In a production of Hamlet, several roles are paired, for example, Gertrude and the Queen in the Play can be played by two actors. Before the performance, they decide by drawing lots which of the two will play the Queen in the Play and which will play Gertrude on that day. Similar lotteries decide for the other pairs as well. Sári has already seen one performance, but she wants to see the Gertrude/Queen in the Play, Claudius/Player King, and Ophelia/Fortinbras roles in the other version as well, although not necessarily in the same performance. How many more performances should she buy tickets for so that there is at least a $90 \%$ probability that she will see all three roles in the other casting as well?
I. solution. The probability that in $n$ performances, the person does not see the role pairing from a pair of roles that they like is $\frac{1}{2^{n}}$, because there are $2^{n}$ ways to assign the roles, and only one case occurs where the role they have already seen is always drawn. Therefore, the probability that they see the desired pairing at least once is $1-\frac{1}{2^{n}}$. Since there are three pairings (and their drawing is independent), the probability that they see all three in the desired role pairing is $\left(1-\frac{1}{2^{n}}\right)^{3}$. This probability must be at least $90\%$: $\left(1-\frac{1}{2^{n}}\right)^{3} \geq 0.9$. Rearranging and solving this:

$$
\begin{aligned}
1-\frac{1}{2^{n}} & \geq \sqrt[3]{0.9} \\
1-\sqrt[3]{0.9} & \geq \frac{1}{2^{n}} \\
2^{n} & \geq \frac{1}{1-\sqrt[3]{0.9}} \approx 28.9766 \\
n & \geq 5
\end{aligned}
$$

Thus, they need to attend 5 more performances.

II. solution. Again, we calculate the probability that in $n$ performances, the person does not see the role pairing from any pair of roles that they like. In this case, one of the role pairings is constant. There are three pairs, so the probability of this is $3 \cdot\left(\frac{1}{2}\right)^{n}$. However, this double-counts the cases where two pairs remain constant, so we need to subtract the probability of this. But then we also subtract the cases where all three pairs remain constant, so we need to add this back:

$$
p=3 \cdot\left(\frac{1}{2}\right)^{n}-3 \cdot\left(\left(\frac{1}{2}\right)^{2}\right)^{n}+\left(\left(\frac{1}{2}\right)^{3}\right)^{n}
$$

Let $\left(\frac{1}{2}\right)^{n}=x$, then

$$
1-\left(3 x-3 x^{2}+x^{3}\right) \geq 0.9, \quad(1-x)^{3} \geq 0.9
$$

From here, the calculations already seen in the first solution follow.