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Consider a binary message in which $0$ has has probability $1/3$ and $1$ has probability $2/3$. What value of $H$ should be assign?

I know that you split up $1$ into two messages $1a$ and $1b$. Then I think you have to use conditional probability here. Am I on the right track? So $$\left[\text{entropy of} \ (0,1a, 1b) \ \text{message} \right] = \left[ \text{entropy of} \ (0,1) \ \text{message} \right] + \text{something}$$

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  • $\begingroup$ What does "split up 1 into two messages 1a and 1b" mean? $\endgroup$
    – Emre
    May 1, 2011 at 4:14
  • $\begingroup$ I agree with Emre. There is no reason to split 1 into two messages 1a and 1b (unless you are looking at a problem which explicitly involves such a split, in which case you should specify it properly in your question.) $\endgroup$
    – svenkatr
    May 1, 2011 at 6:12

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Could you define $H$ so others don't have to look? Wikipedia says $H(x)=-\sum p(x_i)\log(p(x_i)).$ You have the $p(x_i)$ so what more do you need? Entropy is additive.

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