On my puzzle blog earlier today I set this question from Brazil’s State School Mathematics Olympiad

*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.*

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

**Solution**

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**

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?*

**Solution**

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 *n *pieces of string? Once you tie the ends on each side (leaving one end free if *n *is odd), what is the chance all the strings are joined together? Please show your workings for others below the line!

I’ll be back with another puzzle in two weeks.

Meanwhile, if you are yearning for more puzzles before then, my mother recommends that you go out and buy several copies of her son’s latest book, **Puzzle Ninja: Pit Your Wits Against the Japanese Puzzle Masters**, which contains more than 200 hand-crafted, joyously addictive logic puzzles from Japan. (Thanks mum). Here’s a story about the book with examples of some of the puzzles. You can buy it here.

*I set a puzzle here every two weeks on a Monday. **Send me your email i**f you want me to alert you each time I post a new one. **I’m always on the look-out for great puzzles. If you would like to suggest one, email me.*

*Thanks to Marco Moriconi for letting me know about the State School Maths Olympiad.*