On the calculator keyboard, there are digits from 0 to 9 and signs of two operations (see the figure). Initially, the display shows the number 0. You can press any keys. The calculator performs operations in the sequence of key presses. If the operation sign is pressed several times in a row, the calculator will only remember the last press. The Absent-Minded Scientist pressed a lot of buttons in a random sequence. Find the approximate probability that the result of the resulting chain of operations is an odd number?

![](https://cdn.mathpix.com/cropped/2024_05_08_293b4772a418c01a5915g-30.jpg?height=623&width=414&top_left_y=532&top_left_x=822)

#
We will say that the Scientist has performed $n$ steps if he has entered $n$ numbers, and between them, $n-1$ times performed some arithmetic operations. Let $p_{n}$ denote the probability that after $n$ steps, the number on the calculator is odd, and express $p_{n+1}$ in terms of $p_{n}$.

If the last operation was multiplication, the result will be odd only if both factors are odd; the probability of this is $1 / 2 p^{2}$.

If the last operation was addition, the result will be odd if the last addend differs in parity from the penultimate one; the probability of this is $1 / 2$.

The last operation can be either multiplication or addition, and both of these options are equally likely.

Therefore, the formula for the total probability gives:

$p_{n+1}=1 / 2 \cdot 1 / 2 + 1 / 2 \cdot 1 / 2 p_{n} = 1 / 4 + 1 / 4 p_{n}$.

Rewriting this equality as $p_{n+1} - 1 / 3 = 1 / 4 (p_{n} - 1 / 3)$, we see that when $n$ increases by 1, the difference $p_{n} - 1 / 3$ decreases by a factor of 4. Therefore, $p_{n}$ is close to $1 / 3$ for large $n$.

## Answer