## How does one split 3 bits between two persons?

You noticed it: you had problem with you dial-up, you fiddled with your **modem settings**, you thought you knew something about how computers work, then it struck you: **stop bits: 1, 1.5 or 2**

How does one define **half a bit**?

First, let’s say how do we define a bit: **binary digit**. In the binary system, only 0 and 1 are digits. So one bit is either 0 or 1… we’re not getting anywhere, aren’t we?

Let’s go further and dig the Shannon definition: the information gathered from **tossing a coin**, each side having *equal probabilities* to show up.

So **one bit** of information is gathered from a coin with equal probabilities for the sides. How would a coin that emits a **half bit** look like?

What happens if the sides *don’t* have equal probabilities? Let’s say the probabilities are p1 and p2

For therefore the number of bits is bit. Exactly *one bit.*

Question is, *what are the probabilities the coin’s sides should have to transmit a full half-a-bit*?

Given that the two probabilities must sum exactly 1, therefore are and respectively (just like ), the problem of **how does one split one bit in two** becomes

If bits, what’s the value of ?

Using an online newton approximation and guessing 0.25 as the ‘seed’ the solution becomes 0.11002786443836 or roughly **0.11.**

So, each time a **modem** wants to send 1.5 bits, it sends a bit and **it tosses a special coin** that has 0.11 probability to fall on one side and 0.89 probability to fall on the other side.

Now that’s some light shed on modem’s speed bottleneck!

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