17. The Paradox of Three Coins ${ }^{5}$ (from 9th grade, 4 points). Mitya, Vanya, and Dima each flipped a coin $2 n$ times. It is known that Mitya and Vanya got the same number of heads. What is more likely: event $A$ "Dima got exactly $n$ heads" or event $B$ "Mitya got exactly $n$ heads"? If the events are not equally likely, by what factor is one of these events more likely than the other?
Solution. The probability that Mitya and Vanya have the same number of heads is ${ }^{6} C_{4 n}^{2 n} / 2^{4 n}$. Clearly, $\mathrm{P}(A)=C_{2 n}^{n} / 2^{2 n}$. Let's find $\mathrm{P}(B)$:

$$
\mathrm{P}(B)=\mathrm{P}(\text { Mitya has } n \text { heads } \mid \text { Mitya and Vanya have the same number of heads })=
$$

$$
=\frac{\mathrm{P}(\text { Mitya has } n \text { heads and Vanya has } n \text { heads })}{\mathrm{P}(\text { Mitya and Vanya have the same number of heads })}=\left(C_{2 n}^{n} \cdot \frac{1}{2^{2 n}}\right)^{2}:\left(C_{4 n}^{2 n} \cdot \frac{1}{2^{4 n}}\right)=\frac{\left(C_{2 n}^{n}\right)^{2}}{C_{4 n}^{2 n}}
$$

It remains to compare the found probabilities:

$$
\begin{aligned}
\frac{P(B)}{P(A)}= & \frac{C_{2 n}^{n} \cdot 2^{2 n}}{C_{4 n}^{2 n}}=\frac{((2 n)!)^{3} \cdot 2^{2 n}}{(n!)^{2}(4 n)!}=\frac{[(n+1)(n+2) \ldots 2 n]^{2} \cdot 2^{2 n}}{(2 n+1)(2 n+2) \cdot \ldots \cdot 4 n}=\frac{[(2 n+2)(2 n+4) \cdot \ldots \cdot 4 n]^{2}}{(2 n+1)(2 n+2) \cdot \ldots \cdot 4 n}= \\
= & \frac{(2 n+2)(2 n+4) \cdot \ldots \cdot 4 n}{(2 n+1)(2 n+3) \cdot \ldots \cdot(4 n-1)}=\left(1+\frac{1}{2 n+1}\right)\left(1+\frac{1}{2 n+3}\right) \cdot \ldots \cdot\left(1+\frac{1}{4 n-1}\right)>1
\end{aligned}
$$

The found expression shows how many times event $B$ is more likely than event $A$. Using Stirling's formula, we can find an approximate value:

$$
\frac{((2 n)!)^{3} \cdot 2^{2 n}}{(n!)^{2}(4 n)!} \approx \frac{\left(\sqrt{2 \pi \cdot 2 n}(2 n / e)^{2 n}\right)^{3} 2^{2 n}}{\left(\sqrt{2 \pi n}(n / e)^{n}\right)^{2} \cdot \sqrt{2 \pi \cdot 4 n}(4 n / e)^{4 n}}=\sqrt{2}
$$

Answer: The probability of event $B$ is greater than the probability of $A$ by $\frac{C_{2 n}^{n} \cdot 2^{2 n}}{C_{4 n}^{2 n}}(\approx \sqrt{2})$ times.