Solution to the Red and Green Hats (Feb. 13th edition)

A box contains two red hats and three green hats. Azalea, Barnaby and Caleb close their eyes, take a hat from the box and put it on. When they open their eyes they can see each other’s hats but not their own. They do not know which hats are left in the box.

We can assume that all the protagonists are perfect logicians who tell the truth. They know all the information in the above paragraph. In addition, one of them is colourblind. They all know who the colourblind person is.

  • Azalea says: “I don’t know the colour of my hat.”
  • Barnaby says: “I don’t know the colour of my hat.”
  • Caleb says: “I don’t know the colour of my hat.”
  • Azalea says: “I don’t know the colour of my hat.”

Who is the colourblind person, and what colour is their hat?

First consider what would happen if no one was colourblind.

The only way that Azalea could know the colour of her hat was that if both Barnaby and Caleb are wearing red hats, since she would deduce all reds are taken and so her hat must be green.

Since Azalea doesnt know the colour of her hat, we can deduce that she is not seeing two reds on the other two, so she is seeing either two greens, or a red and a green.

Barnaby knows all this. So if Barnaby were to see that Caleb has a red hat, he would be able to deduce that he is wearing green, since he knows Azalea can only see one red. Since Barnaby doesn’t know the colour of his own hat we can eliminate the possibility that Caleb has a red hat.

In other words, once both Azalea and Barnaby have said they don’t know their hats, everyone (including Caleb) knows that Caleb has a green hat.

Now let’s assume that one of them is colourblind.

Already, we know that Caleb cannot be the colourblind one, since if Azalea and Barnaby can see colours, then following the above logic Caleb – even if colourblind – will know the colour of his own hat.

Equally, Azalea cannot be colourblind, since if she was, her first statement is meaningless – of course she doesn’t know the colour of her own hat, she’s colourblind! Then, once Barnaby and Caleb say they don’t know the colour of their hats, everyone (including colourblind Azalea) knows that Azalea has a green hat, following the logic explained above, so she would not say for a second time that she does not know the colour of her hat.

So the colourblind person is Barnaby and – since the other two people don’t know the colour of their hats – he must have a green hat.

Puzzle 2:

Homero is clutching three identical pieces of string in his fist, as illustrated below left. He asks Sofia to tie two ends of the string, chosen at random, at either side of his fist, as illustrated below centre, so that there is one free end at either side.

When Sofia ties two ends either side there are two possibilities: the string is now all joined together in one piece, or one piece is disconnected from the other two.
 When Sofia ties two ends either side there are two possibilities: the string is now all joined together in one piece, or one piece is disconnected from the other two. Illustration: Brazilian maths olympiad

1) What is the probability that all the pieces are joined in one long piece?

Solution

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After Sofia ties two ends on one side, we are left with the situation illustrated left. When Sofia ties two ends on the other side, either she joins A and B, B and C, or A and C. In two out of the three cases, the string is joined in one long piece, so the answer is 2/3.

Now Homero is clutching five identical pieces of string in his fist, in the same way as above. On each side of his fist there are five ends. On each side Sofia ties together free pieces of string, leaving a single end free either side. (So she makes four knots in total.)

2) What is the probability that one piece of string is not connected to any other?

Solution

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After Sofia ties the ends on one side, she is left with the situation illustrated left. It gets a bit tedious working out all the combinations of how to join A, B, C, D and E, so consider the situation in a different way. For each combination of joining the ends, one end is free. The only way for there to be a piece of string not connected to any other is if the end that is free is E. Since the chance of E being free is 1 in 5, the answer to the question is 1/5.

3) What is the probability that all the pieces are joined in one long piece?

From the previous answer, if the chance that E is free is 1/5, then the chance that E is not free, meaning it joins to either A,B, C or D is 4/5. Let’s say E joins to A. There are now three possibilities of the other join: B to C, C to D or B to D. In two of these cases the string is joined in one long piece. We get the same fraction of 2/3 if E was joined to any other end. So the probability of the string being in one long piece is 4/5 x 2/3 = 8/15.

Extra problem: what happens when there are 6, 7, …and pieces of string? Once you tie the ends on each side (leaving one end free if is odd), what is the chance all the strings are joined together? Please show your workings for others below the line!