Five players play ping-pong matches according to the following plan. They write down all possible pairings on identical slips of paper, put the slips into a hat, and before each match, they draw a slip, and the pair on the slip plays. The drawn slip is not returned to the hat.

- What is the probability that the first $N(3 \leq N \leq 10)$ matches will be such that at least 3 players have already played their matches against each other?
We denote the sought probability by $p_{N}$. Since the number of possible pairings of 5 players is $\binom{5}{2}=10$, 10 cards were prepared. Out of these, for the first $N$ games,

$$
V_{N}=10 \cdot 9 \cdot \ldots \cdot(11-N)
$$

ways exist to choose the cards. We denote by $K_{N}$ the number of selections in which all three possible pairings of some three players appear. Then, since the individual selections are equally likely, $p_{N}=K_{N} / V_{N} \cdot-$

The schedule for the first $N$ games can be prepared by first selecting $N$ cards at once, and then arranging the selected cards in a sequence. In the selection step, we choose one out of $v_{N}=\binom{10}{N}$ possibilities, and for each selection, $N!$ different sequences can be made. (Indeed, $V_{N}=v_{N} \cdot N!$). The favorability of a game series does not depend on the order of the games, so if we denote by $k_{N}$ the number of selections (regardless of the order of the games) in which all three possible pairings of some three players appear, then $K_{N}=k_{N} \cdot N!$ and

$$
p_{N}=\frac{K_{N}}{V_{N}}=\frac{k_{N}}{v_{N}}
$$

where, as we have seen, $v_{N}=\binom{10}{N}$, so it is sufficient to determine the value of $k_{N}$.

If $N \geq 6$, then

$$
\begin{aligned}
& p_{6}=1-\frac{\binom{5}{3}}{\binom{10}{4}}=1-\frac{1}{21}=\frac{20}{21} \\
& p_{7}=p_{8}=p_{9}=p_{10}=1
\end{aligned}
$$

As a second solution, we provide a "conversational diagram" without any text. We hope that the interpretation of the diagram will be instructive for the readers.

![](https://cdn.mathpix.com/cropped/2024_05_02_01222ef2d07198d16ca0g-2.jpg?height=1273&width=1185&top_left_y=1109&top_left_x=470)