86. What is the probability that $2^{n}$, where $n$ is a randomly chosen positive integer, ends in the digit 2? What is the probability that $2^{n}$ ends in the digits 12?
86. Writing out all consecutive powers of two

$2,4,8,16,32,64,128,256,512,1024,2048,4096, \ldots$, we can see that the last digits of these numbers repeat periodically with a period of 4. Therefore, out of four consecutive powers of two, exactly one will end in 2; hence, the probability that $2^{n}$ ends in 2 is $\frac{1}{4}=0.25$.

Now let's sequentially write out the last two digits of the powers of two:

$$
\begin{array}{r}
02,04,08,16,32,64,28,56,12,24,48,96 \\
92,84,68,36,72,44,88,76,52,04,08, \ldots
\end{array}
$$

(this is not difficult to do, as it is sufficient to double the last two-digit number in the sequence each time, and if the result is a three-digit number, discard the first digit). Thus, we see that $2^{22}$ ends in the same digits 04 as $2^{2}$, $2^{28}$ ends in the same digits 08 as $2^{3}$, and so on, i.e., the last two digits of the numbers $2^{n}$, starting from the number $2^{2}=04$, also repeat periodically with a period of 20. In this case, out of 20 consecutive powers of two from $2^{2}$ to $2^{21}$, only $2^{12}$ ends in 12. Therefore, out of every 20 consecutive powers of two starting from $2^{k}$, where $k \geqslant 2$, exactly one ends in 12. Hence, the probability that $2^{n}$ ends in 12 is $\frac{1}{20}=0.05$.

Remark. It can be shown that the last $\boldsymbol{k}$ digits of the numbers $2^{n}$, starting from the number $2^{k}$, repeat with a period of $4 \cdot 5^{k-1}$ [see the book by D. O. Shklyarsky, N. N. Chentsov, I. M. Yaglom, Selected Problems and Theorems of Elementary Mathematics, part 1 (\"Library of the Mathematical Circle\", issue 1), problem 243].