Answer: $\frac{1}{720}$.

Solution. The total number of sequences that Vitya could lay out from 12 cards is

$$
C_{12}^{2} C_{10}^{2} C_{8}^{2} C_{6}^{2} C_{4}^{2} C_{2}^{2}=66 \cdot 45 \cdot 28 \cdot 15 \cdot 6 \cdot 1
$$

Let's find how many sequences, after removing six cards according to Vitya's rule, result in 123456. We will restore the removed numbers and highlight the restored numbers in bold. The removed one could only stand before the remaining one:

$$
1123456
$$

The removed two can be placed in three ways:

## $21123456, \quad 12123456$ or 11223456.

In any of these cases, there are exactly 5 equally possible positions for the removed three. Reasoning this way further, we get the total number of ways to restore the numbers:

$1 \cdot 3 \cdot 5 \cdot 7 \cdot 9 \cdot 11$

The desired probability is $\frac{1 \cdot 3 \cdot 5 \cdot 7 \cdot 9 \cdot 11}{1 \cdot 6 \cdot 15 \cdot 28 \cdot 45 \cdot 66}=\frac{1}{720} \approx 0.0014$.
5. Cards (from 7th grade, 2 points). When Vitya was in the first grade, he had a set of digits with 12 cards: two cards with the digit 1, two cards with the digit 2, and so on up to the digit 6. Vitya laid them out in a row on the table in a random order from left to right, and then removed the first one, the first two, the first three, and so on. For example, if Vitya initially had the sequence 434653625112, then the resulting sequence would be 436512. What is the probability that Vitya is left with the sequence 123456 on the table?