Urn problem... probability of drawing 2 balls of same color in urn with 6090 and 435 colors I have looked through the questions in the forum and was not able to spot an answer that resembled by question so I hope you can help me.
I have an urn with 6090 balls in 435 colors (exactly 14 of each). What is the probability that if i take out 6 balls (without replacement) that i get 2 of the same color?
 A: General answer
As the previously given answers are wrong I give here a general solution.
Given is an urn filled with balls with $c$ different colors and $b$ balls per color. The total number of balls is $m=c\cdot b$. The probability that in $n$ draws without replacement all balls have different color is
$$p_1=\prod_{i=1}^{n-1}\frac{m-b\cdot i}{m-i}\tag{1}$$
In case the balls are not all of different colors it follows that at least 2 balls must have the same color. The probability for this case is the complementary set of eq.(1), i.e.
$$p_2=1-\prod_{i=1}^{n-1}\frac{m-b\cdot i}{m-i}\tag{2}$$
Specific answer
For the given problem we have $b=14, c=435, m=b\cdot c=6090, n=6$, and the probability is
$$1-\prod_{i=1}^{5}\frac{6090-14i}{6090-i}\approx0.032$$
For these parameters the plot  shows  (in logarithmic scaling) the probability in dependence of the number of draws. For 50 or more draws it is in principle almost impossible to draw balls that are all differently colored.

Explanation
In the first draw any ball is selected. For the $2^{\text{nd}}$ draw there are $m-b$ balls left that have different colors than the first selected balls. As 1 ball was already  removed, the $2^{\text{nd}}$ ball can be selected out of $m-1$ remaining balls. For the $3^{\text{rd}}$ draw there are $m-2b$ balls left that have different colors than the first 2 drawn balls. As 2 balls were already removed, the $3^{\text{rd}}$ ball can be selected out of $m-2$ balls, and so on.
A: Going by your clarification "$2$ or more are the same color", use the complement.
$1 -$ P(all of different colors) $= 1 - \dfrac{\binom{435}{6}*14^6}{\binom{6090}{6}}= \dfrac{1102112019195239}{34818111237885671}=  \approx 0.0317$
A: The probability that all $6$ balls have a different color is:$$\prod_{k=0}^{5}\frac{6090-435k}{6090-k}$$ 
Then the probability that we will draw at least two balls with equal color is:$$1-\prod_{k=0}^{5}\frac{6090-435k}{6090-k}$$ 
Explanation: 
If the first ball has been drawn then there are $6090-1=6089$ balls left, and $6090-435$ of them have a color that differs from the color of the first. So we have probability $\frac{6090-435}{6090-1}$ that the second ball drawn will have a color that differs from the color of the first ball.
If the first and second ball have been drawn and have different colors then there are $6090-2$ balls left, and $6090-435\times2$ of them have a color that differs from the color of the first and second ball. So under the condition that the first and second ball have different colors we have probability $\frac{6090-435\times2}{6090-2}$ that the third ball drawn will have a color that differs from the color of the first and also the second ball.
If the first, second and third ball have been drawn and have different colors then ... et cetera.
