7. Mothers' Voting (9-11). In the final of the play competition for March 8, two plays made it to the final. The first play featured $n$ students from 5th grade A, and the second play featured $n$ students from 5th grade B. At the play, there were $2 n$ mothers of all $2 n$ students. The best play is chosen by the mothers' voting. It is known that exactly half of the mothers vote honestly for the best play, while the other half always vote for the play in which their child participates.

a) (2 pts.) Find the probability that the best play will win by a majority of votes.

b) (1 pt.) The same question, if more than two plays made it to the final.
# Solution

a) Let's call a mother honest if she votes for the best play in any case. If a mother votes for her child, we will consider her dishonest. It is known that $n$ honest mothers will vote for the best play, so it will not lose for sure (half of the votes are definitely for it). The only case when the best play does not win is a tie, when the worst play also gets half of the votes. This is only possible if all the children of dishonest mothers are in the worst play. Let's find the probability of this event.

There are $C_{2 n}^{n}$ cases of distributing $n$ dishonest mothers among the total number of $2 n$ mothers. And only in one of these cases do all dishonest mothers vote for the worst play. Thus, the probability of a tie is $p=\frac{1}{C_{2 n}^{n}}=\frac{(n!)^{2}}{(2 n)!}$. Therefore, the probability of the event “the best play will get more votes” is $1-p=\frac{(2 n)!-(n!)^{2}}{(2 n)!}$.

Remark. If we assume that $n$ is large, then the factorial $n!$ can be calculated using the approximate Stirling's formula: $n!\approx \sqrt{2 \pi n} \cdot n^{n} e^{-n}$. Let's see what we get if we approximately calculate the probability of a tie using this formula:

$$
p=\frac{(n!)^{2}}{(2 n)!} \approx \frac{2 \pi n \cdot n^{2 n} e^{-2 n}}{\sqrt{2 \pi \cdot 2 n}(2 n)^{2 n} e^{-2 n}}=\frac{\sqrt{\pi n}}{4^{n}} \text {. }
$$

For example, if there were 20 actors in each play, then the approximate calculation would already be justified: $p \approx \frac{\sqrt{20 \pi}}{4^{20}} \approx 7.2 \cdot 10^{-12}$. As we can see, in this case, there is no hope for a tie.

b) Let there be $s>2$ plays. In this case, the best play wins for sure - it gets at least half of the votes, one from each honest mother. The votes of the other mothers are divided among the other plays, each of which gets no more than $\frac{1}{s}<\frac{1}{2}$ of all the votes.