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Anna has 7 pairs of yellow socks, 8 pairs of red socks and 2 pairs of blue socks mixed together in a drawer. All pairs of socks are identical except for their colors. What is the smallest number of socks she has to take out without looking to make sure of having 2 matching pairs of socks?

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    This isn't a question about probability, this is a question about the pigeonhole principle. "If you have more pigeons than pigeonholes, then there must be a pigeonhole with more than one pigeon." – 2017-02-22
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    Do we have to distinguish between left socks and right socks ? I have an anwer assuming that we only need socks with the same colour. – 2017-02-22
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    But even then, we need no probabilities, we only have to consider worst cases and show that a particular number of socks is always successful. – 2017-02-22

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To show that $5$ socks are not sufficient : If we take one yellow, three red socks and one blue sock, we have only one matching pair.

Now , suppose we take $6$ socks. If we have at least two blue socks, the resulting $4$ socks must lead to another match.

If we have at most one blue sock, we have at least two yellow socks or at least two red socks or at least four yellow socks or at least four red socks.

In every scenario, we have two matching pairs.

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    This answers the question both of us think was asked: a matching pair. But the literal request is for two matching pairs - perhaps for Anna and her sister. Care to edit answer that one too? – 2017-02-22
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    I get the pigeonhole principle and how it applies for one pair. What is the generalized formula for two pairs as in the example above? – 2017-02-23
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    Take a triple $(a,b,c)$ meaning that we have $a$ yellow, $b$ red and $c$ blue socks. Now consider the number of the elements $a,b,c$ that are $2$ or bigger : – 2017-02-23
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    If there are at least two such numbers, we have $2$ matching pairs. If we have exactly one such number, then the two other numbers must be $1$ or less. This implies that the largest number must be at least $4$. Again, we have two matching pairs. The case, that no number is $2$ or bigger is impossible because then we would have sum at most $3$. – 2017-02-23
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    This argumentation includes the pigeonhole-principle, but it is much more difficult because we have to distinguish cases (at least I have not found an easier argument). Nevertheless, we can show that $6$ socks are sufficient. – 2017-02-23
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How to explain this?

To a 10 year old?!

Get out some socks!! (... or pieces of different colored paper, or beads, or ...).

And for heaven's sake, don't mention probability, which of course has nothing to do with this!

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    For anyone in my situation, this video does exactly that: https://youtu.be/A2ccjnEFQGU – 2017-02-23
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    @user3919652 That's good, but for a 10 year old I wouldn't bring up the pigeonhole principle until much later (the fact that you are looking for 2 pairs of socks complicates the picture as well...). i think it is really important for the 10 year old to just play with picking socks for a while. And rather than picking socks in the dark, tell the 10 year old to intentionally pick as many socks without making two pairs of socks ... and then have the 10 year old pick one more sock after that. – 2017-02-23
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    that exercise worked - it was also a good opportunity to pick some lego pieces off the floor ;-) – 2017-02-24
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    @user3919652 Ha ha, that's great! :) – 2017-02-24