The Minnesota Twins  face  the  New  York  Mets  in  the  2006  World  Series.  Assuming  the  two  teams are  evenly  matched  (each  has  a  $.5$  probability  of  winning  any  game)  what  is  the  probability  that the  World  Series  (a  best  of  7  series  of  games which lasts  until  one  team  wins  four  games)  will require  the  full  seven  games  to  determine  a  winner?

$\text{(A) }\frac{1}{16}\qquad
\text{(B) }\frac{1}{8}\qquad
\text{(C) }\frac{3}{16}\qquad
\text{(D) }\frac{1}{4}\qquad
\text{(E) }\frac{5}{16}\qquad$

$\text{(F) }\frac{3}{8}\qquad
\text{(G) }\frac{5}{32}\qquad
\text{(H) }\frac{7}{32}\qquad
\text{(I) }\frac{9}{32}\qquad
\text{(J) }\frac{3}{64}\qquad
\text{(K) }\frac{5}{64}\qquad$

$\text{(L) }\frac{7}{64}\qquad
\text{(M) }\frac{1}{2}\qquad
\text{(N) }\frac{13}{32}\qquad
\text{(O) }\frac{11}{32}\qquad
\text{(P) }\text{none of the above}$
To determine the probability that the World Series will require the full seven games, we need to consider the conditions under which this happens. Specifically, the series will go to seven games if and only if both teams win exactly three of the first six games. The seventh game will then determine the winner.

1. **Calculate the number of ways to distribute wins in the first six games:**
   - Each team must win exactly 3 out of the first 6 games.
   - The number of ways to choose 3 wins out of 6 games for one team (say the Twins) is given by the binomial coefficient:
     \[
     \binom{6}{3} = \frac{6!}{3!3!} = 20
     \]

2. **Calculate the total number of possible outcomes for the first six games:**
   - Each game can be won by either team, so there are \(2^6\) possible outcomes for the first six games:
     \[
     2^6 = 64
     \]

3. **Calculate the probability that the series goes to seven games:**
   - The probability that the series goes to seven games is the ratio of the number of favorable outcomes (both teams winning 3 games each in the first six games) to the total number of possible outcomes:
     \[
     P(\text{seven games}) = \frac{\binom{6}{3}}{2^6} = \frac{20}{64} = \frac{5}{16}
     \]

Therefore, the probability that the World Series will require the full seven games is \(\frac{5}{16}\).

The final answer is \(\boxed{\frac{5}{16}}\).