On a standard die one of the dots is removed at random with each dot equally likely to be chosen. The die is then rolled. What is the probability that the top face has an odd number of dots?

$ \textbf{(A)}\ \frac {5}{11} \qquad 
\textbf{(B)} \ \frac {10}{21} \qquad 
\textbf{(C)}\ \frac {1}{2} \qquad 
\textbf{(D)} \ \frac {11}{21} \qquad 
\textbf{(E)}\ \frac {6}{11}$
1. First, we need to determine the total number of dots on a standard die. A standard die has faces numbered from 1 to 6, so the total number of dots is:
   \[
   1 + 2 + 3 + 4 + 5 + 6 = 21
   \]
   Therefore, there are 21 dots in total.

2. Each dot is equally likely to be removed, so the probability of removing any specific dot is:
   \[
   \frac{1}{21}
   \]

3. We need to calculate the probability that the top face has an odd number of dots after one dot is removed. We will consider each face individually and calculate the probability for each case.

4. **Case 1: Removing a dot from the face with 1 dot (face 1)**
   - If the dot on face 1 is removed, face 1 will have 0 dots, which is even.
   - Probability of this happening:
     \[
     \frac{1}{21}
     \]
   - Probability of rolling an odd number of dots (faces 3 and 5):
     \[
     \frac{2}{6} = \frac{1}{3}
     \]
   - Combined probability:
     \[
     \frac{1}{21} \cdot \frac{1}{3} = \frac{1}{63}
     \]

5. **Case 2: Removing a dot from the face with 2 dots (face 2)**
   - If one of the dots on face 2 is removed, face 2 will have 1 dot, which is odd.
   - Probability of this happening:
     \[
     \frac{2}{21}
     \]
   - Probability of rolling an odd number of dots (faces 1, 3, and 5):
     \[
     \frac{3}{6} = \frac{1}{2}
     \]
   - Combined probability:
     \[
     \frac{2}{21} \cdot \frac{1}{2} = \frac{1}{21}
     \]

6. **Case 3: Removing a dot from the face with 3 dots (face 3)**
   - If one of the dots on face 3 is removed, face 3 will have 2 dots, which is even.
   - Probability of this happening:
     \[
     \frac{3}{21}
     \]
   - Probability of rolling an odd number of dots (faces 1 and 5):
     \[
     \frac{2}{6} = \frac{1}{3}
     \]
   - Combined probability:
     \[
     \frac{3}{21} \cdot \frac{1}{3} = \frac{1}{21}
     \]

7. **Case 4: Removing a dot from the face with 4 dots (face 4)**
   - If one of the dots on face 4 is removed, face 4 will have 3 dots, which is odd.
   - Probability of this happening:
     \[
     \frac{4}{21}
     \]
   - Probability of rolling an odd number of dots (faces 1, 3, and 5):
     \[
     \frac{3}{6} = \frac{1}{2}
     \]
   - Combined probability:
     \[
     \frac{4}{21} \cdot \frac{1}{2} = \frac{2}{21}
     \]

8. **Case 5: Removing a dot from the face with 5 dots (face 5)**
   - If one of the dots on face 5 is removed, face 5 will have 4 dots, which is even.
   - Probability of this happening:
     \[
     \frac{5}{21}
     \]
   - Probability of rolling an odd number of dots (faces 1 and 3):
     \[
     \frac{2}{6} = \frac{1}{3}
     \]
   - Combined probability:
     \[
     \frac{5}{21} \cdot \frac{1}{3} = \frac{5}{63}
     \]

9. **Case 6: Removing a dot from the face with 6 dots (face 6)**
   - If one of the dots on face 6 is removed, face 6 will have 5 dots, which is odd.
   - Probability of this happening:
     \[
     \frac{6}{21}
     \]
   - Probability of rolling an odd number of dots (faces 1, 3, and 5):
     \[
     \frac{3}{6} = \frac{1}{2}
     \]
   - Combined probability:
     \[
     \frac{6}{21} \cdot \frac{1}{2} = \frac{3}{21}
     \]

10. Adding up all the probabilities:
    \[
    \frac{1}{63} + \frac{1}{21} + \frac{1}{21} + \frac{2}{21} + \frac{5}{63} + \frac{3}{21}
    \]
    Converting all terms to a common denominator of 63:
    \[
    \frac{1}{63} + \frac{3}{63} + \frac{3}{63} + \frac{6}{63} + \frac{5}{63} + \frac{9}{63} = \frac{27}{63} = \frac{9}{21} = \frac{3}{7}
    \]

The final answer is $\boxed{\frac{11}{21}}$