A bag has 50 green marbles and 50 blue marbles that are drawn randomly, two at a time. If two blue marbles are drawn, the marbles are placed into Box A. If two green marbles are drawn, they are placed into Box B. If one of each color marble is drawn, they are placed into box C. If all the marbles are drawn, what is the probability that Box A and Box B will have the same amount of marbles in them?

Question

ddcamp · Answer

Let's look at a smaller scale version, where there are only 4 green and 4 blue to begin with.

Say you draw 2 green marbles to begin with, you'll have 2 green and 4 blue marbles left in the bag. Now, you either draw another 2 green marbles, 2 blue marbles, or one of each.

If you draw 2 green marbles again, your next two moves you'll be forced to grab two blue marbles, and boxes A and B will each have 4 marbles in them.

If you draw 2 blue marbles, your next options would be: draw mixed, leaving one marble of each color in the bag (forcing you to draw mixed again) resulting in 2 marbles in both A and B; draw 2 blue, forcing you to draw 2 green next turn (4 marbles in both A and B); or draw 2 green, forcing you to draw 2 blue next turn (4 marbles i both A and B).

If you draw mixed, you'll have 1 green and 3 blue left. Your next two moves will be draw mixed again, then draw 2 blue; or draw 2 blue, then draw mixed again (2 marbles in both A and B).


TLDR: Grabbing 2 green will force you to, at some point, grab 2 blue and vice-versa. Boxes A and B will always have the same number of marbles in them.

anonymous · Answer

So is this more of just a logic question and not some sort of probability formula or probability process?

ddcamp · Answer

Yup

anonymous · Answer

Is there anything that clued you in to that? My first thought is just to try number crunching, but obviously that wasnt the way to go.....

ddcamp · Answer

Nothing much, other than they chose a relatively large number (50 of each color marble) for a complex process. Usually (though not always) that means there's a fairly simple way to solve it.

ddcamp · Answer

Also, my big long explanation is probably not the easiest way to think about it, it's just the explanation I thought of first.

anonymous · Answer

No, it works just fine, thank you for taking the time to type all that out and to explain, I appreciate it ^_^

ddcamp · Answer

No problem.