Home > Problem of the Week > Archive List > Detail

<< Prev 10/25/2009 Next >>

Who Has to Do the Dishes? [Part 2]

Before investigating this problem, be sure that you have done Part 1 of the previous week.

In Part 1, the twins, Polly Dent and Pepso Dent, eventually solved their dilemma by creating a mathematical system that was fair in deciding each day who would wash the dishes and who would dry.

Their successful decision process: Three red spoons and one yellow spoon are placed in a bag. Polly would then draw two spoons randomly out of the bag. If the two spoons pulled out were the same color, then Polly would wash and Pepso would dry. But, if the two spoons were different colors, then Pepso would wash and Polly would dry. The process was now fair, with the probability of each option being 50%.

BUT....being good mathematical students, the Dent twins remembered their mathematics teacher's constant admonition: Once you have solved a problem, ask new questions so you have new problems to solve! So....

Question: Assuming that a person always randomly draws two spoons out the bag, what other combinations of red spoons and yellow spoons will keep the decision process fair?


Source: Harris Shultz's "Mix or Match," CMC ComMuniCator, Volume 17#3, p. 40

Hint: Again, use experimentation to help understand and explore the problem. Try various mixes of r red spoons and y yellow spoons, until you have data that supports one or two other "fair" situations.

Now, ask...is there a pattern!


Solution Commentary: The following is an adaptation of the "nice" commentary provided by Harris Shultz (CSU Fullerton), who enjoys both sharing and exploring problems.

Two spoons are randomly selected from a bag containing r red and y yellow spoons (WLOG assume r>w). Then, [r+y]C2 is the number of possible combinations of two spoons that can be withdrawn.

Then, Prob(mixed) = [ry]/[[r+y]C2] and Prob(matched) = [rC2 + yC2]/[[r+y]C2]. You may need to do some playing with the instances of r=2, y=1 or r=2, y=2 or r=3, y=1...in order to see why these relationships make sense. Remember, if there are r red spoons and y yellow spoons, then the possible number of mixed pairs on a draw is ry, while the number of matched pairs would be [rC2 + yC2]. Also, you should confirm the relationship by showing that [[r+y]C2] = rw + rC2 + yC2.

Now, since we want Prob(mixed) = Prob(matched), we know that ry = rC2 + yC2, or ry = [r(r-1)]/2 + [y(y-1)]/2.

With some algebra play, this last expression produces the relationship r+y = (r-y)2. Now the odd twist by Harris...define n = r-y, making r+y = n2.

But, then r = [r+y+r-y]/2 = [n2+n]/2 = [n(n+1)]/2 and y = [r+y-(r-y)]/2 = [n2-n]/2 = [n(n-1)]/2.

Why is this result so special....because it states that in very fair situation involving r red spoons and y yellow spoons, r and y must be consecutive triangular numbers. That is, r and y are adjacent pairs taken from the set 1, 3, 6, 10, 15, 21, ..... Polly and Pepso know that r=3 and y=1 is fair...but then r=15 and y=10 would also be fair, making this a good use of a pile of plastic spoons.


Note: A nice follow-up question. In every situation so far, the Prob(mixed) either equalled or exceeded Prob(matched). Can you find an instance of when Prob(Matched) > Prob(mixed). After some exploration, you should relook at Part 1's solution...and say Dud! It's obvious!