Nine cards can be numbered using positive half-integers $(1/2, 1, 3/2, 2, 5/2, \dots )$ so that  the sum of the numbers on a randomly chosen pair of cards gives an integer from $2$ to $12$ with  the same frequency of occurrence as rolling that sum on two standard dice.  What are the numbers on the nine cards and how often does each number appear on the cards?
To solve this problem, we need to find nine cards numbered with positive half-integers such that the sum of the numbers on any randomly chosen pair of cards gives an integer from 2 to 12 with the same frequency as rolling that sum on two standard dice. 

First, let's recall the frequency of sums when rolling two standard dice:
- Sum of 2: 1 way (1+1)
- Sum of 3: 2 ways (1+2, 2+1)
- Sum of 4: 3 ways (1+3, 2+2, 3+1)
- Sum of 5: 4 ways (1+4, 2+3, 3+2, 4+1)
- Sum of 6: 5 ways (1+5, 2+4, 3+3, 4+2, 5+1)
- Sum of 7: 6 ways (1+6, 2+5, 3+4, 4+3, 5+2, 6+1)
- Sum of 8: 5 ways (2+6, 3+5, 4+4, 5+3, 6+2)
- Sum of 9: 4 ways (3+6, 4+5, 5+4, 6+3)
- Sum of 10: 3 ways (4+6, 5+5, 6+4)
- Sum of 11: 2 ways (5+6, 6+5)
- Sum of 12: 1 way (6+6)

The total number of ways to get these sums is 36, which matches the number of ways to choose 2 cards out of 9 (since \( \binom{9}{2} = 36 \)).

Next, we need to find a set of nine half-integers that satisfy these conditions. By trial and error, we find that the set:
\[ \left\{ \frac{1}{2}, \frac{3}{2}, \frac{5}{2}, \frac{5}{2}, \frac{7}{2}, \frac{9}{2}, \frac{9}{2}, \frac{11}{2}, \frac{13}{2} \right\} \]
works. Let's verify this by checking the sums:

1. Sum of 2:
   - \( \frac{1}{2} + \frac{3}{2} = 2 \) (1 way)

2. Sum of 3:
   - \( \frac{1}{2} + \frac{5}{2} = 3 \) (2 ways, since there are two \(\frac{5}{2}\))

3. Sum of 4:
   - \( \frac{1}{2} + \frac{7}{2} = 4 \) (1 way)
   - \( \frac{3}{2} + \frac{5}{2} = 4 \) (2 ways, since there are two \(\frac{5}{2}\))

4. Sum of 5:
   - \( \frac{1}{2} + \frac{9}{2} = 5 \) (2 ways, since there are two \(\frac{9}{2}\))
   - \( \frac{3}{2} + \frac{7}{2} = 5 \) (1 way)
   - \( \frac{5}{2} + \frac{5}{2} = 5 \) (1 way)

5. Sum of 6:
   - \( \frac{1}{2} + \frac{11}{2} = 6 \) (1 way)
   - \( \frac{3}{2} + \frac{9}{2} = 6 \) (2 ways, since there are two \(\frac{9}{2}\))
   - \( \frac{5}{2} + \frac{7}{2} = 6 \) (2 ways, since there are two \(\frac{5}{2}\))

6. Sum of 7:
   - \( \frac{1}{2} + \frac{13}{2} = 7 \) (1 way)
   - \( \frac{3}{2} + \frac{11}{2} = 7 \) (1 way)
   - \( \frac{5}{2} + \frac{9}{2} = 7 \) (4 ways, since there are two \(\frac{5}{2}\) and two \(\frac{9}{2}\))
   - \( \frac{7}{2} + \frac{7}{2} = 7 \) (1 way)

7. Sum of 8:
   - \( \frac{3}{2} + \frac{13}{2} = 8 \) (1 way)
   - \( \frac{5}{2} + \frac{11}{2} = 8 \) (2 ways, since there are two \(\frac{5}{2}\))
   - \( \frac{7}{2} + \frac{9}{2} = 8 \) (2 ways, since there are two \(\frac{9}{2}\))

8. Sum of 9:
   - \( \frac{5}{2} + \frac{13}{2} = 9 \) (2 ways, since there are two \(\frac{5}{2}\))
   - \( \frac{7}{2} + \frac{11}{2} = 9 \) (1 way)
   - \( \frac{9}{2} + \frac{9}{2} = 9 \) (1 way)

9. Sum of 10:
   - \( \frac{7}{2} + \frac{13}{2} = 10 \) (1 way)
   - \( \frac{9}{2} + \frac{11}{2} = 10 \) (2 ways, since there are two \(\frac{9}{2}\))

10. Sum of 11:
    - \( \frac{9}{2} + \frac{13}{2} = 11 \) (2 ways, since there are two \(\frac{9}{2}\))

11. Sum of 12:
    - \( \frac{11}{2} + \frac{13}{2} = 12 \) (1 way)

Thus, the set of numbers on the nine cards is:
\[ \left\{ \frac{1}{2}, \frac{3}{2}, \frac{5}{2}, \frac{5}{2}, \frac{7}{2}, \frac{9}{2}, \frac{9}{2}, \frac{11}{2}, \frac{13}{2} \right\} \]

The final answer is \( \boxed{ \left\{ \frac{1}{2}, \frac{3}{2}, \frac{5}{2}, \frac{5}{2}, \frac{7}{2}, \frac{9}{2}, \frac{9}{2}, \frac{11}{2}, \frac{13}{2} \right\} } \)