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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?

kind regards

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    Only two of the same color, or more combination? I mean, you want the probability of (2 ball same color + 4 ball of different color) or at least 2 pair of the same color and then you accept (2 or more of the same color or 2 pair of two same ball)? – 2017-02-03
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    Dear Cuoredicervo. The latter: 2 or more are the same color. – 2017-02-03

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Going by your clarification "2 or more are the same color", use the complement.

$1 -$ P(all of different colors) $= 1 - \dfrac{\binom{435}{6}}{\binom{6090}{6}}$

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    Dear True Blue Anil That is an elegant solution and would be what i am looking for... but would you mind showing exactly how to calculate this i like R/Excel and what the answer would be? – 2017-03-01
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    Sorry, have been away for almost 3 months, I'd suppose the problem has been resolved by now ! – 2017-04-30
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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.