I'm struggling with the following exercise on entropy.
Suppose that your friend Alice chooses a number $X$ uniformly at random from $[1,n]$, which she writes down using $\log n$ bits; you can assume that all quantities concerning $\log n$ are integers. Now the following happens: With probability $1/4$, Alice tells you the first $3/4 \log n$ bits of $X$ and with probability $3/4$, she tells you only $1/8 \log n$ bits of $X$. What is your entropy regarding $X$?
The way I understand this, is that, without the second part where Alice tells me some random bits, my entropy $H(X)$ would be $$ H(X) = -\sum_x p(x) \log p(x) = - n (1/n) \log (1/n) = \log n.$$
Regarding the additional information gained from Alice, my intuition tells me that my entropy should be lowered by the bits weighted with their respective probability of me getting them:
$$ H(X) - (1/4) (3/4) \log n - (3/4) (1/8) \log n = 29/36 \log n $$
Is my intuition correct? If yes, how can I formalize it?