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I came across this question as a part of a practice test I'm working on to prepare for a maths exam at school:

An urn contains 4 red, 6 blue, 3 green, 8 brown and 9 white marbles. The urn has a lid and you can't see into it. How many marbles must you take out in order to be certain of getting two of the same colour?

I think this problem would be solved by using probability theory but wasn't able to find a way to do it. My initial thinking was that since 3 is the smallest of the coloured marbles that if I picked 2 then it would give me at least two green marbles but then there's the issue of not getting 2 of green when picking.

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    $\begingroup$ If you have pulled out $5$ marbles, what is the only way that you don't have two of the same color? When they are all different colors, each of the colors represented once a piece. If you are in this scenario and you pull out a sixth marble... what happens? $\endgroup$
    – JMoravitz
    Jan 28 at 6:13
  • $\begingroup$ en.wikipedia.org/wiki/Pigeonhole_principle $\endgroup$
    – JMoravitz
    Jan 28 at 6:13
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This problem is pretty simple. Start out by taking one. Since you don't want them to be the same, next you take out a different color. Continue this until you get 5 marbles of all different colors. After this, no matter what you take, you will get a duplicate. Therefore, 5 is the most you can get with all different, so you would need 6. This is basically the Pigeonhole Principle.

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    $\begingroup$ Thanks @Fatso Boo. Now that you explain it, it is pretty simple. $\endgroup$ Jan 28 at 21:41

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