## Task A-4.3.

A die is rolled three times in a row. Determine the probability that each subsequent roll (after the first) results in a number that is not less than the previous one.
## First Solution.

Three consecutive rolls of a die determine an ordered triplet $(a, b, c)$ of numbers $a, b, c$ from the set $\{1,2,3,4,5,6\}$. The number $a$ is the number that fell on the first roll, the number $b$ is the number that fell on the second roll, and the number $c$ is the number that fell on the third roll of the die. What we need to determine is the probability that $a \leqslant b \leqslant c$.

For each of the numbers $a, b, c$, there are 6 possibilities, so the total number of possible outcomes from rolling the die three times in a row is

$$
6 \cdot 6 \cdot 6=216
$$

Now let's count the favorable outcomes. If the number 1 falls on the first roll, i.e., if $a=1$, then $b$ can be any number on the die, and $c$ must be from the set $\{b, b+1, \ldots, 6\}$.

Thus, if $a=1$, then for $b=1$ we have 6 possibilities for $c$, for $b=2$ we have 5 possibilities for $c$, and so on, for $b=6$ we have 1 possibility for $c$. Therefore, for $a=1$ we have a total of

$$
6+5+4+3+2+1=21
$$

good outcomes.

By similar reasoning, we conclude that for

$$
\begin{array}{ll}
a=2 & \text { we have } 5+4+3+2+1=15 \text { good outcomes, } \\
a=3 & \text { we have } 4+3+2+1=10 \text { good outcomes, } \\
a=4 & \text { we have } 3+2+1=6 \text { good outcomes, } \\
a=5 & \text { we have } 2+1=3 \text { good outcomes, } \\
a=6 & \text { we have } 1=1 \text { good outcome. }
\end{array}
$$

Thus, we have a total of $21+15+10+6+3+1=56$ good outcomes.

Finally, the desired probability is

$$
\frac{56}{216}=\frac{7}{27}
$$

1 point