13. Where is the deviation greater? We will toss a symmetric coin $n$ times. Suppose that heads appeared $m$ times. The number $m / n$ is called the frequency of heads. The number $\frac{m}{n}-0.5$ is called the deviation of the frequency from the probability, and the number $\left|\frac{m}{n}-0.5\right|$ is called the absolute deviation. Note that the deviation and the absolute deviation are random variables. For example, if the coin is tossed 5 times and heads appear twice, then the deviation is $\frac{2}{5}-0.5=-0.1$, and the absolute deviation is $|-0.1|=0.1$.

The experiment consists of two parts: first, the coin is tossed 10 times, and then 100 times. In which of these cases is the expected value of the absolute deviation of the frequency of heads from the probability greater?
Solution. Let the number of heads in a series of 10 tosses be $m_{10}$. Then the deviation of the frequency from the probability is $|\alpha|$, where $\alpha=\frac{m_{10}}{10}-0.5$. Let $\mathrm{E}|\alpha|=a$.

The second sequence, consisting of 100 tosses, can be divided into 10 series of ten tosses each. Each of these 10 small series has its own deviation (non-absolute) $\alpha_{i}$ (where $i$ is the series number), and $\mathrm{E}\left|\alpha_{i}\right|=\mathrm{E}|\alpha|=a$.

Then the deviation in the sequence of 100 tosses is

$$
\beta=\frac{m_{100}}{100}-0.5=\frac{10 \alpha_{1}+10 \alpha_{2}+10 \alpha_{3}+\ldots+10 \alpha_{10}}{100}
$$

Therefore,

$$
|\beta|=\frac{1}{10}\left|\alpha_{1}+\alpha_{2}+\ldots+\alpha_{10}\right| \leq \frac{1}{10}\left(\left|\alpha_{1}\right|+\left|\alpha_{2}\right|+\ldots+\left|\alpha_{10}\right|\right)
$$

Consider the random variable $X=\left|\alpha_{1}\right|+\left|\alpha_{2}\right|+\ldots+\left|\alpha_{10}\right|-10|\beta|$, which is non-negative and equals zero only if all deviations $\alpha_{i}$ have the same sign. The probability of this is very small (at least less than one), so $\mathrm{E} X>0$. Then

$$
0<\mathrm{E} X=\mathrm{E}\left(\left|\alpha_{1}\right|+\left|\alpha_{2}\right|+\ldots+\left|\alpha_{10}\right|\right)-10 \mathrm{E}|\beta|=10 a-10 \mathrm{E}|\beta|
$$

Thus,

$$
\mathrm{E}|\beta|<a
$$

Answer: The expected absolute deviation in a series of 10 tosses is greater than in a series of 100 tosses.