The math teacher suggested changing the voting scheme for the play competition (see problem 65299). In her opinion, out of all $2 n$ mothers, a jury of $2 m$ people should be randomly selected $(2 m \leq n)$. Find the probability that the best play will win under these voting conditions.

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Let's call a mother honest if she votes for the best play.

Since there are $n$ dishonest mothers and as many honest ones, the probability that $q$ dishonest mothers and $2 m-q$ honest ones end up in the jury is $\frac{C_{n}^{q} C_{n}^{2 m-q}}{C_{2 n}^{2 m}}$.

If this event occurs, the fate of the competition depends on the distribution of votes from the dishonest mothers. If $t$ children of dishonest mothers are in the worst play, then this worst play will receive exactly $t$ votes, while all other votes, totaling $2 m-t$, will be given to the best play. Thus, the best play wins if $2 m-t > t$, which means if $t < m$. The probability that the worst play receives exactly $t$ votes is equal to the probability $\frac{C_{q}^{t} C_{2 n-q}^{n-t}}{C_{2 n}^{n}}$ that exactly $t$ children of dishonest mothers in the jury are in this play.

Therefore, the probability of the combination where $q$ dishonest mothers end up in the jury, and the worst play receives fewer votes than the best, is $\frac{C_{n}^{q} C_{n}^{2 m-q}}{C_{2 n}^{2 m}} \cdot \sum_{\substack{t=0, \ldots, q \\ t<m}} \frac{C_{q}^{t} C_{2 n-q}^{n-t}}{C_{2 n}^{n}}$.

Summing these probabilities over all possible distributions of dishonest mothers, we get that the desired probability is

$$
\sum_{q=0}^{2 m}\left(\frac{C_{n}^{q} C_{n}^{2 m-q}}{C_{2 n}^{2 m}} \cdot \sum_{t=0, \ldots, q, t<m} \frac{C_{q}^{t} C_{2 n-q}^{n-t}}{C_{2 n}^{n}}\right)=\frac{1}{C_{2 n}^{n} C_{2 n}^{2 m}} \sum_{q=0}^{2 m}\left(C_{n}^{q} C_{n}^{2 m-q} \cdot \sum_{t=0, \ldots, q, t<m} C_{q}^{t} C_{2 n-q}^{n-t}\right)
$$

## Answer

$\frac{1}{C_{2 n}^{n} C_{2 n}^{2 m}} \cdot \sum_{q=0}^{2 m}\left(C_{n}^{q} C_{n}^{2 m-q} \cdot \sum_{t=0}^{\min (q, m-1)} C_{q}^{t} C_{2 n-q}^{n-t}\right)$.