Jar of Jelly Beans: 20 of Each Color, How Many to Pick?

[yakin]

A jar contains red, blue, orange and green jelly beans. There are 20 of each color. How many jelly beens must be selected to guarantee that at least 3 are the same color?
The teacher got 9 as answer, but i do not know how she got this. Anybody would like to help please?

 

[alyafey22]

A jar contains red, blue, orange and green jelly beans. There are 20 of each color. How many jelly beens must be selected to guarantee that at least 3 are the same color?
The teacher got 9 as answer, but i do not know how she got this. Anybody would like to help please?

I think the question is 2 of each color ?

 

[Deveno]

It is possible that after 8 draws, we have 2 of each of the four colors. Since the next one drawn must be one of the four colors, we will then have 3 of some color.

If, having drawn 8, we only have at most 3 colors of jelly beans, we must already have 3 of a matching color. For 3 pairs of 3 different colors is only 6 jelly beans, and we have two more, and any other combination of 6 jelly beans with 3 colors already includes 3 of one color.

It is also possible that after 8 draws, we have all four colors, but only one of some color. In this case, we have 7 jelly beans with 3 colors, and as we saw above, we can make at most 3 pairs from these, the 7th must match one of the 3 colors.

One can continue this analysis with just two colors: it can be seen that we need a mere 5 jelly beans to ensure 3 of a kind, and with only one color, we need but 3.

So no matter what, the 9th jelly bean will make (at least) the 3rd bean of some color.

 

[alyafey22]

Assuming that we have 4 holes and each one is one of the mentioned colors. If we select a certain beans we place it at its corresponding color. The worst case scenario is having each hole filled with two beans after the 8th draw. The 9th draw will have to be the third bean at any of the holes. Pigeonhole principle.

 

[Deveno]

Yes, one can easily adapt that method to find the minimum draw that must have $k$ of any color, given we have $n$ colors, which is:

$n(k-1) + 1$.

This can be shown by double induction on $k$ and $n$, and in the case at hand (n = 4, k = 3) gives:

4(3 - 1) + 1 = 8 + 1 = 9.

(Assuming of course, we have enough objects to begin with).