Balls and Bins

One of my pervious posts mentioned the problem of the balls and the bins. I got this problem from a source on twitter that I have since forgotten. Regardless of its origin, the question has been a fun one to pose to students and colleagues alike (I even asked my in-laws with some very interesting results). For those of you who haven’t read “Practice What You Preach“, The problem is as follows:

You have 8 balls, and 2 bins. 4 of the balls are Red, and 4 of the balls are White. Your job is to arrange the balls in the two bins however you like, but every ball must be put in one of the bins. (ie. no throwing balls away). I will then choose one of the 2 bins, and then draw a ball from that bin. If I draw a White ball, I win; if I draw a Red ball, you win. Which arrangement gives you the best chance at winning the game?

Ignoring my better judgement, I was immediately went down a fruitless path. What made things worse is that path involved pages of partial derivatives. There is no worse feeling than combining like terms and knowing that your method is going nowhere. In hindsight, my path toward a solution was an important one; the results I came up with surprised me, and many new questions were posed from them. I will explain all of this in due time.
I began with the observation that if all red were put in one bin, and all white were in the other, the chances are 50/50. This gave me base odds to judge my next trials. After I wrapped my head around my own question, I defined a value P(w)–the probability that the ball drawn was white. I also defined two more quantities:
let ‘x’ be the number of white balls in bin #1
let ‘y’ be the number of red balls in bin #1
Using these two variables, I set out to define a function F(x,y). If I could abstract the probability of drawing a white, I could maximize the function and find my best case scenario. For those of us who have been through calculus, maximizing a function comes naturally. It should not be a surprise that I chose this route. It may, however, surprise you to know that taking this route turned out to be less mathematically rich than listing the trials. When I got knee deep in partial derivatives, I lost all orientation of the mathematical meaning of my functions. Once I was presented with a function and knew where my destination should be, the middle–the actual process of mathematics–became a blur. Two pages of calculations and simplifications later, I started over. I re-focused with my function, and decided to try to list possible trials. This process unlocked the secrets to the problem. My function for the probability of drawing a white was:
Probability of drawing a white from bin #1: x/(x+y)
Probability of drawing a white from bin #2: (4-x)/(8-x-y)
Because each bin has the same chance of being chosen:
P(w) = .5[ x/(x+y) + (4-x)/(8-x-y) ]
The limitations of the computer type provides unnecessary complexity, but the function is relatively simple. It is based solely on the laws of multiplicative and additive probability. When I began listing the possible arrangements in both bins, I hit the major realization: having 1W and 3R in bin #1 was exactly the same as having 3W and 1R in bin #1 because the leftovers in Bin #2 would create the opposite arrangement. Each trial had a partner (or so I thought). It doesn’t matter which bin has which organization of balls; the relationships and subsequent probabilities would be the same. Both those arrangements represent the same trial. I then made a master list to find how many trials I needed to plug into my formula.
Bin #1 Left Over
White Red White Red
1. 0 0 4 4
2. 0 1 4 3
3. 0 2 4 2
4. 0 3 4 1
5. 0 4 4 0
11. 2 0 2 4
12. 2 1 2 3
13. 2 2 2 2
14. 2 3 2 1
15. 2 4 2 0
23. 4 2 0 2
24. 4 3 0 1
25. 4 4 0 0
Twenty-five total options, but quickly scanning the two columns reveals what I thought. After case #13, each is a repetition with the “leftovers” mirroring the combination that was in Bin #1. Essentially, 12 of the cases were repeats. Notice that my initial guess that every trial would have a partner was false. Trial #13 turned out to be its own partner. Once this was revealed, it was as easy as plugging in the cases into the formula. The first column in the above table represents ‘x’ and the second, ‘y’:
Case (x,y) P(w)
(0,0) 1/4
(0,1) 2/7
(0,2) 1/3
(0,3) 2/5
(0,4) 1/2
(1,0) 5/7
(1,1) 1/2
(1,2) 7/15
(1,3) 1/2
(1,4) 3/5
(2,0) 2/3
(2,1) 8/15
(2,2) 1/2
From these calculations, we see that the probability of drawing a white is highest when Bin #1 contains 1 White and 0 Red and Bin #2 contains 3 White and 4 Red. This can be rationalized with the thought that half the time white is guaranteed a win (all of Bin #1 is White). If Bin #2 is chosen, there is still a 3/7 chance of drawing a White, almost 50%. I would encourage students to rationalize through each case. They will begin to look at the rule of multiplicity in a different way.
After I solved the problem, I began my favourite process: problem posing. What other questions could we ask about this situation:
1) What if there was a unequal amount of balls?
2) What if there were 3 bins? 4 bins? ‘n’ bins?
3) Which options create a “fair game”? (ie. 50/50)
4. What if you could not leave a bin empty?
There are numerous problems to be posed from this situation. The important part as teachers (and learners) is to not stop posing them. I enjoy posing and solving problems with my students. Divide and conquer! Although I am still sure that calculus could also solve this problem, it remains an open challenge for me. The best method is the one that is the richest in mathematical meaning.

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