136. How many times should a die be thrown so that the probability of the inequality

$$
|m / n-1 / 6| \leqslant 0.01
$$

is not less than the probability of the opposite inequality, where $m$ is the number of times one dot appears in $n$ throws of a die?
Solution. We will use the formula

$$
P\left(\left|\frac{m}{n}-p\right|<\varepsilon\right)=1-2 \Phi\left(\varepsilon \sqrt{\frac{n}{p q}}\right)
$$

According to the condition, the inequality should hold

$$
2 \Phi\left(\varepsilon \sqrt{\frac{n}{p q}}\right) \geqslant 1-2 \Phi\left(\varepsilon \sqrt{\frac{n}{p q}}\right)
$$

or

$$
4 \Phi\left(\varepsilon \sqrt{\frac{n}{p q}}\right) \geqslant 1
$$

From this,

$$
\Phi\left(\varepsilon \sqrt{\frac{n}{p q}}\right) \geqslant 0.25
$$

Using the table in Appendix 2, we find $\Phi(0.67)=0.2486 ; \Phi(0.68)=0.2517$. Performing linear interpolation, we get $\Phi(0.6745)=0.25$.

Considering the relation (*) and taking into account that the function $\Phi(x)$ is increasing, we have

$$
\varepsilon \sqrt{\frac{n}{p q}} \geqslant 0.6745, \text { or } 0.01 \sqrt{\frac{n}{1 / 6 \cdot 5 / 6}}=0.6745
$$

From this, the required number of coin tosses $n \geqslant 632$.