Original Problem
Scott and Letitia determined who would wash the dishes and who would
dry by having Scott pull two spoons out of a bag.  The bag contained
two purple spoons and three green ones.  Of the two spoons Scott
pulled out were the same color, then Scott washed and Letitia dried. 
If the two were different colors, then Letitia washed and Scott dried.

New Problem
Letitia decided she didn't like that system, because it turned out
that she washed the dishes about 60 percent of the time.  Scott
thought that if they found the right number of spoons of each color to
put in the bag, they could make the probability of a match equal to 50
percent.  Find out as much about their choices as you can.

Dad needs help!  Any help would be appreciated.  Thanks.

Hello callawar,

Three spoons of one color and one spoon of a different color would
lead to a 50% probability that Scott would draw matching colors.

Say that Scott and Letitia put in three purple spoons and one green
spoon.  We could call them P1, P2, and P3 ("P" for purple) and G1 ("G"
for green).

If Scott drew P1 first, he would have 2 chances of drawing the same
color as P1(P2 or P3) versus 1 chance of drawing a different color
(G1).

The same would be true if Scott instead drew P2 first: 2 chances of
the same color (P1 or P3), versus 1 chance of a different color (G1).

And likewise for P3: 2 chances of the same color (P1 or P2), versus 1
chance of a different color (G1).

But if Scott selected G1 first, there would be 0 chances of selecting
the same color, versus 3 chances of selecting a different color (P1,
P2, or P3).

In summary, Scott could draw two spoons in the following orders:

1) P1 - P2
2) P1 - P3
3) P1 - G1
4) P2 - P1
5) P2 - P3
6) P2 - G1
7) P3 - P1
8) P3 - P2
9) P3 - G1
10) G1 - P1
11) G1 - P2
12) G1 - P3

There are 6 possibilities of different colors (3, 6, 9, 10, 11, 12 in
this list) and 6 possibilities of the same color (1, 2, 4, 5, 7, 8 in
the list).
Since there are 6 possibilities of "different color" and 6
possibilities of "same color," both are equally likely.

Incidentally, here is the list for three purple spoons and two green spoons:  

1) P1 - P2
2) P1 - P3
3) P1 - G1
4) P1 - G2
5) P2 - P1
6) P2 - P3
7) P2 - G1
8) P2 - G2
9) P3 - P1
10) P3 - P2
11) P3 - G1
12) P3 - G2
13) G1 - P1
14) G1 - P2
15) G1 - P3
16) G1 - G2
17) G2 - P1
18) G2 - P2
19) G2 - P3
20) G2 - G1

There are 12 possibilities of "different color" (3, 4, 7, 8, 11, 12,
13, 14, 15, 17, 18, 19) and 8 possibilities of "same color" (1, 2, 5,
6, 9, 10, 16, 20), which explains why Letitia was washing the dishes
60% of the time (12 chances out of 20).

I hope that this information is helpful.  Please let me know if you
need any clarification.

- justaskscott (unrelated to the Scott in this question!)

---

Clarification of Answer by justaskscott-ga on 25 May 2005 10:38 PDT

After considering crythias's alternative answer, it struck me that the
two answers thus far involved 4 spoons (2 squared) and 9 spoons (3
squared).  I wondered whether 16 spoons (4 squared) would work.

And indeed, 10 spoons of one color and 6 spoons of another color
result in a 50% probability of selecting two same-color spoons.  This
is shown by substituting the relevant number into the calculation
below:

There will be 16*15=240 total ways to pull 2 spoons.
There will be 10*9=90 ways to pull 2 purple spoons.
There will be 6*5=30 ways to pull 2 green spoons.
(90+30)/240=1/2=50%

So here is the pattern:

4: 1 + 3
9: 3 + 6
16: 6 + 10

Notice that the high number for one set of spoons becomes the low
number for the next set of spoons.

The pattern works for 5 squared and 6 squared -- again, you'll find a
50% probability of same-color spoons.

25: 10 + 15
36: 15 + 21

I presume that this pattern continues successfully for the square of 7, 8, 9 ....

Thanks for your help!

