me- even when I phrase the question like this, I am not seeing how to relate this to the sample points which vary in terms of how many boxes it took to obtain the $i$th new coupon and which coupons is the $i$th, etc. So why does $\mathbb{P}[X_i=j]= (\frac 1 n)^{j-1} \cdot (\frac {n-i +1} 1)$ here?
If we can express the probability $P(X_i = j)$ as $P(Z_i = j | \mbox{ I have already found the coupons $j_1,\dots,j_{i-1}$})$ where $Z_i$ is again the number of boxes I have to open before finding the $i$-th coupon (where $i\in\{1,\dots,n\}\setminus\{j_1,\dots,j_{i-1}\}$) the problem is almost straigthforward. In fact, define $I$ as: $$ I = \{1,\dots,n\}\setminus\{j_1,\dots,j_{i-1}\} \implies |I| = n-i+1 $$ So the distribution of $Z_i$ given that we've already found coupons $j_1,\dots,j_{i-1}$ is geometric and in particular: $$ P(Z_i = j | j_1,\dots,j_{i-1}) = P(Y_1\in\{j_1,\dots,j_{i-1}\})\cdots P(Y_{j-1}\in\{j_1,\dots,j_{i-1}\})P(Y_j\in I) $$ where $Y_k$ is the coupon we find in the $k$-th box. Finally: $$ P(Z_i = j | j_1,\dots,j_{i-1}) = \Bigr(\frac{i-1}{n}\Bigl)^{j-1}\Bigl(\frac{n-i+1}{n}\Bigr) $$