Albert and Bob and Charlie are each thinking of a number. Albert's number is one more than twice Bob's. Bob's number is one more than twice Charlie's, and Charlie's number is two more than twice Albert's. What number is Albert thinking of?
$\mathrm{(A) \ } -\frac{11}{7} \qquad \mathrm{(B) \ } -2 \qquad \mathrm {(C) \ } -1 \qquad \mathrm{(D) \ } -\frac{4}{7} \qquad \mathrm{(E) \ } \frac{1}{2}$
1. Let the number Albert is thinking of be \( A \), Bob's number be \( B \), and Charlie's number be \( C \).

2. According to the problem, we have the following relationships:
   \[
   A = 2B + 1
   \]
   \[
   B = 2C + 1
   \]
   \[
   C = 2A + 2
   \]

3. Substitute \( B \) from the second equation into the first equation:
   \[
   A = 2(2C + 1) + 1
   \]
   Simplify:
   \[
   A = 4C + 2 + 1
   \]
   \[
   A = 4C + 3
   \]

4. Substitute \( C \) from the third equation into the equation \( A = 4C + 3 \):
   \[
   A = 4(2A + 2) + 3
   \]
   Simplify:
   \[
   A = 8A + 8 + 3
   \]
   \[
   A = 8A + 11
   \]

5. Rearrange the equation to solve for \( A \):
   \[
   A - 8A = 11
   \]
   \[
   -7A = 11
   \]
   \[
   A = -\frac{11}{7}
   \]

Thus, the number Albert is thinking of is \(\boxed{-\frac{11}{7}}\).