\section*{Exercise 4 - 051214}

Klaus and Dieter agree on the following game:

Klaus takes 6 threads of the same length in one hand, so that six thread ends protrude from each side of the fist. Dieter is invited to tie the ends on each side in pairs. If it turns out that the threads form a single ring when the hand is opened, Dieter wins; otherwise, Klaus wins.

Who of the two has the better chances of winning? Consider the following:

a) How many different ways \(m\) are there to tie the thread ends together?

b) In how many cases \(r\) does a single ring result?

c) What is the probability \(w\) that a single ring will form?

Note: \(w\) is defined as \(\frac{r}{m}\), where \(m\) and \(r\) are explained in a) and b).
}

a) Let the ends be connected on one side according to the problem statement, e.g., w.l.o.g. 1 with 2, 3 with 4, 5 with 6. On the other side, the first end has 5 possible connections. Then, one of the four free ends can be connected to any of the three other ends. For the remaining two ends, only one possibility remains. The number of possible cases is thus \(n=5 \cdot 3 \cdot 1=15\).

b) Connection on one side as in a). A single ring can only result if, on the other side, 1 is connected to 3, 4, 5, or 6.

Assume 1 is connected to 3. In this case, a single ring is formed exactly when 2 is connected to 5 or 6. Correspondingly, in each of the other cases, namely when 1 is connected to 4, 5, or 6, there are exactly two ways to form a single ring. The number of favorable cases in which a single ring is formed is thus \(r=4 \cdot 2=8\).

c) The probability is

\[
w=\frac{r}{n}=\frac{8}{15}=0.533 \ldots
\]

Therefore, Dieter has the better chances of winning.

Adapted from \([5]\)

\subsection*{9.7.2 Second Round 1965, Grade 12}