# Channel capacity of a coin flip

I'm having some issue in understanding the channel capacity.

$C=max_{p(x)}I(X, Y)$

In particular the practical side. For example (an exercise), if I toss a fair coin and I transmit the result in a binary channel. What's the channel capacity?

I'm trying to imagine the channel graph. I believe is something like this.

H --1--> H

T --1--> T

A noiseless channel like this has $C=log|X|$ so $C=log2=1bit$. Why? I could represent it like this:

x --0.5--> H

x --0.5--> T

And the channel capacity becomes $C=log1=0$ because $|X|=1$.

Please someone help I'm quite confused.

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## 1 Answer

If you channel is that you can decide if the cion shows head or tail and the pass the coin to your friend with that side up, then the channel has capacity of one bit. This corresponds to your first computation.

If you channel is that you throw the coin to you friend on such a way that it is head with 50% probability and tail with 50% probability, then you cannot send any information over the channel. This corresponds to your second computation.

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Thanks for your reply. But how can I distinguish from a case to another? –  Andrea Aug 27 '12 at 8:44