# ETOOBUSY ðŸš€ minimal blogging for the impatient

# A 4-faces die from a 6-faces die

**TL;DR**

How can we simulate a die with 4 faces with a die that has 6?

Some games require rolling a 4-sided die to get some random outcome in a specific phase. This is usually accomplished usingâ€¦ a 4-sided die (also known as D4), usually a regular tetrahedron (which has, indeed, 4 faces).

What can we do if we only have a more commong 6-sided die instead, i.e. a cube or D6 in playing jargon? Letâ€™s take a look.

# Warm up: a coin from a D6

Letâ€™s start simple and try to generate the outcome of a fair two-sided coin from a fair D6 die. Our goal is to generate Heads with probability $\frac{1}{2}$ and Tails with probability $\frac{1}{2}$:

We have a (hopefully fair!) die at our disposal, i.e. a device that is capable of generating each of its faces with probability $\frac{1}{6}$:

Letâ€™s take any three faces of the die (e.g. 1, 2, and 3) and ask: what is the probability of a single roll to give us one of them? Itâ€™s the sum of the respective probabilities, i.e. $\frac{1}{6} +\frac{1}{6} + \frac{1}{6} = \frac{1}{2}$.

Weâ€™re definitely onto something, because at this point we can do the following association:

- if the D6 roll gives us either 1, or 2, or 3, then our
*virtual coin*says Heads - otherwise, our
*virtual coin*says Tails.

and we are sure that our *virtual coin* is fair (as long as the D6 is
fair, or course!).

Any three-faces subset of the D6 will do, of course: you might choose
e.g. to take odd values for Heads and even values for Tails, or whatever
else you like. The important thing is that you decide this *before*
rolling the die!

# One step ahead: a D3 from a D6

Itâ€™s easy to extend the previous section to also enable generating a
3-sided die from a D6: just group the D6 faces in pairs, and each of
them will have probability $\frac{1}{3}$ to come out from a single roll,
thus giving us a *virtual D3*. Again, itâ€™s important to decide
*beforehand* what is the *virtual D3* value associated to each pair,
e.g. like in the following picture.

# OK, what about a D4?

The two examples above were easy because itâ€™s easy to combine faces with
an outcome probability of $\frac{1}{6}$ into the needed output
probabilities (respectively of $\frac{1}{2}$ for the *virtual coin* and
$\frac{1}{3}$ for the *virtual D3*). At the end of the day, it depends
on $6$ being divisible by both $2$ (for the *virtual coin*) and $3$ (for
the *virtual D3*).

On the other hand, the D4 is a different beast because $6$ is not divisible by $4$, so we cannot recombine $\frac{1}{6}$-sized pieces to obtain $\frac{1}{4}$ pieces instead. Itâ€™s either too little or too much.

One powerful tool at our disposal is the *rejection method*. Letâ€™s keep
rolling the die until we get a number between `1`

and `4`

included,
*rejecting* the roll every time itâ€™s a `5`

or a `6`

.

If this gives out anything, it will surely a valid outcome for a D4,
because only admissible values `1`

to `4`

are allowed. Is it fair
though?

The overall probability of the event *value 1 comes out from a roll of
the D6* is still $\frac{1}{6}$, of course. Considering that

`1`

is also
an admissible value, this is also the same probability as event *value*. The same applies to

`1`

comes out of a roll of the D6 AND the outcome of the D6 roll is an
admissible value`2`

, `3`

, and `4`

of course.This probability of the union of the two events * 1 comes out of the
D6 roll* and

*the D6 roll gives out an admissible value*can be expressed in terms of conditional probability, as in

*the roll was a*, which translates into the following formula:

`1`

, subject to the fact that the roll was admissibleThe conditional probability $P_{1 | {1, 2, 3, 4}}$ is exactly what we
are after for establishing the fairness of our *virtual D4*, because the
rejection method *ensures* us that we got an admissible value (hence,
the condition applies). Itâ€™s easy to invert the formula, then:

The probability at the numerator is $\frac{1}{6}$. This was easy.

The probability at the denominator is simply the sum of the
probabilities of having a `1`

, a `2`

, a `3`

, or a `4`

; in the case of a
fair die, itâ€™s $4 \cdot \frac{1}{6} = \frac{4}{6}$.

So we end up with this:

The same reasoning can of course be applied to `2`

, `3`

, and `4`

with
equal conclusions, hence our *virtual D4* is, indeed, a fair one
assuming that the starting D6 is fair. Yay!

# Conclusions

As we saw, itâ€™s possible to simulate a *virtual D4* from a real D6,
although there might be some additional rolling involved when we have to
reject a specific roll. This happens with non-trivial probability
$\frac{1}{3}$ for a single roll, i.e. on average we will have to re-roll
one out of three times and still not be sure it will be enough!

Could we end up trapped in a nightmare of rolling over and over without getting anything meaningful for a long time? Sure we could, but the distribution of this is geometric with an average of $1.5$ rolls needed to get something meaninful out, so there should not be too much to worry.

Or should there be?