compute H(X|Y) ( conditional probablity) Can someone help me on this?
X = {$X_1,X_2,X_3,X_4$}
Y = {$Y_1,Y_2,Y_3,Y_4$}
Suppose p($X_i$) = p($Y_j$) = 1/4 (each X and each Y equally likely)
$1 \leq i, 4 \ge j$
and now suppose 
$Y_1 : X_1 or X_2$
$Y_2 : X_3 or X_4$
$Y_3 : X_2 or X_3$
$Y_4 : X_1 or X_4$
what is $H(X|Y)$?

I tried to create a table, like
$$ \left[
    \begin{array}{ccc}
   & y1 & y2 & y3 & y4 \\
x1 &    & 0  & 0  &    \\
x2 &    & 0  &    & 0  \\
x3 & 0  &    &    & 0  \\
x4 & 0  &    & 0  &   
    \end{array}
\right] $$
I know p(X) = 1/2, because only two choices. 
And p(Y)=1/4 is same.
Thank you

$\sum_{y}\sum_{x}p(y)p(x|y)log_2(p(x|y))$
$(1/4)p log_2 p + (1/4)((1/4)-p) log_2 ((1/4)-p)$
$4* [(1/4)p log_2 p + (1/4)((1/4)-p) log_2 ((1/4)-p)]$
$p log_2 p + ((1/4)-p) log_2 ((1/4)-p)$
 A: Let $p\mathop{:=}\mathsf P(X_1,Y_1)$.   Then since each column and each row must total $\tfrac 14$, we complete your table as follows:
$$\boxed{\begin{array}{|c|c:c:c|}\hline
   & Y_1 & Y_2 & Y_3 & Y_4 \\ \hline
X_1 & p   & 0  & 0  &  \tfrac 14-p  \\ \hdashline
X_2 & \tfrac 14-p   & 0  &  p  & 0  \\ \hdashline
X_3 & 0  &  p  & \tfrac 14-p & 0  \\ \hdashline
X_4 & 0  & \tfrac 14-p & 0  & p \\ \hline
\end{array}}$$
The information you provided is insufficient to determine what $p$ actually is.
If the choice between each of the two $X_k$ for each $Y_h$ were unbiased, then $p=\tfrac 18$.   However, nothing indicates this is so.
But since there are the same two conditional probabilities for the two $X_k$ any given $Y_h$ (although which event they match differ), we can calculate the conditional entropy as:
$$\begin{align}\mathsf H(X\mid Y_h) & = - \sum_{k=1}^4 \mathsf P(X_k\mid Y_h)\log_2(\mathsf P(X_k\mid Y_h)) & \big[h\in\{1,2,3,4\}\big]
\\ & = -\big( 4p \log_2 (4p) + 4(1-p)\log_2(4(1-p)) + 0\log 0 + 0\log 0\big)
\\ & = -4\big( p \log_2(p) + (1-p) \log_2(1-p)+2 ) \big)
\end{align}$$
