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This is a variation of the combinatorial problem considered in section 5 of Some new aspects of the coupon collector’s problem (2003).

The $t$ singleton sides (sides which appear once) can be chosen as an ordered sequence in $t! {s\choose t}$ ways; this sequence can appear among the $n$ rolls in ${n\choose t}$ ways and the remaining $n-t$ rolls constitute an ordered partition of $n-t$ elements into $s-t-k$ classes, $k=0,1,\ldots s-t-1$, no class having fewer than two elements, which can be chosen in $\sum_{k=0}^{s-t-1}(s-t-k)!\begin{Bmatrix} n-t\\ s-t-k \end{Bmatrix}_2$ ways. Multiply these together and divide by $s^n$, the number of $n$ sequences, to obtain the desired probability

$$p=\frac{s!}{(s-t)!s^n} {n\choose t}\sum_{k=0}^{s-t-1}(s-t-k)!\begin{Bmatrix} n-t\\ s-t-k \end{Bmatrix}_2$$

The generating function for the coefficients $\begin{Bmatrix} n\\ k \end{Bmatrix}_2$ is $$\sum_{n\geq 0}\begin{Bmatrix} n\\ k \end{Bmatrix}_2\frac{x^n}{n!}=\frac{1}{k!}\left(e^x-1-x\right)^k$$

This is a variation of the combinatorial problem considered in section 5 of Some new aspects of the coupon collector’s problem (2003).

The $t$ singleton sides (sides which appear once) can be chosen as an ordered sequence in $t! {s\choose t}$ ways; this sequence can appear among the $n$ rolls in ${n\choose t}$ ways and the remaining $n-t$ rolls constitute an ordered partition of $n-t$ elements into $s-t-k$ classes, $k=0,1,\ldots s-t-1$, no class having fewer than two elements, which can be chosen in $\sum_{k=0}^{s-t-1}(s-t-k)!{{s-t}\choose k}\begin{Bmatrix} n-t\\ s-t-k \end{Bmatrix}_2$ ways. (The integer $k$ counts the number of classes that do not appear at all.) Multiply these together and divide by $s^n$, the number of $n$ sequences, to obtain the desired probability

$$p=\frac{s!}{s^n} {n\choose t}\sum_{k=0}^{s-t-1}\frac{1}{k!}\begin{Bmatrix} n-t\\ s-t-k \end{Bmatrix}_2$$

The generating function for the coefficients $\begin{Bmatrix} n\\ k \end{Bmatrix}_2$ is $$\sum_{n\geq 0}\begin{Bmatrix} n\\ k \end{Bmatrix}_2\frac{x^n}{n!}=\frac{1}{k!}\left(e^x-1-x\right)^k$$

This is a variation of the combinatorial problem considered in section 5 of Some new aspects of the coupon collector’s problem (2003).

The $t$ singleton sides (sides which appear once) can be chosen as an ordered sequence in $t! {s\choose t}$ ways; this sequence can appear among the $n$ rolls in ${n\choose t}$ ways and the remaining $n-t$ rolls constitute an ordered partition of $n-t$ elements into $s-t$ classes, no class having fewer than two elements, which can be chosen in $(s-t)!\begin{Bmatrix} n-t\\ s-t \end{Bmatrix}_2$ ways. Multiply these together and divide by $s^n$, the number of $n$ sequences, to obtain the desired probability

$$p=\frac{s!}{s^n} {n\choose t}\begin{Bmatrix} n-t\\ s-t \end{Bmatrix}_2$$

The generating function for the coefficients $\begin{Bmatrix} n\\ k \end{Bmatrix}_2$ is $$\sum_{n\geq 0}\begin{Bmatrix} n\\ k \end{Bmatrix}_2\frac{x^n}{n!}=\frac{1}{k!}\left(e^x-1-x\right)^k$$

This is a variation of the combinatorial problem considered in section 5 of Some new aspects of the coupon collector’s problem (2003).

The $t$ singleton sides (sides which appear once) can be chosen as an ordered sequence in $t! {s\choose t}$ ways; this sequence can appear among the $n$ rolls in ${n\choose t}$ ways and the remaining $n-t$ rolls constitute an ordered partition of $n-t$ elements into $s-t-k$ classes, $k=0,1,\ldots s-t-1$, no class having fewer than two elements, which can be chosen in $\sum_{k=0}^{s-t-1}(s-t-k)!\begin{Bmatrix} n-t\\ s-t-k \end{Bmatrix}_2$ ways. Multiply these together and divide by $s^n$, the number of $n$ sequences, to obtain the desired probability

$$p=\frac{s!}{(s-t)!s^n} {n\choose t}\sum_{k=0}^{s-t-1}(s-t-k)!\begin{Bmatrix} n-t\\ s-t-k \end{Bmatrix}_2$$

The generating function for the coefficients $\begin{Bmatrix} n\\ k \end{Bmatrix}_2$ is $$\sum_{n\geq 0}\begin{Bmatrix} n\\ k \end{Bmatrix}_2\frac{x^n}{n!}=\frac{1}{k!}\left(e^x-1-x\right)^k$$

This is a variation of the combinatorial problem considered in section 5 of Some new aspects of the coupon collector’s problem (2003).

The $t$ singleton sides (sides which appear once) can be chosen as an ordered sequence in $t! {s\choose t}$ ways; this sequence can appear among the $n$ rolls in ${n\choose t}$ ways and the remaining $n-t$ rolls constitute an ordered partition of $n-t$ elements into $s-t$ classes, no class having fewer than two elements, which can be chosen in $(s-t)!\left\{{n-t}\over{s-t}\right\}_2$ ways. Multiply these together and divide by $s^n$, the number of $n$ sequences, to obtain the desired probability

$$p=\frac{s!}{s^n} {n\choose t}\left\{{n-t}\over{s-t}\right\}_2$$

This is a variation of the combinatorial problem considered in section 5 of Some new aspects of the coupon collector’s problem (2003).

The $t$ singleton sides (sides which appear once) can be chosen as an ordered sequence in $t! {s\choose t}$ ways; this sequence can appear among the $n$ rolls in ${n\choose t}$ ways and the remaining $n-t$ rolls constitute an ordered partition of $n-t$ elements into $s-t$ classes, no class having fewer than two elements, which can be chosen in $(s-t)!\begin{Bmatrix} n-t\\ s-t \end{Bmatrix}_2$ ways. Multiply these together and divide by $s^n$, the number of $n$ sequences, to obtain the desired probability

$$p=\frac{s!}{s^n} {n\choose t}\begin{Bmatrix} n-t\\ s-t \end{Bmatrix}_2$$

The generating function for the coefficients $\begin{Bmatrix} n\\ k \end{Bmatrix}_2$ is $$\sum_{n\geq 0}\begin{Bmatrix} n\\ k \end{Bmatrix}_2\frac{x^n}{n!}=\frac{1}{k!}\left(e^x-1-x\right)^k$$