Mit der Hypergeometrischen Verteilung erhalte ich die Verteilung
\(\scriptsize \left(\begin{array}{rrrrrr}0& \left\{ 3, 0, 0 \right\} &\frac{1}{22}& \left\{ \left(\begin{array}{r}5\\3\\\end{array}\right), \left(\begin{array}{r}3\\0\\\end{array}\right), \left(\begin{array}{r}4\\0\\\end{array}\right) \right\} &/&\left(\begin{array}{r}12\\3\\\end{array}\right)\\1& \left\{ 2, 1, 0 \right\} &\frac{3}{22}& \left\{ \left(\begin{array}{r}5\\2\\\end{array}\right), \left(\begin{array}{r}3\\1\\\end{array}\right), \left(\begin{array}{r}4\\0\\\end{array}\right) \right\} &/&\left(\begin{array}{r}12\\3\\\end{array}\right)\\2& \left\{ 2, 0, 1 \right\} &\frac{2}{11}& \left\{ \left(\begin{array}{r}5\\2\\\end{array}\right), \left(\begin{array}{r}3\\0\\\end{array}\right), \left(\begin{array}{r}4\\1\\\end{array}\right) \right\} &/&\left(\begin{array}{r}12\\3\\\end{array}\right)\\3& \left\{ 0, 3, 0 \right\} &\frac{1}{220}& \left\{ \left(\begin{array}{r}5\\0\\\end{array}\right), \left(\begin{array}{r}3\\3\\\end{array}\right), \left(\begin{array}{r}4\\0\\\end{array}\right) \right\} &/&\left(\begin{array}{r}12\\3\\\end{array}\right)\\4& \left\{ 1, 2, 0 \right\} &\frac{3}{44}& \left\{ \left(\begin{array}{r}5\\1\\\end{array}\right), \left(\begin{array}{r}3\\2\\\end{array}\right), \left(\begin{array}{r}4\\0\\\end{array}\right) \right\} &/&\left(\begin{array}{r}12\\3\\\end{array}\right)\\5& \left\{ 0, 2, 1 \right\} &\frac{3}{55}& \left\{ \left(\begin{array}{r}5\\0\\\end{array}\right), \left(\begin{array}{r}3\\2\\\end{array}\right), \left(\begin{array}{r}4\\1\\\end{array}\right) \right\} &/&\left(\begin{array}{r}12\\3\\\end{array}\right)\\6& \left\{ 0, 0, 3 \right\} &\frac{1}{55}& \left\{ \left(\begin{array}{r}5\\0\\\end{array}\right), \left(\begin{array}{r}3\\0\\\end{array}\right), \left(\begin{array}{r}4\\3\\\end{array}\right) \right\} &/&\left(\begin{array}{r}12\\3\\\end{array}\right)\\7& \left\{ 1, 0, 2 \right\} &\frac{3}{22}& \left\{ \left(\begin{array}{r}5\\1\\\end{array}\right), \left(\begin{array}{r}3\\0\\\end{array}\right), \left(\begin{array}{r}4\\2\\\end{array}\right) \right\} &/&\left(\begin{array}{r}12\\3\\\end{array}\right)\\8& \left\{ 0, 1, 2 \right\} &\frac{9}{110}& \left\{ \left(\begin{array}{r}5\\0\\\end{array}\right), \left(\begin{array}{r}3\\1\\\end{array}\right), \left(\begin{array}{r}4\\2\\\end{array}\right) \right\} &/&\left(\begin{array}{r}12\\3\\\end{array}\right)\\9& \left\{ 1, 1, 1 \right\} &\frac{3}{11}& \left\{ \left(\begin{array}{r}5\\1\\\end{array}\right), \left(\begin{array}{r}3\\1\\\end{array}\right), \left(\begin{array}{r}4\\1\\\end{array}\right) \right\} &/&\left(\begin{array}{r}12\\3\\\end{array}\right)\\\end{array}\right) \)
da lässt sich dann alles ablesen