A dice is rolled $20$ times, with the possible results $\left\{1,2,3,4,5,6\right\}$.
Let $X$ be the number of results, out of the possible 6, which were chosen only once during the 20 rolls.
Calculate $P\left\{X\right\}$
I find it hard to identify the kind of variable it is. It isn't bio nominal nor hyper geometric.
I understand I have to choose 4 rolls out of the 20, and the combination between them is $4!$, giving me –
$$ \frac{\binom{20}{4} \times \binom{6}{4} \times 4!}{6^{20}} $$
For the chosen "results", the chosen "rolls' and the inner combination between them. But how about the other "rolls"? Something is missing here.
Best Answer
Why would it be anything with a nice name?
I have no idea where this came from, but it is wrong.
To actually answer your question, we'll first find a (reverse) cumulative density function for $P$, then calculate the actual values from that.
So, what's the probability that $X$ is at least $k$ (for $1 \leq k \leq 5$? That means that there are at least $k$ elements of $\{1,2,3,4,5,6\}$ (with $\left(\array{6\\k}\right)$ choices) that don't come up, and $k$ of our rolls take the appropriate values, one at a time ($\left(\array{20\\k}\right)k!$ possibilities), while the other $20-k$ take values the remaining $6 - k$ possible values ($(6-k)^{20-k}$ possibilities).
Our numbers $N(X=k)$ of possibilities for $X$ to take a value no smaller than each $k$ is therefore given by:
\begin{array}{c|c|c}X&N(X\geq k)&N(X\geq k)\mbox{ simplified}\\\hline \\0&6^{20}&6^{20} \\1&\left(\array{6\\1}\right)1!\left(\array{20\\1}\right)(6-1)^{20-1}&24(5)^{20} \\2&\left(\array{6\\2}\right)2!\left(\array{20\\2}\right)(6-2)^{20-2}&1425(4)^{19} \\3&\left(\array{6\\3}\right)3!\left(\array{20\\3}\right)(6-3)^{20-3}&15200(3)^{19} \\4&\left(\array{6\\4}\right)4!\left(\array{20\\4}\right)(6-4)^{20-4}&218025(2)^{19} \\5&\left(\array{6\\5}\right)5!\left(\array{20\\5}\right)(6-5)^{20-5}&11162880 \\6&0&0\end{array}
The actual counts for each value of $X$ are then given by the differences between these: $N(X = k) = N(X \geq k) - N(X \geq k+1)$:
\begin{array}{c|c|c}X&N(X= k)&N(X=k)\mbox{ evaluated}\\\hline \\0&6^{20}-24(5)^{20}&1367340080687976 \\1&24(5)^{20}-1425(4)^{19}&1897117341979800 \\2&1425(4)^{19}-15200(3)^{19}&374034643096800 \\3&15200(3)^{19}-218025(2)^{19}&17552066407200 \\4&218025(2)^{19}-11162880&114296728320 \\5&11162880&11162880 \\6&0&0\end{array}
And our probabilities are, therefore, given by dividing these by $6^{20}$:
\begin{array}{c|c|c|c}X&P(X= k)&P(X=k)\mbox{ evaluated}&\mbox{approx.}\\\hline \\0&1-24\left(\frac{5}{6}\right)^{20}&\frac{56972503361999}{152339935002624}&0.37398 \\1&24\left(\frac{5}{6}\right)^{20}-1425\left(\frac{2}{3}\right)^{19}&\frac{79046555915825}{152339935002624}&0.51888 \\2&1425\left(\frac{2}{3}\right)^{19}-15200\left(\frac{1}{2}\right)^{19}&\frac{3896194198925}{38084983750656}&0.10230 \\3&15200\left(\frac{1}{2}\right)^{19}-218025\left(\frac{1}{3}\right)^{19}&\frac{20314891675}{4231664861184}&0.0048007 \\4&218025\left(\frac{1}{3}\right)^{19}-\frac{11162880}{6^20}&\frac{5511995}{176319369216}&0.000031261 \\5&\frac{11162880}{6^{20}}&\frac{1615}{528958107648}&0.0000000030532\end{array}