Suppose Alice and Bob agree that she will send him a sequence of $4$ bits of data which are not all $1$s. This is a simple example of a communication protocol as defined, for example, in Wikipedia. The result is that Alice can make a free choice among 15 options and send that choice to Bob. So this protocol effectively sends slightly fewer than $4$ bits of information: to be precise, $\log_2 15$ bits. Similarly, for any integer $n >= 1$, there's an obvious protocol which effectively sends $\log_2 n$ bits of information.

Are there any other possibilities here? That is, for what real values $r$ is there a protocol which effectively sends $r$ bits?

In particular, is there a protocol which sends more than zero bits but less than one bit? If there were, then we would expect that Alice cannot use it to send one bit, but can use several repetitions of the same protocol to send one bit.

I suspect the answer is no. I'd appreciate any thoughts on how to prove it.

I don't have a definition of "protocol" or "effectively sends", but the idea seems pretty clear. Here are some more details. (1) We're assuming that the basic communication channel has no noise. (2) If she sends the forbidden $4$-bit sequence, we consider that the protocol was not completed; it's much the same as if she sent $3$ bits and then stopped. (3) We should allow a protocol to specify multiple messages back and forth between Alice and Bob.

Edit: expanded for clarity.

  • $\begingroup$ I just don't understand the question, there seeems to be some conceptual problems here. What is a "protocol"? "This is a protocol for sending slightly fewer than 4 bits of information" (?) For me, that's just a noiseless channel with 15 symbols. I cannot make sense of all this. $\endgroup$
    – leonbloy
    Aug 22, 2016 at 15:59
  • $\begingroup$ @leonbloy, sorry, please see if the edits helped. $\endgroup$
    – Hew Wolff
    Aug 24, 2016 at 15:27
  • $\begingroup$ @igael, where would I read a little more about these "fidel bits"? $\endgroup$
    – Hew Wolff
    Aug 24, 2016 at 15:27

2 Answers 2


It seems you are only considering cases where Alice sends a complete message to Bob. A trivial case would be that the protocol says Alice will always send a $1$ bit. There is then no information in a message so Alice has transmitted zero bits. Otherwise, if Alice has at least two choices of message which are regarded as equiprobable at the outset, she transmits at least one bit. But suppose she transmits $1$ if the sun rose in the morning and $0$ if the sun did not rise. One could argue that a message $1$ carries much less than $1$ bit of information based on our a priori assessment of the probabilities.

  • $\begingroup$ That's true, but I think there is a reasonable concept of the information transmitted which is inherent in the protocol. That's the concept I have in mind. $\endgroup$
    – Hew Wolff
    Feb 23, 2020 at 19:04

First, you need to distinguish between source entropy and the amount of information per bit.

Note that by "slightly fewer than 4 bits of information" or $\log_2(15)$ you are referring to the source entropy with an implicit assumption that all words are equally likely (assume the source as a dictionary).

Here you send $4$ actual bits. If your dictionary had two equally likely words and you still used $4$ bits, then $\frac{\log_2(2)}{4}=0.25$ bits of information per actual bit was communicated.

You could also consider a different source where the word probabilities are such that the source entropy (information) is less than one bit. In such case, regardless of what "protocol" you use, no matter how many actual bits it uses, the amount of information you communicate is equal to the source entropy which is less than one.

Edit: Here is an example for more clarity. The symbol probabilities of two binary sources $S_1$ and $S_2$ with 15 symbols as well as their entropies are as follows:

Symbol  p (S1)    p (S2)

0000    0.0667    0.9414
0001    0.0667    0.0039
0010    0.0667    0.0039
0011    0.0667    0.0039
0100    0.0667    0.0039
0101    0.0667    0.0039
0110    0.0667    0.0039
0111    0.0667    0.0039
1000    0.0667    0.0039
1001    0.0667    0.0039
1010    0.0667    0.0039
1011    0.0667    0.0039
1100    0.0667    0.0039
1101    0.0667    0.0039
1110    0.0667    0.0078

H       3.9069    0.5429

where entropy is $H=-\sum_i p_i\log2(p_i)$

$S_1$ is the source assumed in the question whose entropy is $3.875$. However, the entropy of source $S_2$ is $0.5429$.

You can see it is not the "protocol" that governs the amount of information but the source itself.

  • $\begingroup$ This is a good point, but I'm interested in the information transmitted in the protocol itself, with no assumptions about the frequency of the various cases. That is, "information" in the sense of the example I gave. There might be a clearer or more standard way to put it. $\endgroup$
    – Hew Wolff
    Feb 23, 2020 at 19:02

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