# To Mock A Mockingbird: Flower Garden (Math Puzzle)

Question:

In a certain flower garden, each flower was either red, yellow, or blue, and all three colours were represented. A statistician once visited the garden and made the observation that whatever three flowers you picked, at least one of them was bound to be red. A second statistician visited the garden and made the observation that whatever three flowers you picked, at least one was bound to be yellow. Two logic students heard about this and got into an argument. The first student said: "It therefore follows that whatever three flowers you pick, at least one is bound to be blue, doesn't it?" The second student said: "Of course not!"

Which student was right, and why?

A few sample Questions from To Mock a Mocking Bird can be found here.

The first student was right, and here is why.

Problem

I have to disagree with the authors answer. You can satisfy both the first and second statistician with combinations of two yellow and one red (YYR) or two red and one yellow (RRY)

To generate all 27 possibilities I wrote a quick Haskell list comprehension:

--ghci version
--let r = 'R'; y = 'Y'; b = 'B'; fl = r:y:b:[] in putStr $unlines [z|z<-[i:j:k:[]|i<-fl,j<-fl,k<-fl]] let r = 'R' y = 'Y' b = 'B' fl = r:y:b:[] in putStr$ unlines [i:j:k:[]|i<-fl,j<-fl,k<-fl]


The list of all 27 possibilities is:

RRR
RRY
RRB
RYR
RYY
RYB
RBR
RBY
RBB
YRR
YRY
YRB
YYR
YYY
YYB
YBR
YBY
YBB
BRR
BRY
BRB
BYR
BYY
BYB
BBR
BBY
BBB


But if I filter where at least one flower is yellow and least one flower is red I get:

--ghci version
--let r = 'R'; y = 'Y'; b = 'B'; fl = r:y:b:[] in  putStr $unlines [i:j:k:[]|i<-fl,j<-fl,k<-fl,i==r||j==r||k==r,i==y||j==y||k==y] let r = 'R' y = 'Y' b = 'B' fl = r:y:b:[] in putStr$ unlines [i:j:k:[]|i<-fl,j<-fl,k<-fl,
i==r||j==r||k==r,
i==y||j==y||k==y]


This list of 12 possibilities including at least one red and yellow are:

RRY
RYR
RYY
RYB
RBY
YRR
YRY
YRB
YYR
YBR
BRY
BYR


What am I misunderstanding?

I confess that I don't understand what it is you are doing with your list of all possible choices or the filter.

But: While a garden with three flowers, two red and one yellow, would satisfy the first two statistician's observations, it would not satisfy the original conditions of the problem, which requires that the garden contain at least one flower of every color (as it contains no blue flowers). Similarly with a garden with three flowers, two yellow and one red.

We know the garden contains at least one red flower, at least one yellow flower, and at least one blue flower.

The first statistician's observation amounts to saying that the garden cannot contain three or more non-red flowers: because if there were at least three flowers that are not red in the garden, then you could pick those three, and not get any red flowers. But you know there are at least two non-red flowers: at least one yellow, and at least one blue; so there cannot be more than one yellow flower, nor more than one blue flower. That means that you must have exactly one yellow flower, exactly one blue flower, and an indeterminate number of red flowers in the garden.

The second statistician's observation, symmetrically, amounts to saying that the garden cannot contain three or more non-yellow flowers. Buy the same argument, that means that the garden contains exactly one red flower, exactly one blue flower, and an indeterminate number of yellow flowers.

Putting the two observations together tells you that the garden contains exactly three flowers, one of each color. So the first student is correct: picking "any" three flowers (i.e., picking all of the flowers in the garden) guarantees that at least one of the picked flowers is blue.

The list of "all possibilities" are just all possible ways to pick 3 flowers, assuming that there are at least three of each color. This doesn't really tell you much. Likewise, your second list seems to be all possible ways in which you can pick three flowers, with at least one yellow and at least one red, assuming that there are "enough" flowers of each color (where "enough" means here 'at least two of each color'). But in the "actual garden" not all these possibilities need be possible; indeed, a garden with exactly one blue, exactly one red, and exactly one yellow flower only allows for one possible way to pick three flowers, RYB, and the other "abstract possibilities" are irrelevant.

• I was interpreting as "If I pick any three flowers at random" – snmcdonald Aug 25 '11 at 19:18
• @snmcdonald: I still don't understand. What the first statistician is saying is that any choice of three flowers from among those in the garden will always include a red flower; you can, if you wish, interpret this as a "probability": the probability that, if you pick three flowers at random from among those in the garden, you get at least one red flower is $1$. But the key here ...from among those in the garden... – Arturo Magidin Aug 25 '11 at 19:21
• @snmcdonald: Also, the statisticians are not describing a single sample taken from the garden, but a property that all samples taken from the garden share. – Arturo Magidin Aug 25 '11 at 19:23
• That most recent comment is the main misunderstanding. – The Chaz 2.0 Aug 25 '11 at 19:27
• Sorry but I still don't get this. This is how I understand the question: There is a garden that contains flowers. There is at least one red, one blue and one yellow flower in that garden. Now if you pick any three flowers at any time then there must be a yellow and a red among them. How does that force the last flower to be blue. How did you conclude that the garden contains only three flowers ? Couldn't it have more than one flower and thus the original OP question ? Thank you. – Ibrahim Najjar Jun 29 '15 at 13:01

I think you're simply missing the part where it says "all three colours were represented" — i.e., there is at least one red flower, one blue flower, and one yellow flower in the garden. If the garden consisted of two reds and one yellow (or vice versa), it would satisfy both statisticians' statements but not the original description of the garden.

• ouch, was it that obvious?! – snmcdonald Aug 25 '11 at 19:14