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 Then, _{[r+y]}C_{2} is the number of possible combinations of two spoons that can be withdrawn.

Then, Prob(mixed) = [ry]/[_{[r+y]}C_{2}] and Prob(matched) = [_{r}C_{2} + _{y}C_{2}]/[_{[r+y]}C_{2}]. 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 [_{r}C_{2} + _{y}C_{2}]. Also, you should confirm the relationship by showing that [_{[r+y]}C_{2}] = rw + _{r}C_{2} + _{y}C_{2}.

Now, since we want Prob(mixed) = Prob(matched), we know that ry = _{r}C_{2} + _{y}C_{2}, 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 = n^{2}.

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

Wow!

**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!*