Bags containing Balls, Here we go again

Relevant wiki: Pigeonhole Principle

The maximum number you can take out without having four matching is eight:

• $3$ red
• $2$ green
• $3$ blue

The next ball you take out, though, is guaranteed to give you four that match.

So, the answer is $\boxed{9}$

The answer is min(6.3) + min(2,3) + min(8,3) + 1

In general if there are a Red colored balls, b Green Colored Balls and c Blue colored balls. and we are wanting to get k balls of the same color,

We know that one of a,b, or c has to be at least equal to k, otherwise it is impossible to have k balls of the same color.

Then the minimum number of balls required is

min(a, k-1) + min(b, k-1) + min(c, k-1 ) + 1

We add one because, we can pick min(a, k-1) + min(b, k-1) + min(c, k-1 ) and still not have k balls of the same color. Adding any ball will guarantee that we now have k balls of the same color.

A trivial example is as follows:

If the bag contains 2 red, 2 green, and 2 blue balls and k=2 Then we need to choose minimum 4 balls out of bag such that 2 balls are of same color.