What's interesting about this problem is that the "intuitive" answer
-- 2P and 2G -- isn't 1/2...

HOWEVER, the answer posed an interesting paradox, if you will. If it
is *known* that only one green spoon exists in the bag of 4, the
actual number of draws is reduced. There is zero chance of getting a
pair if green is drawn first. This can be seen as removing some of the
available draws, -- what is the point of drawing the second time if
green is chosen first? -- leaning toward 4 invalid draws and 6 valid
draws.

6P and 3G work:
There will be 9*8=72 total ways to pull 2 spoons.
There will be 6*5=30 ways to pull 2 purple spoons.
There will be 3*2=6 ways to pull 2 green spoons.
This is (30+6)/72=1/2=50%

Thanks, crythias!  I agree with that six spoons of one color and three
spoons of a different color is another correct answer.

Here's another possibility...
I was throwing some numbers around and 
The first column is the number of one color of spoons (or, all
spoons). The second column is the number of ways to draw two spoons
based upon the first column (column 1)*[(column 1) minus 1] or (column
1)P2, where #P2 is the number of permutations that 2 can be chosen
from # spoons.
The Third column is half (50%) of the second column. It hit me that
any consecutive two numbers in the third column can answer the
problem. We've already demonstrated  that (1 and 3) and (3 and 6)
work.  Column 2 for 6 and 10 sum to 120, which is Column 3 value for
(6+10=16).

#  #P2  half
2  2    1
3  6    3
4  12   6
5  20   10
6  30   15
7  42   21
8  56   28
9  72   36
10 90   45
11 110  55
12 132  66
13 156  84
16 240  120

Ack! I've duplicated justaskscott-ga's answer (much easier to follow).

Note that these are the so-called "triangular" numbers, the ones you
get by adding up the first n positive integers (1, 3=1+2, 6=1+2+3,
etc.) The general form of these is n(n+1)/2.

Here's the analysis of why these numbers work and no others:

Suppose we have A spoons of one colour and B spoons of another colour.
There are (A+B) (A+B-1) different choices we can make (including
order). Of these, there are A (A-1) + B (B-1) draws with the same
colour and AB + BA draws with different colours. We want these values
to be equal.

So we want A^2 - A + B^2 - B = 2AB (using ^2 to indicate "squared").
We can rewrite this as
B^2 - (2A+1) B + (A^2-A) = 0, looking at it as a quadratic equation in B.

This equation has the solution
B = [(2A+1) +/- sqrt ((2A+1)^2 - 4(A^2-A))] / 2
  = [(2A+1) +/- sqrt (4A^2 + 4A + 1 -4A^2 + 4A)] / 2
  = [(2A+1) +/- sqrt (8A+1)] / 2

Now for this to be an integer we need sqrt (8A+1) to be an odd
integer, i.e. 8A+1 must be the square of an odd integer. The general
formula for an odd integer is 2k+1, where k is any integer; in this
case 8A+1 is greater than 1 (since A>0) so k > 1 also.

So let 8A+1 = (2k+1)^2 = 4k^2 + 4k + 1. Then A = (k^2+k)/2 = k(k+1)/2.
This is a triangular number in the form given (substitue k for n).

Also sqrt(8A+1) = 2k+1, so for B we get
B = [(2A+1) +/- sqrt (8A+1)] / 2
  = [(k^2 + k + 1) +/- (2k + 1)] / 2
  = (k^2 - k) / 2 or (k^2 + 3k + 2) / 2
  = (k-1)k / 2 or (k+1)(k+2)/2.
These answers are respectively the next lowest and the next highest
triangular number after A (replace n by (k-1) and by (k+1) in the
original formula to see this). So in general the only answer that will
work is a consecutive pair of triangular numbers, and any such pair
will work.

Also note that consecutive pairs of triangular numbers sum to a square
number (and that all squares can be produced this way), as
justaskscott noticed:

Take the (k-1)-th and k-th triangular numbers and add them together:
(k-1)k/2 + k(k+1)/2
= (k^2 - k + k^2 + k) / 2
= (2k^2)/2
= k^2.