In a domino game with 45 tiles, we draw 3 tiles one after another. What is the probability that the second tile can be placed next to the first tile, and the third tile can also be attached to either of the first two? How many attempts do we need to make at least so that the probability of the occurrence is at least 0.9?
A domino tile whose both ends are equal is commonly referred to as a "double."

The desired chain can be created in three ways:

a) Neither the first nor the second draw is a double.

b) The first draw is not a double, but the second is.

c) The first draw is a double. (Then the second draw cannot be a double.)

Let the probabilities of the above three cases be denoted by $v_{a}, v_{b}$, and $v_{c}$, respectively. Since the events in a), b), and c) are mutually exclusive, the probability we are looking for is, according to the addition theorem,

$$
v=v_{a}+v_{b}+v_{c}
$$

Let's calculate $v_{a}$. The probability that the first draw is not a double: $\frac{36}{45}$, since among the $C_{9}^{1,2}=\frac{9 \cdot 10}{1 \cdot 2}=45$ domino tiles, there are 9 doubles. If the first tile drawn is not a double, the probability that the second tile drawn is a connectable non-double: $\frac{14}{44}$, because 7-7 non-double tiles can be attached to each end of the first tile.

The probability that the third tile can be attached to the first two tiles thus connected: $\frac{15}{43}$, since 8-8 tiles can be attached to each tile, but one of them can be attached to both sides.

Therefore, according to the multiplication theorem,

$$
v_{a}=\frac{36}{45} \cdot \frac{14}{44} \cdot \frac{15}{43}=\frac{2 \cdot 7 \cdot 3}{11 \cdot 43}=\frac{42}{473}
$$

With similar reasoning,

$$
\begin{aligned}
& v_{b}=\frac{36}{45} \cdot \frac{2}{44} \cdot \frac{15}{43}=\frac{2 \cdot 1 \cdot 3}{11 \cdot 43}=\frac{6}{473} \\
& v_{c}=\frac{9}{45} \cdot \frac{8}{44} \cdot \frac{15}{43}=\frac{1 \cdot 2 \cdot 3}{11 \cdot 43}=\frac{6}{473}
\end{aligned}
$$

Therefore, the probability we are looking for is

$$
v=\frac{42}{473}+\frac{6}{473}+\frac{6}{473}=\frac{54}{473} \approx 0.114
$$

The probability that the connection does not succeed is, therefore, $1-\frac{54}{473}=\frac{419}{473}$. The probability that it does not succeed in $x$ attempts is $\left(\frac{419}{473}\right)^{x}$, and the probability that it succeeds at least once in $x$ attempts is $1-\left(\frac{419}{473}\right)^{x}$. Therefore, $x$ must be determined such that

$$
1-\left(\frac{419}{473}\right)^{x} \geq 0.9
$$

or

$$
\left(\frac{419}{473}\right)^{x} \leq 0.1
$$

Solving the exponential equation $\left(\frac{419}{473}\right)^{x}=0.1$,

$$
x=\frac{\lg 0.1}{\lg 419-\lg 473}=\frac{1}{\lg 473-\lg 419} \approx 18.9
$$

which implies that at least 19 attempts are needed.

Fuchs Tamás (Bp., II. Rákóczi g. III. o. t.)