76. a) In some rural areas of Rossin, there once existed the following divination. A girl holds six blades of grass in her hand so that the ends of the blades stick out at the top and bottom; a friend ties these blades of grass in pairs at the top and bottom separately. If, as a result, all six blades of grass are tied into one ring, it was supposed to mean that the girl would get married in the current year. What is the probability that the blades of grass will form a ring when tied at random?

b) Solve the same question for the case of $2 n$ blades of grass.
76. a) Six ends of blades of grass can be paired with each other in $5 \cdot 3 \cdot 1=15$ different ways (the first end can be tied to any of the five remaining; after this, one of the four remaining can be tied to any of the three others; and after this, the two remaining ends can be tied together in only one way). Since the upper and lower ends can be paired independently in 15 different ways, the total number of equally probable outcomes of the experiment is $15 \cdot 15=225$.

Now let's count the number of favorable outcomes. Suppose the upper ends are paired in one of the fifteen possible ways; for example, the end of the first blade is tied to the end of the second, the end of the third to the end of the fourth, and the end of the fifth to the end of the sixth. For a single ring to form, the lower end of the first blade must be tied to the lower end of the third, fourth, fifth, or sixth blade, so we have four possibilities when tying the lower end of the first blade. Further, if, for example, the lower end of the first blade is tied to the end of the third, then the lower end of the second blade must be tied to the end of either the fifth or sixth blade—here we have only two possibilities. After this, only two unconnected ends remain to be tied together. Combining all possibilities, we see that for each of the 15 ways of tying the upper ends, there are exactly $4 \cdot 2=8$ ways of tying the lower ends that lead to a favorable outcome of the experiment; hence, the total number of favorable outcomes is $15 \cdot 8=120$.

Thus, the desired probability is $\frac{15 \cdot 8}{15 \cdot 15}=\frac{8}{15} \approx 0.53$.

b) Reasoning as in the solution to part a), we find that for $2n$ blades of grass, the total number of outcomes of the experiment is $[(2n-1)(2n-3)(2n-5) \ldots 1]^2$, and the number of favorable outcomes is

$$
[(2n-1)(2n-3)(2n-5) \ldots 1][(2n-2)(2n-4) \ldots 2]
$$

Thus, in this case, the probability is given by

$$
p_{n}=\frac{(2n-2)(2n-4) \ldots 2}{(2n-1)(2n-3) \ldots 3 \cdot 1}=\frac{2n-2}{2n-1} \cdot \frac{2n-4}{2n-3} \ldots \frac{2}{3}
$$

Note. It should be noted that the seemingly simple answer to part 76b) becomes very inconvenient for large values of $n$, as in this case, many fractions need to be multiplied. However, using Stirling's formula

$$
n! \approx \sqrt{2 \pi n} n^n e^{-n}
$$

(see problems 160-161 and the text following them on pages 80-81), a simple approximate expression can be obtained for this probability, which is very accurate precisely for large $n$. Indeed, noting that $(2n-2)(2n-4) \ldots 2 = 2^{n-1}(n-1)(n-2) \ldots 1 = 2^{n-1}(n-1)!$, and multiplying the numerator and denominator of the answer to part 76b) by this expression, we find that

$$
\begin{aligned}
p_{n}=\frac{\left[2^{n-1}(n-1)!\right]^2}{(2n-1)(2n-2)!} & \approx \frac{\left[2^{n-1} \sqrt{2 \pi (n-1)} (n-1)^{n-1} e^{-(n-1)}\right]^2}{(2n-1) \sqrt{2 \pi \cdot 2(n-1)} \left[2(n-1)^{2(n-1)} e^{-2(n-1)}\right]} = \\
& = \frac{\sqrt{\pi (n-1)}}{2n-1} \approx \frac{\sqrt{\pi}}{2 \sqrt{n}},
\end{aligned}
$$

since for large $n$

$$
\frac{\sqrt{n-1}}{2n-1} = \frac{\sqrt{n} \sqrt{1-\frac{1}{n}}}{2n \left(1-\frac{1}{2n}\right)} \approx \frac{\sqrt{n}}{2n} = \frac{1}{2 \sqrt{n}}
$$

Here $\pi$, as usual, is the ratio of the circumference of a circle to its diameter: $\pi \approx 3.14$. The error in the approximate formula

$$
p_{n} \approx \frac{\sqrt{\pi}}{2 \sqrt{n}}
$$

will be smaller the larger $n$ is.