Naz N.

# please show me how to solve it or what will be the outcome

In a giant bin, we stir together 2019 red jelly beans and 2019 green jelly beans. We pull 3 jelly beans out of the bin at a time. There are two possibilities – they are all the same color or there are 2 of one color (and one of the other). In the first case (3 of the same color), we eat all 3 jelly beans. In the second case (2 of one color), we put 1 jelly bean back of the majority color and eat the other 2.

If we repeat this process until there are fewer than 3 jelly beans left in the bin, there are these possibilities

1.    No jelly beans.

2.    One red jelly bean.

3.    One green jelly bean.

4.    Two red jelly beans.

5.    Two green jelly beans.

6.    One red and one green jelly bean

Which of these outcomes is most likely? Which is least likely?

Al P.

Sorry I don't have a solution, but it seems to me that choices 2-5 are just distractions as there should be no bias for one color over the other. To end with {R,G}, the previous set MUST be {R,R,G,G}, {R,R,R,R,G} (where RRR is selected) or {G,G,G,G,R} (where GGG is selected). To end with {null}, the previous set must be {R,R,R} or {G,G,G}. About all I can say. I may revisit if I have time this weekend.
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09/19/19

Naz N.

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09/19/19

Al P.

I wrote a small C program to simulate this scenario, running 1000 trials I get: No jelly beans: 969 times One red and one green: 31 times Other cases: 0 times A second run of 1000 trials: No jelly beans: 967 One red and one green: 33 Other cases: 0 The simulation uses a true random number generator.
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09/20/19

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