How to best obtain bit sequences from throwing normal dice? Throwing normal dice, one can get sequences of digits in [0,5]. In practice, which is the best procedure to transform such sequences into a desired number of bit sequences?
Thomas' first procedure produces 2 bits per roll with a probability of $\frac23$, i.e. it produces $\frac43 \doteq 1.333$ bits on the average. This can be improved as he describes, but it gets quite complicated soon.

Producing a single bit per roll by taking the result mod two leads to $1$ bit per roll, which is not much worse. Combining the two simple methods like

1 -&gt; 00
2 -&gt; 01
3 -&gt; 10
4 -&gt; 11
5 -&gt; 0
6 -&gt; 1


is quite simple and produces $1+\frac23 = \frac53 \doteq 1.667$ bits per roll. AFAIK, this is the maximum you can get without aggregating the rolls.

You can apply this idea to two rolls and obtain $5$ bits with probability of $\frac{32}{36}$ and $2$ bits otherwise thus getting

$$2 + (5-2) \cdot \frac{32}{36} = 2 + \frac{3 \cdot 8}9 = 2 + \frac83
= \frac{14}3 = 4\frac23$$

bits for two rolls, i.e., $2\frac13 \doteq 2.333$ bits per roll, which is quite close to the theoretical maximum of $\log_2 6 \doteq 2.585$.

For a simple "implementation without computer", note that you can extract 1 bit directly from each roll and need to process the "rest" only (e.g. you can get the bit via $n \bmod 2$ and the rest via $\lfloor \frac{n-1}3 \rfloor$). This makes even combining 3 or 4 rolls using a small table easy.

The exact procedure for two rolls

For simplicity, let's assume, each roll produces a number $n_i$ from the set $\{0, 1, \ldots, 5\}$.


Extract one bit from each roll as $n_i \bmod 2$.
Compute $3 \cdot \lfloor \frac {n_1}2 \rfloor + \lfloor \frac {n_2}2 \rfloor$, which is number $x \in \{0, \ldots, 8\}$
In case of $x = 8$, you're out of luck and get nothing but the two bits from the first step.
Otherwise, use $x \in \{0, \ldots, 7\}$ as the next three bits.


So, most of the time you get five bits. On the average, you get

$$\frac19 \cdot 2 + \frac89 \cdot 5 = \frac{42}9 = 4\frac23$$

bits as stated above.

There are nearly 5.170 bits in two rolls, but you can't extract 5 bits from them on the average without further aggregation.