16. Draws (from 9th grade. 6 points). Two hockey teams of equal strength agreed to play until the total score reaches 10. Find the mathematical expectation of the number of times a draw occurred.
Solution. If $2n$ is the maximum total score, then the game can be viewed as a random walk of length $2n$: at each step, the score difference either increases or decreases by one.

![](https://cdn.mathpix.com/cropped/2024_05_06_853fe0fb72821c0a3a48g-18.jpg?height=394&width=596&top_left_y=1596&top_left_x=1324)

Let $I_{2k}$ be the indicator of a tie at the $2k$-th step:

$$
I_{2k} \sim \left(\begin{array}{cc}
0 & 1 \\
1-\frac{C_{2k}^{k}}{2^{2k}} & \frac{C_{2k}^{k}}{2^{2k}}
\end{array}\right)
$$

The random variable $X$ "the number of times a tie occurred" is the sum of all indicators. The start of the game will not be considered as a "tie occurrence." Therefore, $X = I_{2} + I_{4} + \ldots + I_{2n}$. This gives the solution:

$$
\mathrm{E} X = \sum_{k=1}^{n} \mathrm{E} I_{2k} = \sum_{k=1}^{n} \frac{C_{2k}^{k}}{2^{2k}}
$$

IX Olympiad in Probability Theory and Statistics for Schoolchildren. Correspondence Round.

© Laboratory of Probability Theory and Statistics, MCCME, 2015-2016

For $n=5$, we get: $\mathrm{E} X = \frac{2}{4} + \frac{6}{16} + \frac{20}{64} + \frac{70}{256} + \frac{252}{2048} \approx 1.707$.

Answer: approximately $1.707$.

Remark. If we use Stirling's formula $n! \approx \sqrt{2 \pi n} (n / e)^{n}$, then after transformations, we get:

$$
\mathrm{E} X \approx \sum_{1}^{n} \frac{1}{\sqrt{\pi k}} = \frac{1}{\sqrt{\pi}} \sum_{1}^{n} \frac{1}{\sqrt{k}}
$$

For $n=5$, this approximate formula gives 1.823 (all digits are correct). Even for such a small $n$, the relative error is about $4\%$, and it will decrease with increasing $n$.

Further simplification (also approximate) can be made by replacing the sum with a suitable integral. For example:

$$
\frac{1}{\sqrt{\pi}} \sum_{1}^{n} \frac{1}{\sqrt{k}} \approx \frac{1}{\sqrt{\pi}} \int_{0}^{n} \frac{1}{\sqrt{x + \pi / 4}} \, dx = \frac{2}{\sqrt{\pi}} \left( \sqrt{n + \pi / 4} - \frac{\sqrt{\pi}}{2} \right) = \sqrt{\frac{4n}{\pi} + 1} - 1
$$

For large $n$, the 1 under the square root is almost negligible, so it can be dropped, yielding an even simpler formula:

$$
\mathrm{E} X \approx 2 \sqrt{\frac{n}{\pi}} - 1
$$

Verify using a computer that both of these formulas give an acceptably small error.

[^0]: ${ }^{1}$ An octahedron is an eight-faced polyhedron with six symmetrically located vertices.

    IX Olympiad in Probability Theory and Statistics for Schoolchildren. Correspondence Round.

    © Laboratory of Probability Theory and Statistics, MCCME, 2015-2016

[^1]: 2 The law is named after the American linguist George Kingsley Zipf, although it was first described long before Zipf by the French stenographer Jean-Baptiste Estou in the work "The Range of Stenography" in 1908. Justifications for Zipf's law appeared recently, in the 21st century, but it is still unclear whether Zipf's law is more based on the nature of the quantities it governs or on the properties of random sequences.