(5) Divide the positive integers $1,2,3,4,5,6,7$ into two groups arbitrarily, with each group containing at least one number, then the probability that the sum of the numbers in the first group is equal to the sum of the numbers in the second group is $\qquad$
(5) $\frac{4}{63}$ Hint: Only consider all scenarios of the first set of numbers, the second set of numbers will then be determined.
All possible cases are $2^{7}-2=126$. Note that
$$
1+2+3+4+5+6+7=28,
$$

When the sums of the two sets of numbers are equal, the value of their sum is 14. Therefore, at this time, the first set of numbers must contain at least one of the numbers $5,6,7$. Thus,
When the first set of numbers contains 5, there are 4 cases where the sum equals 14:
$$
\{2,3,4,5\},\{1,2,5,6\},\{3,5,6\},\{2,5,7\} \text {; }
$$

When it does not contain 5 but contains 6, there are 2 cases where the sum equals 14:
$$
\{1,3,4,6\},\{1,6,7\} \text {; }
$$

When it does not contain 5 and 6 but contains 7, there are 2 cases where the sum equals 14:
$$
\{1,2,4,7\},\{3,4,7\} \text {. }
$$

So there are a total of 8 cases where the sum equals 14, hence the required probability is $\frac{8}{126}=\frac{4}{63}$.